The Analysis of Biological Data - Whitlock and Schluter Solutions to practice problems Chapter 1 1. (a) Ordinal (b) Nominal (c) Nominal (d) Ordinal 2. (a) Observational (cats were observed to fall a given number of stories ! the the researchers did not assign the number of stories fallen to the cats). (b) The number of floors fallen. (c) The number of injuries per cat. 3. (a) Discrete (b) Continuous (c) Discrete (d) Continuous 4. (a) Collectors do not usually sample randomly, but b ut prefer rare and unusual specimens. Therefore, the moths are probably not a random sample. (b) A sample of convenience. (c) Bias (rare color types are likely to be over-represented ove r-represented in the sample compared with the p opulation). 5. Accuracy. 6. (a) Categorical (b) No random sampling procedure was followed so the survey is not random; the respondents were volunteers. (c) Volunteer bias (those most interested in the issue were more likely to respond). 7. (a) US army personnel stationed in Iraq. (b) Yes. The number of subjects interviewed was a random sample of only 100 from the population; therefore the responses of the particular individuals who happened to be sampled will differ from the population of interest by chance. (c) The advantage of random sampling is that it minimizes bias and allows the precision of estimates of stress levels to be measured. 8. (a) The population being estimated is all the small mammals of Kruger National Park. (b) No, the sample is not likely to be random. In a random sample, every individual has the same chance of being selected. But some small mammals might be easier to trap than others (for example, trapping only at night might miss all the mammals active only in the day time). In a random sample individuals are selected independently. Multiple animals ca ught in the same trap might not be independent if they are related or live near one another (this is harder to judge). (c) The number of species in the sample might underestimate the number in the Park if sampling was not random (e.g., if daytime mammals were missed), or if rare species happened to avoid capture. In this case m would be a biased estimate of M of M . 9. Chicks were not sampled randomly. The multiple chicks from the same nest are not independent. Chicks from the same nest are likely to be more similar to one another in their measurements than chicks chosen randomly from the population. Ignoring this problem will lead to e rroneous measurements of the precision of survival estimates. sampled randomly from the population.
Chapter 2 1. (a) 25% (b) 33% (c) 33%
2. (a) Skewed right (b) Modes: 7.5"8, 2.5"3, 11"11.5 (c) The bimodal distribution in this example is left skewed.
3. The histogram on the right is correct. Histograms use area, not height, to represent frequenc y.
4. 5. Contingency table.
Survived Perished Total
Source of fry Hatchery Wild 27 51 3973 3949 4000 4000
Total 78 7922 8000
6. (a) The following table orders the taxa from those with the most endangered species to those with the fewest endangered species. (Other orderings could make sense, such as by phylum) Taxon Number of species Plants 745 Fish 115 Birds 92 Mammals 74 Clams 70 Insects 44 Reptiles 36 Snails 32 Amphibians 22 Crustaceans 21 Arachnids 12 (b) Frequency table. (c) Bar graph, because data are categorical.
2. (a) Skewed right (b) Modes: 7.5"8, 2.5"3, 11"11.5 (c) The bimodal distribution in this example is left skewed.
3. The histogram on the right is correct. Histograms use area, not height, to represent frequenc y.
4. 5. Contingency table.
Survived Perished Total
Source of fry Hatchery Wild 27 51 3973 3949 4000 4000
Total 78 7922 8000
6. (a) The following table orders the taxa from those with the most endangered species to those with the fewest endangered species. (Other orderings could make sense, such as by phylum) Taxon Number of species Plants 745 Fish 115 Birds 92 Mammals 74 Clams 70 Insects 44 Reptiles 36 Snails 32 Amphibians 22 Crustaceans 21 Arachnids 12 (b) Frequency table. (c) Bar graph, because data are categorical.
(d) The baseline should be 0 so that bar height and area correspond to frequency.
7. (a) The variables are year (or famine stage) and incidence of schizophrenia. (b) Year (and famine stage) is categorical – ordinal. Incidence of o f schizophrenia is categorical – nominal. (c) Contingency table.
Schizophrenic Non-schizophrenic Total
1956 (pre famine) 483 58605 59088
1960 (famine) 192 13556 13748
1965 (pos (postt fami famine ne)) 695 82841 83536
Tota Totall 1370 155002 156372
(d) Proportions: 1956: 0.0082, 1960: 0.0140, 1965: 0.0083. Other variations of the line graph below are valid (e.g., famine state could be the explanatory variable instead of year)
Pattern revealed: incidence of schizophrenia highest in 1960, the famine year. 8. (a) Grouped histograms. (b) Groups are disease types (diseases transmitted directly from one individual to another and diseases transferred by insect vectors). (c) The variable is the virulence of the disease. It is a continuous numeric variable. (d) Relative frequency, the fraction of diseases occurring in each interval of virulence. (e) Directly transmitted diseases tend to be less virulent than insect-transmitted diseases. Human diseases transmitted directly more frequently have low virulence, and less frequently have high virulence, than diseases transmitted by insect vector. 9. (a) Map. (b) Date of first appearance of rabies (measured by the number of months following 1 March 1991). (c) Geographic location - the township.
10. (a) Launch temperature is the explanatory variable, number of O-ring failures is the response variable. Because we wish to predict failures from temperature.
(b) There is a negative association. (c) Low temperature is predicted to increase the risk of Oring failure.
11. (a) The distribution is left skewed. (b) The mode is between 65 and 80 degrees (if you used an interval width of 10 degrees instead of 5 degrees then the mode is between 70 and 80 degrees. 12. (a) Husbands of adopted women resemble a woman's adoptive father, more than they resemble the women themselves or the women’s adoptive ad optive mother. (b) Inappropriate baseline. The eye is receptive to bar height and area, but because of the baseline height and area do not depict magnitude. A baseline of 0 would probably be best (adding a horizontal line at resemblance = 25 might also be a good idea to indicate the random expectation). (c) No, the graph is depicting a measurement (resemblance) in each of several categories, not frequency of occurrence. 13. (a) Mosaic plot. The explanatory variable is fruit set in previous years. The response va riable is fruiting in the given year. Both variables are categorical. (b) Line graph. The explanatory variable is year. The response variable is the density of the wood. Both variables are numerical. (c) Grouped cumulative frequency distributions. The explanatory variable is the taxonomic group (butterflies, birds, or plants). The response variable is the percent change in range size. Group is a categorical variable. Range change is numerical.
Chapter 3 1. (a) The number of doctors studied who sterilized a given percentage of patients under age 25. (b) The median would be more informative. The mean would be sensitive to the presence of the outlier (sterilizing more than 30% of female patients under 25), whereas the median is not affected. 2. (a) Box plot. (b) Median body mass of the mammals in each group. (c) The first and third quartiles of body mass in each group. (d) “Extreme values”, those lying farther than 1.5 times the interquartile range from the box edge. (e) Whiskers. They extend to the smallest and largest values in the data, excluding extreme values (those lying farther than 1.5 times the interquartile range from the box edge). (f) Living mammals have the smallest median body size (with a log body mass of about 2), and mammals that went extinct in the last ice age had the largest median size (around 5.5). Mammals that went extinct rece ntly were intermediate in size, with a median of around 3.2. (g) The body size distribution of living mammals is right skewed (long tail toward larger values), whereas the size frequency distribution in mammals that went extinct in the last ice age is left skewed. The frequency distribution of sizes is nearly symmetric in mammals that went extinct recently. (h) One way to answer this is to use the interquartile range as a measure of spread, in which case the mammals that went extinct in the last ice age have the lowest spread. Another way to compare spread is to calculate the standard deviation of log body size in each group, but we are unable to calculate this without the raw data. (i) It is likely that extinctions have reduced the median body size of mammals. 3. (a) Grouped histograms. (b) Explanatory variable: sex. Response variable: Number of words per day. (c) Men: 8,000-12,000 words per day. Women: 16,000-20,000 words per day. (d) Women. (e) Men 4. (a) The averages are 7.5 g for the crimson-rumped waxbill (CW), 15.4 g for the cutthroat finch (CF), and 37.9 g for the white-browed sparrow weaver (WS). (b) Standard deviations are 0.6 g for CW, 1.2 g for CF, and 3.1 g for WS. The standard deviation is greater when the mean is greater. (c) Coefficients of variation:8.3% for CW, 8.0% for CF, an d 8.2% for WS. The coefficients of variation are much more similar than the standard deviations. (d) Compare coefficients of variation to compare variation relative to the mean. Beak length: 3.8%, Body mass: 8.2%. Body mass is most variable relative to the mean. 5. Use the cumulative frequency distribution to estimate the median (0.50 quantile, approximately 1.4%), the first and third quartiles (approximately 0.7% and 2.4%). The interquartile range is about 2.4 " 0.7 = 1.6% and 1.5 times this amount is about 2.4%, which sets the maximum length of each whisker. The smallest value is about "1.1% and the largest value is about 4.3%. Both these values are within 1.5 times the interquartile range, so the whiskers extend from "1.1% and 4.3%. The boxplot is:
6. (a) cm/s.
(b) (c) Asymmetric. The range of changes in speed is much greater above the median than below the rd median. This is true whether we look at the 3 quartile or the total range. (d) The span of the box indicates where the middle 50% of the measurements occur (first quartile to third quartile). (e) Mean is 1.19 cm/s, which is greater than the median (0.94). The distribution is right skewed, and 2 the large values influence the mean more than the median. (f) 1.15 (cm/s) (g) The interval from Y " s and Y + s is 0.11 to 2.26. This includes 11 of the 16 observations, or 69%.
7. (a) Increase the mean by k . (b) No effect. 8. (a) Increase Y by 10 times (11.9 mm/s) (b) Increase s by 10 times (10.7 mm/s) (c) Increase median by 10 times (9.4 mm /s) (d) Increase interquartile range by 10 times (16.8 mm/s) (e) No 2 2 effect. (f) Increase s by 100 times (115.3 (mm/s) 9. (a) 5.7 hours. (b) 2.4 hours. (c) 83/114 = 0.73 (73%). (d) Median = 5 hours, which is less than the mean. The distribution is right skewed, and the large values influence the mean more than the median.
Chapter 4 1. (a) 0.22 hours. (b) The sampling distribution of the estimates of the me an time to rigor mortis. (c) Random sampling from the population of corpses. 2. The median will probably be smaller than the mean. The distribution is right skewed. Extreme values have a greater effect on the mean, pulling it upwards, than on the median. 3. Sample size.
4. (a) False ( p is just an estimate based on a sample). (b) True (c) True (d) False, this fraction is a constant. (Leaving aside the possibility of gene copy number variation between people) (e) True ˆ
5. (a) The fraction of all Canadians who agree with the statement. (b) The sample estimate is 73%. (c) The sample size is 1641 people. (d) The 95% confidence interval for the population fraction. 6. (a) 95.9 msec. The population mean flash duration. (b) No, because the calculation was based on a sample rather than the whole population. By chance, the sample mean will differ from that of the population mean. (c) 1.9 msec. (d) The spread of the sampling distribution of the sample mean. (e) Using the 2SE rule, 95.9+1.9 msec yields 92.2 < µ < 99.7. (f) The interval represents the most plausible values for the population mean µ . In roughly 95% of random samples from the population, when we compute the 95% confidence interval the interval will include the true population mean. 7. The population mean is likely to be small or zero.
Chapter 5 1. (a) 5/8 (b) 1/4 (c) 7/8 (either in this case means pepperoni or anchovies or both) (d) No. (some slices have both pepperoni and anchovies) (e) Yes. Olives and mushrooms are mutually exclusive (f) No. Pr[mushrooms] = 3/8; Pr[anchovies] = 1/2; if indepe ndent, Pr[mushrooms & anchovies] = Pr[mushrooms] # Pr[anchovies] = 3/16. Actual probability = 1/8. Not independent. (g) Pr[anchovies | olives] = 1/2 (two slices have olives: one of these two has anchovies) (h) Pr[olives | anchovies] = 1/4 (four slices have an chovies, one of these has olives) (i) Pr[last slice has olives] = 1/4 (two slices of the eight have olives; you still get one slice - it doesn't matter whether your friends pick before you or after you). (j) Pr[two slices with olives] = Pr[first slice has olives] # Pr[second slice has olives | first slice has olives] = 2/8 # 1/7 = 1/28. (k) Pr[slice without pepperoni] = 1 - Pr[slice with pepperoni] = 3/8 (l) Each piece has either one or no topping. 2. Pr[successful hunting) = Pr[find prey] # Pr[captures prey | finds prey] = 0.8 # 0.1 = 0.08. 3. 45 of 273 trees have cavities, so the probability of choosing a tree with a cavity is 45/273 = 0.165. 4. (a) Pr[vowel] = Pr[A] + Pr[E] + Pr[I] + Pr[O] + Pr[U] = 8.2% + 12.7% + 7.0% + 7.5% + 2.8% = 38.2%. (b) Pr[five randomly chosen letters from an English text wou ld spell "STATS"] = Pr[S] # Pr[T] # Pr[A] # Pr[T] # Pr[S] = 0.063 # 0.091 # 0.082 # 0.091 # 0.063 = 2.7 # 10-6. (Each draw is independent, but all must be successful to satisfy the conditions, so we must multiply the probability of each independent event.) (c) Pr[2 letters from an English text = "e"] = 0.127 # 0.127 = 0.016. 5. (a) Pr[A1 or A4] = Pr[A1] + Pr[A4] = 0.1 + 0.05 = 0.15 (b) Pr[A1 A1) = Pr[A1] # Pr[A1] = 0.1 # 0.1 = 0.01. (This is only true because the individuals mate at random. If we did not know this, we could not calculate the probability). (c) Pr[not A1 A1] = 1 - Pr[A 1 A1] = 1 - 0.01 = 0.99 (d) Pr[two individuals not A1A1] = Pr[not A1 A1] # Pr[not A1 A1] = 0.99 # 0.99 = 0.9801. (e) Pr[at least one of two individuals chosen at random has genotype A1 A1] = 1 - Pr[neither individual is A1 A1] = 1 - 0.9801 = 0.0199. (f) Pr[three randomly chosen individuals have no A2 or A3 alleles] 6 = Pr[six randomly chosen alleles are not A2 or A3] = Pr[one allele is neither A 2 nor A3] ;
Pr[one allele is neither A2 nor A3] = 1 - Pr[individual has either A2 or A3] = 1 - [Pr[A2) + Pr[A3)] 6 = 1 - (0.15 + 0.6) = 0.25. Pr[all three lack these alleles] = 0.25 = 0.00024. 6.
Pr[sum to 7] = 1/6 7. (a) Pr[no venomous snakes] = Pr[not venomous in the left hand] # Pr[not venomous in the right hand]= 3/8 # 2/7 = 6/56 = 0.107 (b) Pr[bite]= Pr[bite| 0 venomous snakes] Pr[0 venomous snakes] + Pr[bite| 1 venomous snakes] Pr[1 venomous snakes] + Pr[bite| 2 venomous snakes] Pr[2 venomous snakes] Pr[0 venomous snakes] = 0.107 (from part a) Pr[1 venomous snake] = (5/8 # 3/7) + (3/8 # 5/7) = 0.536 Pr[ 2 venomous snakes] = 5/8 # 4/7 = 0.357 Pr[bite| 0 venomous snakes] = 0 Pr[bite| 1 venomous snake] = 0.8 2 Pr[bite| 2 venomous snakes] = 1 – (1–0.8) = 0.96 Putting these all together: Pr[bite]= (0 # 0.107) + (0.8 # 0.536) + (0.96 # 0.357) = 0.772 Pr[no bite | defanged] Pr[defanged] (c) Pr[defanged | no bite] Pr[no bite] Pr[no bite | defanged] = 1; Pr[defanged] = 3/8; Pr[no bite] = Pr[defanged] Pr[no bite |defanged] + Pr[venomous] Pr[no bite | venomous] = (3/8 # 1)+ (5/8 # (1– 0.8)) = 0.5 So, [defanged | one snake did not bite] = (1.0 # 3/8) / (0.5) = 0.75. =
8. (a) Pr[all five researchers calculate 95% CI with the true value]? Each one has a 95% chance, all 5 samples are independent, so Pr = 0.95 = 0.774 (b) Pr[at least one does not include true parameter] = 1 - Pr[all include true parameter] = 1 - 0.774 = 0.226. 9. (a) Pr[cat survives seven days] = Pr[cat not poisoned one day]7 = 0.997 = 0.932 (b) Pr[cat 365 365 survives a year] = Pr[cat not poisoned one day] = 0.99 = 0.026 (c) Pr[cat dies within year] = 1 - Pr[cat survives year] = 1 - 0.26 = 0.974 10. (a) Pr[single seed lands in suitable habitat] = 0.3 (b) Pr[two independent seeds land in suitable habitat] = 0.32 = 0.09. (c) Pr[two of three independent seeds land suitable habitat] = 3 # (0.7 # 0.3 # 0.3) = 0.189
11 (a) False positive rate = Pr[positive test | no HIV] = 8 out of 4000 = 0.002 (b) False negative rate = Pr[negative test | HIV] = 20 out of 1000 = 0.02 (c) 988 tested positive, of which 980 had HIV; 980/988= 0.9919; 12. Each win has probability 0.09, each loss probability 0.91. 5 -6 5 -6 (a) Pr[WWLWWW] = 0.09 # 0.91 = 5.4 # 10 (b) Pr[WWWWWL] = 0.09 # 0.91 = 5.4 # 10 5 -6 3 3 -4 (c) Pr[LWWWWW] = 0.09 # 0.91 = 5.4 # 10 (d) Pr[WLWLWL] = 0.09 # 0.91 = 5.5 # 10 3 3 -4 6 -7 (e) Pr[WWWLLL] = 0.09 # 0.91 = 5.5 # 10 (f) Pr[WWWWWW] = 0.09 = 5.3 # 10 13. (a)
The probability that the match lasts two sets is 1/4 + 1/4 = 1/2. The probability that it lasts three games is 1/8+1/8+1/8+1/8=1/2.
(b)
The probability of the weaker player winning is (0.45
2
# 0.55) + (0.452 # 0.55) + 0.452 = 0.42525
14. Pr[next person will wash his / her hands] = Pr[wash | man] # Pr[man] + Pr[wash | woman] # Pr[woman] = 0.74 # 0.4 + 0.83 # 0.6 = 0.794
15.
Total probability: 4/36 = 1/9 = 0.11
16. (a) Pr[one person not blinking] = 1 - Pr[person blinks] = 1 - 0.04 = 0.96 (b) Pr[at least one blink in 10 people] = 1 - Pr[no one blinks] = 1 - 0.9610 = 0.335
Chapter 6 1. (a) True. (b) False. 2. (a) H 0: The rate of correct guesses is 1/6. (b) H A: The rate of correct guesses is not 1/6. (We would argue that a two-tailed test is more appropriate than a one-tailed test ! who can say that ESP would work the way we want it to?).
3. (a) Alternative hypothesis. (b) Alternative hypothesis. (c) Null hypothesis. (d) Alternative hypothesis (e) Null hypothesis. 4. (a) Lowers the probability of committing a Type I error. (b) Increases the probability of committing a Type II error. (c) Lowers power of a test. (d) No effect. 5. (a) No effect. (b) Decreases the probability of committing a type II error. (c) Increases the power of a test. (d) No effect. 6. (a) P = 2 # (Pr[15] + Pr[16] + Pr[17]+Pr[18]) = 0.0075. (b) P = 2 # (Pr[13] + Pr[14] + …. + Pr[18]) = 0.096. (c) P = 2 # (Pr[10] + Pr[11] + Pr[12] + … + Pr[18]) = 0.815. (d) P = 2 # (Pr[1] + Pr[2] + Pr[3] + … + Pr[7]) = 0.481. 7. False. The hypotheses aren’t variables subject to chance. Rather, the P value measures how unusual the data are if the null hypothesis is true. 8. Failing to reject H 0 does not mean H 0 is correct, because the power of the test might be limited. The null hypothesis is the default and is either rejected or not rejected. 9. Begin by stating the hypotheses. H 0: Size on islands does not differ in a consistent direction from size on mainlands in Asian large mammals (i.e., p = 0.5); H A: Size on islands differs in a consistent direction from size on mainlands in Asian large mammals (i.e., p $ 0.5), where p is the true fraction of large mammal species that are smaller on islands than on the mainland. Note that this is a 2-tailed test. The test statistic is the observed number of mammal species for which size is smaller on islands than mainland: 16. The P -value is the probability of a result as unusual as 16 our of 18 when H 0 is true: P = 2# (Pr[16] + Pr[17] + Pr[18]) = 0.00135. Since P < 0.05, reject H 0. Conclude that size on islands is usually smaller than on mainlands in Asian large mammals. 10. (a) Not correct. The P -value does not give the size of the effect. (b) Correct. H 0 was rejected, so we conclude that there is indeed an effect. (c) Not correct. The probability of committing a Type I error is set by the significance level, 0.05, which is decided beforehand. (d) Not correct. The probability of committing a type II error depended on the effect size, which wasn’t known. (e) Correct.
Chapter 7 1. (a) 91 / 220 had cancer, so the estimated probability of a cast or crew member developing cancer is 0.414. (b) The standard error of the proportion is 0.033. [The square-root of (0.414)(1 0.414)/(220-1). This quantity measures the standard deviation of the sampling distribution of the proportion. (c) Using the Agresti-Coull method to generate confidence intervals, we first calculate p' = ( x+2)/(n+4) = (91 + 2) / (220 + 4 ) = 0.415. The lower bound for the 95% confidence interval is 0.415 - 1.96 The upper bound is 0.415 + 1.96
(
p! 1 " p! n + 4
)
(
p! 1 " p! n + 4
)
= 0.351
= 0.480. The confidence intervals do not bracket the
typical 14% cancer rate for the age group. It is unlikely that 14% is the true cancer rate for this group. 2. (a) 46/50 bills had cocaine, so the estimated proportion is 0.92. (b) The 99% confidence interval, calculated using the Agresti-Coull method: p' = 48 / 54 = 0.89. Z for 99% = 2.58. The bounds of
the confidence interval are 0.779 and 0.999. (c) We are 99% confident that the true proportion of US one-dollar bills that have a measurable cocaine content lies between 0.779 and 0.999. 3. (a) No; the probability of drawing a red card changes depending on the cards that have been drawn. (b) Yes: because the sampling is with replacement, the probability of success remains constant. (c) No, the probability of drawing a red ball will change with each draw. (d) No, the number of trials is not known. (e) Yes, the number of red-eyed flies in a sample from a large population can be described using the binomial distribution. (Because the population sampled is large, we will assume that the sampling of each individual has a negligible effect on the probability of drawing a red fly on the next draw). (f) No, the individuals within different families may have different probabilities of having red ey es due to shared genetic differences. 4. (a) This estimate pools numbers from two different groups. Women with rosacea are not a random sample of the population. (b) If the control group women are a random sample, 15/16 have mites, 0.938. (c) p'= 17 / 20 = 0.85. The 95% confidence interval is 0.69 to 1.00. (Proportions cannot be larger than 1, so the confidence interval is truncated at 1). (d) For the women with rosacea, p' = 18/20 = 0.90. The 95% confidence interval is: 0.769 < p < 1.0. 5. (a) Pr[male is eaten] = 21/52 = 0.404. p'=23/56 = 0.411. The 95% confidence interval is from 0.282 to 0.540. (b) This estimate is consistent with a 50% capture of males, but not with a 10% capture of males. (c) The larger sample would not change the estimate of the proportion (the fraction is exactly the same), but it would reduce width of the confidence interval. (The sample size is in the denominator of the confidence interval equations.) 6. Estimated probability of moving north is 22/24 = 0.917. To test the hypothesis that north and south movements of ranges are equally probable, we need to calculate the probability of 22, 23, or 24 species moving northward if p = 0.5. We will make this a two-tailed test, just to satisfy the skeptics, so we'll multiply this probability by two. ' 24 $ 22 -5 Pr[22] %% " 0.5 (1 ! 0.5) 2 = 1.6 # 10 " & 22 # =
Doing the same for 23 and 24, summing all three, and multiplying by two, we find that P = 3.6 # -5 10 . We can confidently reject the null hypothesis. 7. (a) The first study with the narrower confidence interval probably had the larger sample size. (b) The estimate with the smaller confidence interval is the more believable, in the sense that the true value is likely to be somewhere near the estimate. (c) The differences are what we might expect due to random sampling: the second confidence interval overlaps the estimate of the first study. 8. (a) This is evidence that males' choices may be affected by female positioning of the females: the probability of seeing 19 of 24 choose the M0 female with null hypothesis that there is no preference ( p = 0.5). With a binomial test,
{ [ ]
[ ]
[ ]
[ ] = 2{0.00253 + 0.0006 + 0.0001 + ...}
[ ]
[ ]}
P = 2 Pr 19 + Pr 20 + Pr 21 + Pr 22 + Pr 23 + Pr 24
=
0.0066.
Therefore P < 0.05, and we can reject the null hypothesis that the males have no preference. (b) However, the males may have preferred one female over the other for a number of reasons. If the females were sisters, this would reduce the number of differences due to genes and maternal
environment, and so would be more likely to test for fetal position. Ideally, this would be done with 24 sets of sisters so that each trial is independent. 9. (a) In 20 samples, the average proportion of A will be 30%. (b) We would expect the number of A strain bacteria in samples to conform to the binomial distribution. (c) The standard deviation of the proportion of A cells is the standard error: ! pˆ
=
0.3(1 " 0.3) 15
=
0.118. (d) 95% of the
technicians should construct a confidence interval for the proportion of A cells that includes 0 .3. 10. (a) 1/6 of the 12 dice should have "3", or 2 dice. (b) Pr[no three out of 12 rolls] = (1 - Pr[3 on ! 12$ 12 12 one roll]) = 5/6 = 0.112. (c) Pr[3 threes out of 12] = Pr[3] # & 0.167 3 (1 ' 0.167) 9 = 0.197. " 3 % =
(d) Each die has six faces, each with probability 1/6 of showing. The average value for the die will be the sum of each value times its probability, or (1 # 1/6) + (2 # 1/6) + . . . (6 # 1/6) or 1/6 # 21 = 3.5. For twelve dice, the average sum of the numbers showing will be 12 # 3.5 = 42. (e) Pr[all dice show 1 or 6] = (Pr[1]+Pr[6])12 = 1.88 # 10-6.
11. (a) 6052 out of 12028 of the deaths occurred in the week before Christmas, or 0.503. (b) Using the Agresti-Coull confidence interval approximation for proportions, p' = (X + 2) / (n + 4), where X is the number of successes (= 'died before Christmas') and n is the total number of trials (= 'died within a week of Christmas') = 6054 / 12032 = 0.503. Then the confidence interval is 0.494 to 0.512. (c) This interval includes 50%, which is what is expected by chance. There is no statistical support for the belief that the living hang on until that special day. 12. (a) We can calculate the P -value by summing the probabilities of 13, 14, 15, or 16 resin clades ' 16 $ 13 having more species, then multiplying by two. Pr[13] %% " 0.5 (1 ! 0.5) 3 = 0.0085. Summing " & 13 # =
the probabilities for 13 through 16, then multiplying by two, P = 0.021. (b) This is an observational study, not an experimental study. We did not randomly assign clades to have latex/resin. It is possible that clades with resin have higher diversity not due to the resin, but due to some third factor that affects both resin and the number of species.
Chapter 8 1. (a) There would be six categories, one for each outcome, and so five degrees of freedom, since no parameters are estimated. (b) There are 11 categories (from 0 heads to 10 heads), and ten degrees of freedom, since no parameters are estimated. (c) There are now nine degrees of freedom, since p must be estimated from the data. (d) There would be five categories (0 to 4 insect heads per sample), and three degrees of freedom, since the mean number of heads per sample would need to be estimated from the data. 2. No, the tickets purchased on a given day are not all independent; for example a single person could buy multiple tickets at one purchase. Therefore this is not a random sample as presented. 3. To test whether nematodes are distributed at random among fish, we can see whether the data fit the Poisson distribution. The mean number of parasites per fish is X = 103(0) + 72 (1) + 44 (2) + 14 (3) + 3( 4 ) + 1(5) + 1(6) = 0.94538 .
The expected values from a Poisson distribution with this mean are given below, from the proportion expected from the Poisson distribution times the total sample size, n= 238: Number of Observed Expected parasites 0 103 92.47 1 72 87.42 2 44 41.32 3 14 13.02 4 3 3.08 5 1 0.58 6 1 0.09 The expected values for 5 and 6 are below one, so we will combine 4 through 6 into one category. Number of Observed Expected (Observed ! Expected ) 2 parasites Expected 0 103 92.47 1.20 1 72 87.42 2.72 2 44 41.32 0.17 3 14 13.02 0.07 5 %4 3.75 0.42 2 &2 = 4.58. There are five categories, one parameter estimated, so df = 3. ! 0.05,3
=
7.81. 4.58 <
7.81, so we cannot reject the random distribution of parasites on fish. 4. (a) If pine seedlings are distributed randomly, the frequencies should fit the Poisson distribution. (b) The mean number of seedlings per quadrat was 1.325. Number of Observed Expected seedlings 0 47 21.26 1 6 28.18 2 5 18.67 3 8 8.24 14 3.65 %4 The expected counts for 5-7 were below 1, so 4 through 7 were combined. & = 88.03, with df = 2 2 3 (five categories, minus one, minus one parameter estimated). ! 0.001,3 16.27 , and & is greater than this critical value, so P < 0.001. (c) Seedlings are not distributed at random. (The variance, 3.49, is much greater than the mean here, indicating that some quadrats are much more likely than others to have seedlings: seedlings are clumped.) 2
=
5. (a) The mean number of goals per side is 1.26. (b) Fitting this to a Poisson and testing with a goodness of fit test, we must again combine the four goals and above categories to avoid expected values less than one. Number of Observed Expected
goals 0 1 2 3 %4
37 47 27 13 4
&2 = 0.44, which is less that the critical value
36.3 45.7 28.8 12.1 5.0 2
! 0.05,1
=
3.84 , so we cannot reject the null
hypothesis of a Poisson distribution of goals per side pe r game. 6. (a) Pr[8]
=
' 15 $ 8 %% " 0.5 (1 ! 0.5) 7 = 0.19638. (b) We test the fit of these data to the binomial " & 8 #
distribution, combining categories where the expected value is less than one. This leads to the following table:
Total wins 0-2 3 4 5 6 7 8 9 10 11 12 13-15
# wrestlers 32 56 112 190 290 256 549 339 132 73 56 47
expected proportion
expected number
0.0037 0.0139 0.0417 0.0916 0.1527 0.1964 0.1964 0.1527 0.0916 0.0417 0.0139 0.0037
7.85 29.52 88.56 194.84 324.73 417.51 417.51 324.73 194.84 88.56 29.52 7.85
test statistic:
(Observed ! Expected )
2
Expected
74.3 23.8 6.2 0.1 3.7 62.5 41.4 0.6 20.3 2.7 23.8 195.3 454.6
There are twelve categories and no estimated parameters, so eleven degrees of freedom. The test 2 statistic greatly exceeds the critical value for & with df = 11 for P = 0.001, so the data do not fit the binomial distribution.
(c) These data show fewer than expected wrestlers with seven wins and more than expected with eight wins. In a match when one wrestler has already one eight rounds, cheating so as to lose to an opponent who has won seven rounds would tend to inflate the number of wins at eight. (d) The binomial distribution of wins assumes that each match is independent, and that each wrestler has the same probability of winning each match. Motivation and stamina might well change from match to match. A wrestler with only seven wins would have a lot more motivation to win another match than would one who had already won eight rounds.
7. (a) number heads Observed 0 6 1 32 2 105 3 186 4 236 5 201 6 98 7 33 8 103 Total 1000
binomial expectation 0.0039 0.0313 0.1094 0.2188 0.2734 0.2188 0.1094 0.0313 0.0039
(Observed ! Expected ) Expected 3.91 31.25 109.38 218.75 273.44 218.75 109.38 31.25 3.91 1000
2
Expected 1.12 0.02 0.18 4.90 5.13 1.44 1.18 0.10 2513.81
The expected values for 0 and 8 are less than 5, so we combined (0 and 1) and (7 and 8) to produce: 2 (Observed ! Expected ) number heads Observed 0 or 1 38 2 105 3 186 4 236 5 201 6 98 7 or 8 136
Expected 35.16 109.38 218.75 273.44 218.75 109.38 35.16
Expected 0.23 0.18 4.90 5.13 1.44 1.18 289.2
Total
1000
&2 =302.27
1000
A & goodness-of-fit test to the binomial on these data with the null hypothesis that the proportion of heads is p = 0.5, gives &2 = 302.3. Reject the null hypothesis with P < 0.001 (df = 6). (b) Graphing this, we can see very easily that there is an excess of coins yielding eight heads. (c) We would expect the observed number of coins yielding 8 heads to be roughly equal to the number of coins yielding 0 heads, or around 4. This means that there is an excess are two-headed coins, so roughly 103 – 4 = 99. 2
8. We will use a goodness of fit test to see whether the data differ from the 0.5 expectation if cancer deaths were randomly distributed around Christmas. 2 (Observed ! Expected ) observed die before Xmas die after Xmas total deaths
6052 5976 12028
Expected
expected 6014 6014
0.24 0.24
&2 = 0.48, and df = 1. 0.48 < 3.84, the critical value for P = 0.05, so we do not reject the hypothesis of a random distribution of cancer d eaths around Christmas. 9.
(Observed ! Expected ) observed cancer death non-cancer death total
2
Expected
14% expected 91
30.8
117.67
129
189.2
19.15 test stat: 136.82
220
220
2
df = 1 ; & =136.82 >> 10.83, the critical value for P = 0.001, so we can reject the hypothesis that the cast and crew suffered the population-wide cancer mortality rate, P < 0.001. 10. The average admissions are 20 per night. What is the probability of five or fewer admissions? Assume that each admission is independent (no riots, please). Then, they should be Poisson distributed. To find the probability of a qu iet night, we sum the probabilities of 0 to 5 admissions. admissions 0 1 2 3 4 5
probability 2.06E-09 4.12E-08 4.12E-07 2.75E-06 1.37E-05 5.50E-05 -5
Total probability = 7.2 # 10 . Best to keep that coffee brewing: you'll need to work 58 years on average before you'll have a quiet night!
Chapter 9 1. (a)
(Observed ! Expected ) white blue total
observed capture escape 9 92 92 10 101 102
total 101 102 203
expected capture white 50.25 blue 50.75 total 101
2
Expected escape 50.75 51.25 102
total 101 102 203
white blue total
capture 33.86 33.53
escape 33.53 33.20
total
134.13
2
& = 134.13 >> 10.83 (critical value for P = 0.001 for 1 df ), so P < 0.001. The two types of pigeons differ in their rate of capture. (b) Odds of capture for white rumped: (9 / 101) / (92 /101) = 0.098. Odds of capture for blue rumped: (92 / 102) / (10 / 102) = 9.2 Odds ratio: 0.098 / 9.2 = 0.01 For the 95% confidence interval, we use the log odds ratio, ln(0.01) = -4.54. ˆ R)] = 1 + 1 + 1 + 1 = 0.48 .For the 95% confidence interval, we use Z = 1.96. SE [ln(O 9 92 92 10 !4.54 ! 1.96(0.48) < ln(OR) < !4.54 + 1.96(0.48)
!5.49 < ln(OR) < !3.6 0.004 < OR < 0.027
2.
(Observed ! Expected ) observed one multi 69 20 157 16 226 36
malar healthy total
Expected
expected total 89 173 262
malar healthy total
2
one 76.77 149.23 226
multi 12.23 23.77 36
total 89 173 203
one 0.79 0.40
malar healthy total
multi 4.94 2.54
total
8.67
& = 8.67 > 7.88, the critical value for P = 0.005 for 1 df , (but less than 10.83, the critical value 2
for P = 0.001), so P < 0.005. Infected and uninfected mosquitoes differ in the probability of multiple blood meals. 3. (a) The odds of the second suitor being accepted if the first was eaten are 0.5, while the odds of the second suitor being accepted if the first escaped are 22.0. There is a much higher probability of the second suitor being rejected if the first was eaten. (b) There are only four cells, so if any 2 cell has an expected frequency of less than five, the assumptions of the & contingency test will be violated. In this case, it seems likely that one cell will less than five, so it would be necessary to use Fisher's Exact Test in this case. 4.
(Observed ! Expected ) observed
leave stay total
fire 18 2 20
Expected
expected rev fire 6 14 20
white 0 20 20
total 24 36 60
leave stay total
fire 8 12 20
2
rev fire 8 12 20
white 8 12 20
total 24 36 60
fire leave stay total
12.5 8.33 20.8
rev fire 0.5 0.333 0.833
white 8 5.33 13.3
total 21 14 35
&2 = 35.0, for 2 df , (since there are two rows, three columns, so (2-1)(3-1) = 2 df . 35.0>13.82, the critical value for P = 0.001 for 2 df , so P < 0.001. Yes, there is evidence that reed frogs react consistently to the sound of fire. 5. observed juv m juv f adult m adult f
1 6 7
expected juv m juv f
total
11 4 15
12 10 22
adult m adult f total
3.82 3.18 7
total
8.18 6.82 15
12 10 22
(a) Two of the expected values are less than 5, so we cannot use a & contingency analysis. Fisher's exact test is a good alternative. (b) The effects is in the expected direction. 2
6. (a) Proportions of kids in each TV viewing class having violent records eight years later, and confidence intervals (using Agresti-Coull calculations from chapter 7) 1hr 1-3 hr 3+ hr
record 5
total 88
proportion 0.057
p' 0.0761
low CI 0.022
high CI 0.130
87
386
0.225
0.2282
0.187
0.270
67
233
0.288
0.2911
0.233
0.349
(b) Contingency table for test of independence of TV watching and later violent record. 2 (Observed ! Expected )
class 1hr 1-3 hr 3+ hr
observed no record record 83 5 299 166 548
87 67 159
total 88 386 233 707
class 1hr 1-3 hr 3+ hr
expected no record record 68.21 19.79 299.2 180.6 548
86.81 52.4 159
Expected total 88
class 1hr
no record 3.21
record 11.05
386 233 707
1-3 hr 3+ hr
0.0001 1.18
0.0004 4.068
total
19.5
& = 19.5, for 2 df , (since there are three rows, two columns, so (3-1)(2-1) = 2 df . 35.0>13.82, the 2
critical value for P = 0.001 for 2 df , so P < 0.001. There is a relationship between watching TV and future violence: those that watched less than one hour TV were less likely to have a record, while those that watched three or more hours were more likely to have a record. (c) No, this does not prove that TV watching causes aggression in kids. This was an observational study, not experimental, so we do not know if kids watching more TV have other factors in common (e.g. lower parental supervision, etc.). 7.
(Observed ! Expected ) abstainers drinkers total
observed arrest healthy 12 197 9 192 21 389
Expected
expected total 209 201 410
abstainers drinkers total
2
arrest 10.70 10.30 21
healthy 198.30 190.70 389
total 209 201 203
abstainers drinkers total
arrest 0.16 0.16
healthy 0.01 0.01
total
0.34
&2 = 0.34, for 1 df << 3.84 (critical value for P = 0.05), so we cannot reject the null hypothesis that there is no difference in heart attack risk between drinkers and non-drinkers. (b) No, this does not imply that drinking has no effect on cardiac arrest, only that this study found no evidence for a connection. This study was observational: non-drinkers at the age of 40 may have been heavy drinkers at the age of 20, or may have never consumed alcohol. drinking may have an effect on heart attacks, but with too small an effect to be detected by this sample size. 8. (a) 9.7% of women with vaginal deliveries became depressed, as compared to 12.3% of women with C-sections. (b) The odds for vaginal are 0.11 vs. 0.14 for C-section, yielding a ratio of 0.76. (c) 2 (Observed ! Expected ) observed depressed vaginal C-section total 2
Expected
expected not
total
depressed
1025
9520
10545
48
341
389
1073
9861
10934
not
total
1034.83
9510.17
10545
C-section
38.17
350.83
total
1073
9861
vaginal
depressed
not
vaginal
0.09
0.01
389
C-section
2.53
0.28
203
total
2.91
& = 2.91, for 1 df , which is less than 3.84 (critical value for P = 0.05), so we cannot reject the null hypothesis that there is no difference in depression risk for different methods of childbirth. 9. (a) The odds of a migraine sufferer having a cardiac shunt are 44:49, or 0.90, while the odds for a non-sufferer are 16:77, or 0.21. The odds ratio is 4.32. (b) To find the confidence interval, we first take the log of the odds ratio, ln(4.32) = 1.46 and calculate the SE of the odds ratio, 1 44 + 1 49 + 1 16 + 1 77 = 0.34. The confidence interval for the log odds ratio is 1.46 1.96(0.34) to 1.46 + 1.96(0.34), or 0.79 to 2.14. Taking the exponential of these to find the confidence interval of the odds ratio, we get exp(0.79) = 2.2, and exp(2.14) = 8.5. 2.2 < OR < 8.5. 10. (a) The proportion of babies born with finger defects is 6522 out of 6,839,854, or 0.001. The 95% confidence interval (using the Agresti-Coull method) is from 0.0009 3 to 0. 00097. It is very likely that the proportion of babies born with finger defects is between 0.00093 to 0. 00097. (b) 2 (Observed ! Expected )
smoke no smoke
observed defect normal 805 1280 4366 5171
9062 10342
Expected
expected total 2085 13428 15513
smoke no smoke total
defect 695.00
normal 1390.00
total 2085
4476.00 5171
8952.00 10342
13428 203
smoke no smoke total
total
defect 17.41
normal 8.71
2.70
1.35
total
30.17
For 1 df , the test statistic exceeds the critical value for P = 0.001, so P < 0.001. There is an association between finger defects and smoking during pregnancy. (c) The odds of a smoking mother having a child with a finger defect are 0.62, while the odds of a non-smoking mother having a child with a finger defect are 0.48. The odds ratio is 1.3 (smoking mothers are 1.3 times as likely to have a child with a finger defect). For the confidence interval, we first calculate the ln odds ratio = 0.27. We then calculate the SE of the ln odds ratio, the square root of the sum of reciprocals of the frequencies, or 0.05. The confidence interval for the log odds ratio is +1.96 #
0.05, so from 0.172 to 0.362. Taking the exponent of the confidence interval for the ln odds ratio, we find that the 95% confidence interval for the odds ratio is from 1.19 to 1.44.
Chapter 10 1. From statistical table B. (a) 0.09012 (b) 0.09012 (c) Pr[Z > -2.15] = 1 - Pr[Z < -2.15] = 1 - Pr[Z > 2.15] = 1-0.01578 = 0.9842. (d) Pr[Z < 1.2] = 1 - Pr[z > 1.2] = 1- 0.11507 = 0.8849 (e) Pr[0.52 < Z < 2.34] = Pr[0.52 < Z] - Pr[2.34 < Z], because the first value is the area under the curve from 0.52 to infinity, and the second value is the area under the curve from 2.34 to infinity. The difference will be the area under the curve from 0.52 to 2.34. 0.30153 - 0.00964 = 0.29189. (f) Pr[-2.34 < Z < -0.52] = 0.29189: the curve is symmetrical on either side of 0. (g) Pr[Z <-0.93] = Pr[z > 0.93] = 0.17619 (h) Pr[-1.57 < Z < 0.32] = (1 - Pr[Z > 1.57]) - Pr[Z > 0.32] = (1 0.05821) - 0.37448 = 0.56731 2. (a) To determine the proportion of men excluded, we convert the height limit into standard normal deviates. (180.3 - 177.0) / 7.1 = 0.46. Pr[Z > 0.46] = 0.32276 - roughly one third of British men are excluded from applying. (b) (172.7 - 163.3) / 6.4 = 1.47. Pr[Z > 1.47] = 0.07078 - which is the proportion of British women exclude d. 1 - 0.0708 = 0.9292 proportion of British women acceptable to MI5. (c) (183.4 - 180.3) / 7.1 = 0.44 standard deviation units above the height limit. 3. (a) B is most like the normal distribution. A is bimodal, while C is skewed. (b) All three would generate approximately normal distributions of sample means due to the Central Limit Theorem. 4. (a) Pr[weight > 5 kg]? Transform to standard normal deviate: (5 - 3.339) / 0.573 = 2.90. Pr[2.90 < Z] = 0.00187. (b) Pr[3 < birth weight < 4]? Transform both to standard normal deviates, and subtract probabilities of the Z values. (3 - 3.339)/0.573 = -0.59. (4-3.339)/0.573 = 1.15. Pr[-0.59 < Z] = 1 - Pr[0.59 < Z] = 1 - 0.2776 = 0.7224. Pr[1.15 < Z] = 0.12507. 0.7224 - 0.12507 = 0.59733. (c) 0.06681 babies are more than 1.5 standard deviations above, with the same fraction below, so 0.13362 of babies are more than 1.5 standard deviations either way. (d) First, transform 1.5 kg into normal standard dev iates: 1.5 / 0.573 = 2.62 standard deviations. Pr[2.62 < Z] = 0.0044. Since the distribution is symmetric, we multiply this by two to reflect the probability of being 2.62 standard deviations above or below the mean: 0.0044 * 2 = 0.0088. (e) The standard error is the same as the standard deviation of the mean. It is equal to the standard deviation divided by the square-root of n, or 0.573 / 10 = 0.18 kg. To find the probability that the mean of a sample of 10 babies is greater than 3.5 kg, we transform this mean into a Z score. (3.5 - 3.339) / 0.18 = 0.89 standard deviations of the mean = 0.18673. 5. (a) The right-hand graph has the higher mean (ca. 20 vs. 10), the left-hand has the higher standard deviation (ca. 2 vs. 1). (b) The right hand graph has the higher mean (ca. 15 vs. 10), and the higher standard deviation (ca. 5, vs. 2.5). 6. The standard deviation is approximately 10, as the region within one standard deviation from the mean will contain roughly 2/3 of the data points. 7. (a) In a normal distribution, the modal value occurs at the mean, so the mode is 35mm. (b) A normal distribution is symmetric, so the middle data point is the mean, 35 mm. (c) Twenty percent of the distribution is less than 20 mm in size. (Why? Normal distributions are symmetric,
so if 20% of the distribution is 15 mm or larger than the mean, 20% must be 15 mm or smaller than the mean.) 8. (a) The right-hand distribution (II.) would more have sample means that had a more normal distribution, because the initial distribution is closer to normal. Both distributions would converge to a normal distribution if the sample size were sufficiently large. (b) The distribution of the sums of samples from a distribution will be normally distributed, given a sufficiently large number of samples. 9. (a) The probability of sampling haplotype A is p = 0.3. We expect a sample of 400 fish to have 0.3 # 400 A individuals on average, or 120. The standard deviation is approximately
(
)
n p 1! p
=
400( 0.3)(0.7)
=
9.17 .
To find the probability of sampling 130 or more of
haplotype A, we convert this into a standard normal deviate: (130 - 120) / 9.17 = 1.09; Pr[1.09 < Z] = 0.13786. (b) Pr[at least 300 B haplotype out of 400]? Average number of B sampled = (1-p) # 400 = 280. Convert to standard normal: (300 - 280) / 9.17 = 2.18; Pr[2.18 < Z] = 0.01463. (c) Pr[115 < freq A < 125]? Convert to standard normal deviates, find the probabilities, and calculate the difference. (115-120)/9.17 = -0.55. (125 - 120)/ 9.17 = 0.55. Pr[-0.55 < Z] = 1 Pr[0.55 < Z] = 1 - 0.29116 = 0.70884. Pr[0.55 < Z] = 0.29116. 0.70884 - 0.29116 = 0.41768. 10. Average expected number of cancer victims = 0.14 # 220 = 30.8. Standard deviation =
(
)
n p 1! p
=
220(0.14)( 0.86)
=
5.15 .
Convert actual number of victims, 91, into standard
normal deviate: (91 - 30.8) / 5.15 = 11.7. This is off the chart: P < 0.00002. 11. SE = s n . Z = ( y - mean) / SE. Mean SD y SE 10 Z 10 14 15 -23 72
5 3 4 50
15 15.5 -22 45
1.58 0.95 1.26 15.81
0.63 0.53 0.79 -1.71
Pr(Y > y)
SE30 Z 30
Pr(Y > y)
0.26435 0.29806 0.21476 0.95637
0.91 0.54 0.73 9.18
0.13567 0.18141 0.08534 0.99846
1.10 0.91 1.37 -2.96
Chapter 11 1. The 99% confidence interval must be larger than the 95% confidence interval, in order to have a higher probability of capturing the true mean. 2. (a) The sample mean is 10.32 cm and the standard error of the mean is 0.056 cm. If we sampled this wolf population repeatedly, 68% of the estimates of the mean would lie between 10.32 ± 0.056. (b) The 95% confidence interval for the mean is 10.32 ± (0.056 t0.05(2),34 ). t = 2.03, so the 95% confidence interval is: 10.21 < µ < 10.44 cm. (c) The variance is 0.11005. There are 35 2 individuals, so 34 df . The & critical values are looked up in Statistical Table A for ' = 0.01 / 2 and ' = 1 - (0.01 / 2) are 58.96 and 16.50. We now calculate the lower bound of the confidence interval: 34 (0.11005) / 58.96 = 0.063. The upper bound is 34 (0.11005) / 16.5 = 0.227. 0.063 < ( 2< 0.227. (d) The 99% confidence interval of the standard deviation of the sample is found by taking the square-root of the variance confidence interval: 0.252 < ( < 0.476, around the estimated standard deviation of 0.332.
3. No, using n = 70 would assume that the right distance was independent of the left distance. This isn't likely, as most animals are symmetric, so we would e xpect similar distances on each side of the mouth. 4. (a) To test whether the mean fitness changed, we will test whether the mean fitness after 100 generations was significantly different from 0. For this, we will use a t -test. We need the sample mean, the sample standard error, and the sample size. Our hypothesized value from the null hypothesis of no change in fitness, µ0, is 0. The mean is 0.2724, the standard deviation is 0.600, and the sample size is 5. From this, we calculate the SE = s / n = 0.6 / 5 = 0.268, and we calculate the t -statistic: t = (mean - µ0 ) / SE = (0.2724 - 0) / 0.268 = 1.02. This is a two-tailed test, so we look up the critical value of '(2) = 0.05 for 4 df , 2.78. t < t crit = 2.78, so we do not reject the null hypothesis of no change in fitness. (b) The 95% confidence interval for the mean fitness change is 0.2724+2.78(0.268) : -0.47 < µ < 1.02. (c) We are assuming that fitness is normally distributed and that the five lineages are random and independent. 5. (a) The mean testes area for monogamous lines is 0.848, while the mean area for polyandrous lines is 0.950. The standard deviation was 0.031 and 0.034 for the two sets. (b) The standard error of the mean is 0.015 for the monogamous lines and 0.017 for the polyandrous lines. (c) The 95% confidence interval for the mean in the polyandrous line is 0.95 ± 0.017t 0.05(2),3 = 0.95 ± 3.18 (0.017), or 0.90 < µ < 1.00. (d) The 99% CI for the standard deviation of testes area among monogamous lines requires the sample variance (0.001), the degrees of freedom (3), and 2 2 the critical values of the & distribution for the appropriate ' and df . The & critical values are looked up in Statistical Table A for ' =0.01 / 2 and 1 - (0.01 / 2) are 0.07 and 12.84. Then, the 99% CI for the variance is 3 (0.001) / 0.07 and 3 (0.001) / 12.84, or 0.0002 to 0.04. To get the 99% CI for the standard deviation, we take the square root of each of these, or 0.015 < ( < 0.20. 6. (a) The 95% CI for the discontinuity score requires the estimated mean (0.183), the SE (calculated as the s divided by n , or 0.051), and the critical t value for '(2) = 0.05 for 6 df , 2.45. Then the CI is 0.18 ± (0.051 * 2.45) , or 0.058 < µ < 0.308. (b) To test whether the sample mean continuity score of 0.183 is consistent with the predicted mean, µ0 = 0, we use the t -test. t = (mean - µ0) / SE = (0.183 - 0) / 0.051 = 3.6. The critical value for '(2) = 0.05 for 6 df is 2.45. t > t crit, so we reject the null hypothesis that the mean discontinuity score is zero. 7. In this case, we want to see if rats do better than the chance performance, so we are comparing their scores with µ0 = 0.5. The mean is 0.684, the standard deviation is 0.071. There were seven rats, so SE = 0.071 / 7 = 0.027. t = (0.684 - 0.5) / 0.027 = 6.82. For '(2) = 0.05, t crit for six df = 2.45; t > t crit so we reject the null hypothesis that rats were doing as expected by chance. (Moreover, we can show that P < 0. 002.) (For this we assumed that the seven rats were random, independent samples and that their performance scores were normally distributed). 8. Mean coconut weight is 3.2 kg, upper 95% CI bound is 3.5 kg. The confidence interval for the mean is symmetric, so the lower bound must be 0.3 kg below the mean, or at 2.9 kg. 9. The confidence interval of the variance is the square of the bounds for the standard deviation.
Standard deviation 2.22 < ( < 4.78 20.6< <26.1 36.4 < ( < 59.6 13.63< <16.70
Variance 2 4.93< <22.85 2 425.4 < ( < 678.8 2 1325.0< <3552.2 2 185.8 < ( < 279.0
10. In this case, we wish to test whether the relative swimming speed in the syrupy pool is different from µ0 = 1. We calculate t from the mean (1.0117) and the standard error (0.00997): t = (1.0117 - 1) / 0.00997 = 1.17. The critical value for '(2) = 0.05 and17 df is 2.11, so we do not reject the null hypothesis that swimming speed remained constant in the two environments. (b) The 99% confidence interval for the relative swimming speed in syrup is 1.01 ± 0.01 t 0.01(2),17 = 1.01 ± 0.01 (2.9), or 0.98 < µ < 1.04.
Chapter 12 1. (a) Paired t -test; (b) Paired t -test; (c) Two-sample t -test; (d) Paired t -test; (e) Two-sample t -test; (f) Paired t -test; (g) Two-sample t -test; (h) Paired t -test. 2. (a) On average there are 25 more injuries on No Smoking Day than on the prior Wednesday. (b) The 99% confidence interval is calculated as the mean difference (25) + SE of the difference times t0.01(2),9 df . t = 3.25, SE = 10.22, so the confidence interval is: -8.2 < µ d < 58.2. (c) If we had data from 100 different samples and we calculated the mean difference for no smoking day and a normal day each time, we expect that 99 of these samples would include the true mean difference. (d) We can test whether the difference is different from 0 by using the paired t-test: t = (25 - 0) / 10.22 = 2.45, for 9 df . t > 2.26, the critical value for '(2) = 0.05, so we conclude that the accident rate does differ between no smoking day and a normal day (and is higher on no smoking day). P < 0.05. 3. The change in mean length is -5.81 mm, the standard deviation is 19.5 mm, and the sample size is 64. The SE is 19.5 / 64 = 2.44. The confidence interval for the difference is the mean difference + 2.44 * t, where t0.05(2),63 = 2.0, so the confidence interval is: -9.69 mm < µ d < -1.935 mm. This assumes that the distribution of change in length is a normal distribution and that the iguanas are a random sample from the population. (b) The 95% confidence interval for the SD of the change in length is derived using the formula for the confidence interval of the variance in 2 chapter 11. The variance of the difference is 380.25. We need the df (63) and the appropriate & for 0.025 and 0.9725, which are 86.83 and 42.95. Then, the lower confidence interval is (380.25 2 2 * 63) / 86.83 = 275.9 mm , and the high is (380.25 * 63) / 42.95 = 557.8 mm . The standard deviation 95% confidence interval is found by taking the square-root of the variance confidence interval, or 16.6 mm < ( < 23.6 mm. (c) No change in length implies a difference of zero, which is excluded from the confidence interval in part (a). This implies that a hypothesis test using these data would reject the null hypothesis. of no difference. (d) t = -2.38, for 63 df . We reject the null hypothesis of no mean difference with P < 0.05, since t > 2.0, the cutoff for '(2) = 0.05 for 63 df . 4. (a) To calculate the 95% confidence intervals for the mean number of dung beetles captured, we must assume that the number captured is normally distributed. So, for the dung-addition
treatment, the confidence interval is 4.8 + (2.26 # 1.03): 2.47 < µ < 7.13. For the control, the confidence interval is 0.51 + (2.26 # 0.28), or -0.13 < µ < 1.15. (b) The two-sample t -test cannot be used test for differences in the means, since the standard deviations are more than three-fold different between the two groups. Instead, a Welch's approximate t -test is appropriate. 5. Because these are not paired samples, we will analyze the difference of the means, not the mean 2 of the differences. The monogamous flies had a mean testes size of 0.8475 mm , the polyandrous 2 2 0.95 mm , for a difference of 0.1025 mm . The 95% confidence interval for this difference requires finding the SEY1bar - Y2bar ,= )(s p2(1/n1 + 1/n2), where n1 and n2 = 4, and s p2 = (df 1s12 + 2 2 2 df 2s2 )/ (df 1 + df 2), where df 1 = df 2 = 3, and s1 and s2 = 0.0010 and 0.0011 respectively. Then, SEY1bar - Y2bar = 0.023. The confidence interval for the difference is the standard error times t0.05(2),6 = 2.45, so 0.1025+ 2.45 # 0.23, or 0.046 to 0.159. (b) The null hypothesis is that there is no difference between the monogamous and polyandrous treatments in testes size, so (µ1 - µ2)0 = 0. t = the difference in means, 0.1025, over the SE of the difference, 0.23. t = 4.48, for 6 df . t crit = 3.71 for P = 0.01, so P < 0.01. The mean testes sizes are significantly different. 6. (a) On average, 33% more of the male bodies were covered if they emitted pheromones. s p = 26.5%; SE Y !Y = 6.02%. df 1 = 48 and df 2 = 31, so df = 79. t0.05(2), 79 df = 1.99, so the confidence 1
2
interval is: 21% < µ 1 ! µ 2 < 45%. (b) Using a two-sample t -test, we will assume that the percent coverage is normally distributed, that each snake is independent, and that the standard deviations are not different (they are not more than threefold different). The null hypothesis is that there is no difference between the males emitting pheromones and those not, so (µ1 - µ2)0 = 0. t = 0.33 / 0.0602 = 5.47 > 3.9, the critical value for '(2) = 0.0002 for 79 df , so we can reject the null hypothesis, with P < 0.0002. 7. (a) The PLFA levels between the control and addition plots before cicada death were not significantly different (two sample t -test: t = -0.19, df = 42, P > 0.10). (b) After the cicada addition, the PLFA levels differed significantly: t = 2.89, df = 42, P < 0.01. (c) No: we are interested in the effect of the cicada addition, so we need to look at whether the change in PLFA levels differed between the treatments. The correct c omparison is the mean change in the control plots compared to the mean change in the addition plots. It is possible to have a situation in which test b was significant and test a was not, but the effect of cicada addition was not significant. For instance, the control plots may hav e been non-significantly lower in PLFA prior to cicada addition, and significantly lower after cicada addition, but the difference might not be significant. 8. As described, this test assumes that the eight "o pen water" samples were independent of the eight "near shore" samples, as it uses a two-sample t-test (and so would have 14 df ). Differences in growth rate could be due to differences between lakes, so the two samples within each lake are not independent. The paired t-test would better reflect this (and would have 7 df ). 9. (a) Since we assume that the distributions are normal, we can use the F -test to compare the variances. The ratio of the variances (the larger variance is always in the numerator) is F = 2 2 (0.1582) /(0.0642) = 0.025 / 0.004 = 6.07. The degrees of freedom are 32 and 19. F 0.05(1),32,19 = 2.05 (between 2.07 for 30 df in the numerator and 2.03 for 40 df ), so we reject the null hypotheses that the variances are equal. (b) The variances are not equal, but the difference in the standard deviations is not greater than threefold. The refore a two-sample t -test could be used: t =
0.28, df = 51, P > 0.05. You could use Welch's approximate t test rather than the two-sample ttest. t = 0.13, with 46 df , which is not significant. 10. (a) Remember to multiply each value of flower length by the number of flowers in that category, when calculating the mean and variance (the zeros are dropped in the equation below): 4 (55)
(58) + 41(61) + 75(64 ) + 40(67) + 3( 70)
+ 10
4
4 (55 ! 63.5)
2
+
10(58 ! 63.5)
2
+
+ 10 + 41 + 75 + 40 + 3
41(61 ! 63.5) (4
+
2
+
75(64 ! 63.5)
2
+
= 63.5
40(67 ! 63.5)
2
+
3( 70 ! 63.5)
2
10 + 41 + 75 + 40 + 3) ! 1
(b) For the simplest test, we will assume that distributions are normal. Then we can use the F test to compare the variances. 42.4 / 8.6 = 4.93; 443, 172 df . Looking this up (roughly) we find that the critical value is 1.21 (for 1000,200) and 1.31 (for 200,100) for '(1) = 0.05, so we conclude that the variance for the F2 is significantly greater than the variance for the F1.
11. Paired t-test: mean difference is 22.3, which is significantly different from 0 ( P <0.01, t = 4.5, df =7). 12. (a) We would use Welch's approximate t-test for this comparison since the standard deviations differ between Pteronotus and the vampires, by more than threefold. (b) The average strength was higher for Pteronotus, which is in the opposite direction as predicted by the model. 13. (a) We can use a paired t -test in this case, as the body temperature measurements are taken on the same individuals as the brain temperature measurements. To do this, we calculate the difference in temperature between brains and bodies for each ostrich, find the mean difference (0.648 degrees C) and the standard error of the difference (0.116 degrees C). µ0 = 0: the null hypothesis is that the brain temperature does not differ from the body temperature. t = 0.648 / 0.116 = 5.6, with 5 df , which is greater than the critical value for '(2) = 0.01, 4.03, so P < 0.01. We reject the null hypothesis of no difference between brain and body temperature. (b) While our test is significant, the deviation is the opposite predicted from observations of mammals in similar environments: brains are hotter than bodies in ostriches, not cooler. 14. (a) B. (b) B. (c) From B, we can still be fairly confident that the groups are different, but we need to mentally double the size of the error bars to make this determination. (d) With sample sizes of 100, the standard errors will be on e tenth as great as the standard deviations. Graphs A, B, and C will be significantly different.
Chapter 13 1. (A) The point fall on a curve, not a line, so the distribution is unlikely to be normal. (B) There is some curvature to this distribution, more than would be expected for a normal distribution. (C) This is very close to the straight line distribution with the p oints more densely clustered at the center that you would expect from a normal distribution. (D) This quantile plot has two distinct curves. The middle of the plot has relatively few points, not the highest de nsity that you would expect with a normal distribution. 2. (a) I. No, this is a uniform distribution, not a normal distribution. II. No, this plot is bimodal, not normal.
=
8.6
III. No, this plot is right skewed. It looks log-normal rather than normal. IV. Yes, this is a normal distribution. (b) I. The sign test would be best for these data, as they are unlikely to transformed into anything resembling normal. II. The sign test would probably be best for this distribution as well. Bimodal distributions are tough to transform into anything else. III. These data could probably be tested by a one-sample t-test after transformation (probably a log transformation, as it is right-skewed). IV. These data could be tested by a one-sample t-test as they are. (c) I: B (this explains why there is such a constant density of points in the quantile plots). II: D. (The two peaks correspond to the two curves in the quantile plots; these are also the areas of the highest density in the quantile plots). III. A. (The density is highest on the left, with a few points on the right scattered over a large range on the x-axis). IV. C. (Sparse points at the extremes of the range; points falling along a straight line a s expected for a normal distribution). 3. (a) mean 2.75; 95% CI: -3.28 < µlog[x] < 8.78. (b) mean 1.86; 95% CI: -0.02 < µlog[x] < 3.74. (c) Not possible: cannot use ln transformation on negative values. (d) mean 4.23; 95% CI: -2.04 < µlog[x] < 10.5. (e) Here, we must use Y' = ln(Y + 1). Mean: 0.98; 95% CI: -0.23 < µlog[x] < 2.2. 4. After applying the arcsine to the square-root of the raw data, the mean is 1.39 and the SD is 0.09. 5. The simplest analysis is a sign test. We will score e ach lobster as "+" if it is pointing more towards home than away (so -90 to 90) and "-" if pointing more away than towards home (more than 90 or less than -90). Lobsters exactly at 90 or -90 will not be scored. Of the 14 scored
! ! 14$ 2 # ( # & 0.5 " " x % 14
lobsters, 11 were "+" and 3 were "-".
P
'
14 x
=
x 11 =
x
0.5
$ & %
=
0.057 .
Using the usual
standard of ' = 0.05, we cannot reject the null hypothesis with these data. 6. Mann-Whitney U test : U1 = 39. U2 =6, so U = 39. The critical value for n1 = 5 and n2 = 9 is 38. U is larger, so we reject the null hypothesis that there is no difference in the time until reproduction due to accidental death or infanticide. 7. A non-parametric approach to this problem is the Mann-Whitney U test, noting that there will be many ties, so our test will be less powerful than it might be. We assign a rank of "3" to the zeros (the average of 1-5), a rank of 11.5 to the 1's (average of 6-17), a rank of 19.5 to the 2's (average of 18-21), and a rank of 23 to the 3's (average of 22-24). Then we sum the ranks for the blind group: R1 = 140.5. We then calculated U1 = n1n2 + n1(n1+1)/2 - R 1 = 81.5. U2 = n1n2 - U1 = 144 81.5 = 62.5. U = 81.5. The critical value for n1 = 12 and n2 = 12 is 107. U is smaller, so we do not reject the null hypothesis that there is no difference in the number of gestures used by the blind vs. sighted humans. 8. (a) Benton: mean 0.67, variance 0.075; Warrenton: mean: 0.24, variance 0.007. The standard deviations are more than 3 fold different, so a two-sample t -test is inappropriate. (b) (1) Try a log transformation, followed by a two-sample t -test. (2) Welch's t -test on the non-transformed data, which does not require equal variances. (3) Mann-Whitney U test. (c) A log transformation might be appropriate because the population with the larger mean has the larger variance. The two-sample t -test on the transformed data rejects the null hypothesis: t = 0.961 / 0.249 = 3.87,
with 10 df . P < 0.01. (d) For Benton county, -0.94 < µlog[x] < -0.04. For Warrenton county, -1.84 < µlog[x] < -1.06. (e) The transformed confidence interval for Benton county is 0.39 to 0.96; for Warrenton county the interval is 0.16 to 0.35. If we found the means from many samples, 95% of the confidence intervals from these samples would co ntain the true mean. 9. Two-sample t-tests and confidence intervals are robust to violations of equal standard deviations so long as the sample sizes of the two groups are roughly equal and the standard deviations are within 3 times of one another. 10. (A) two-sample t -test (B) These distributions are skewed right, so a log-transformation could be tried. (C) This one is tough: neither distribution is normal, but they deviate in different ways, so that a transformation that would improve the fit for one would not help with the other. Also, the standard deviations will be very different. The d ifferences in shape make the rank-sum test inappropriate. A randomization test (chapter 19) is best. (D) two-sample t -test. (E) two-sample t test. 11. (a) The null hypothesis was that the spermatids of males and hermaphrodites did not differ in their mean size, while the alternative hypothesis is that they do differ in size. (b) U = 35910, n1 = 211, n2 = 700. Use the normal approximation: Z =
2U ! n1n 2
(
n1n 2 n1
)/3
+ n 2 + 1
= -11.32. This is highly
significant ( P < 0.00002). (c) It is not clear whether the variables are distributed in the same way for the hermaphrodites and the males: if not, the Mann-Whitney U test may be inaccurate. 12. Mann-Whitney U tests are insensitive to outliers and may be used appropriately in this case. 13. The differences are not normally distributed, but skewed right. We need a sign test. There are 6 "-" and 22 "+", where a "+" means more species in the dioecious group.
! ! 28$ 2 # ( # & 0.5 " " x % 28
P
=
x
=
22
'
28 x
x
0.5
$ & %
=
0.0037,
so we reject the null hypothesis.
14. First, we transform the data by taking the natural log of each weight. The mean ln-transformed weights are -1.37 mg for females and -1.76 mg for males. Since the transformed weights are normally distributed, we can use the two-sample t-test. t = 3.51, with df = 18. The critical value for '(2) = 0.01 for 18 df is 2.88, so P < 0.01, and we reject the null hypothesis. Female mosquitoes weight more that males. 15. No, the differences are probably not normally distributed: two values less than 500, three values above 24,000 would not be expected from a normal distribution. (b) We can use a sign test: we have five pairs where there are more species feeding on angiosperms than on gymnosperms. The probability of five out five in the binomial distribution, assuming equal probabilities of either outcome, is 0.5^5, or 0.031. However, for a two-tailed test, we must double this, so P = 0.062. We are unable to reject the null hypothesis with the usual significance level. (c) The number of species that feed on angiosperms were often two orders of magnitude greater, and all pairs had higher number of species in the angiosperm group. It is impossible to get a P -value under 5% with only 5 data points in a sign test, even with all the data in a consistent direction.
Chapter 14 1. (a) Limit sampling error. (b) Reduce bias. (c) Reduce sampling error. (d) Reduces bias. (e) Reduce bias. 2. (a) [answers will vary] T, H, H, H, H, T, T, H (b) [answers will vary] no (c) 1 " Pr[exactly 4 heads and 4 tails in 8 tosses] = 1 "
& 8 # $$ ! ! 0.5 % 4 "
8
= 0.727. (d) [answers will vary] Assign a random
number between 0 and 1 to each unit. Assign the first treatment to the units with the 4 smallest random numbers and the second treatment to the remaining units. 3. Use a randomized block design, where each block is a position along the moisture gradient. Place three plots in each block, one for each of the three fertilizer treatments (call the fertilizers A, B, and C). Within each block, randomly assign the three fertilizers plots. The figure below illustrates the design for 6 blocks.
4. The researchers planned their sample size assuming a significance level of 0.05 and an 80% probability of rejecting a false null hypothesis (for a specified difference between treatment means). 5. Observational study. The treatments, presence and absence of brook trout, were not randomly assigned to the units (streams) ! the trout were already present in the streams prior to the study. Potential confounding factors (water temperature, stream depth, food supply) might differ between streams with and without brook trout, and randomization was not used to break their association with the treatment variable. 6. (a) Decrease bias (reduces effects of confounding variables); decrease sampling error (by grouping similar units into pairs) (b) No effect on bias; reduce sampling error. (c) Decrease bias (corrects for effects of age, a possible confounding variable); no effect on sampling error. (d) No effect on bias; reduce sampling error. 7. (a) No. In an experimental study the experimenter assigns two or more treatments to subjects. Here, there was only one treatment. (b) Have two treatments: the salt infusion and a placebo control (e.g., distilled water infusions). Assign treatments randomly to a sample of severe pneumonia patients. Ensure equal numbers of patients in each treatment. Keep patients unaware of which treatment they are receiving. A clinician unaware of which patient received which treatment should record their subsequent condition. 8. (a) Observational study: cancer treatments were not assigned rando mly to subjects. (b) Yes: it compares marijuana use of cancer patients and non-cancer (control) patients. (c) Reducing bias.
Age and sex might be confounding variables, affecting both marijuana use and cancer incidence. Using only subjects similar in age and sex reduces the effect of these confounding variables on the association between marijuana use and cancer. (d) First, it is not wise to “accept” the null hypothesis because the study might not have had sufficient power to detect an effect. The 95% confidence interval for the odds ratio ranged from 0.6 to 1.3, which still includes the possibility of a moderate effect. Second, observational studies such as this one cannot decide causation because of confounding variables. Perhaps marijuana use does increase cancer risk, but marijuana users have lifestyle differences that diminish cancer risk, offsetting any marijuana effect. 9. (a) The stings were applied to two volunteers, which means that the reactions to the 40 stings were probably not independent. Treating the sample size as 40 is a case of pseudo-replication. (b) More conservatively and appropriately, this study should be treated as a paired design with two samples. The mean difference in swelling would be found for each subject, and then the average of the two subjects would be tested to see if it was significantly different from zero. (c) Since each subject adds just one data point, there is no need to inflict 20 stings on each subject. Fewer stings per subject would be less cruel, and more subjects would allow more replicates an d so more power. 10. Extreme doses increases power and so enhanced the probability of detecting an effect. If an effect is detected, then studies of the effects of more realistic doses would be the next step. 2
11. (a) 13 plots per treatment. The "uncertainty" is 0.4/2 = 0.2, so n = 8 (0.25 / 0.2) = 12.5, round up 2 to 13. (b) 50 plots per treatment. The "uncertainty" is 0.2/2 = 0.1, so n = 8 (0.25 / 0.1) = 50. (c) A greater total sample size would be needed if the design were not balanced. For a given total sample size, the expected width of a confidence interval for the difference between two means increases as the design is more imbalanced (because the precision of the treatment mean having the lower sample size is greatly reduced). Achieving the same confidence interval width as a balanced design will therefore require a greater total sample size. (d) Because environmental differences between the normal-corn plot and the Bt-corn plot would be confounding variables. In effect, such a design would lack replication because plants in the same plot are not independent. 2
12. 16 plots per treatment. n = 16 (0.25 / 0.25) = 16.
Chapter 15 1. (a) Population isolation. (b) Experimental study: treatments were assigned to plants by the experimenters. (c) We assume that the variable has a normal distribution with equal variances in every population. (We also assume random samples.) Treatment Isolated Medium Long Continuous
Mean 9.3 14.5 10.8 12.8
95% CI 5.3 < µ < 13.2 11.5 < µ < 17.5 8.0 < µ < 13.5 8.2 < µ < 17.3
(d) (answers may vary). In the figure below, open circles are the data (offset where needed to minimize overlap). Means are filled circles. Vertical lines indicate 95% confidence intervals..
2. (a) H0: the mean persistence times of the four isolation treatments are equal (µ 1 = µ 2 = µ 3 = µ 4) HA: at least one of the means µ i is different (b) Source of variation Sum of squares df Mean squares F -ratio P Groups (treatments) 63.188 3 21.063 3.996 0.035 Error 63.250 12 5.271 Total 126.438 15 (c) F -distribution with 3 and 12 degrees of freedom. (d) The probability of obtaining an F ratio statistic as large or larger than the value observed when the null hypothesis is true. (e) The total sum of squares is the sum of the deviations squared between each observation and the grand mean. The error sum of squares is the sum of the squared deviations between each observation and its group mean. The group sum of squares is the sum of the squared deviations between the 2 group mean for each individual and the grand mean. (f) Use R , the ratio of the group sum of 2 squares and the total sum of squares. (g) R = 63.188 / 126.438 = 0.50.
3. (a) H0: The mean of group i equals the mean of group j, HA: The mean of group i does not equal the mean of group j, for all pairs of means i and j. (b) These are unplanned comparisons, because they are intended to search for differences among all pairs of means. Planned comparisons must be fewer in number and identified as crucial in advance of gathering and analyzing the data. (c) Failure to reject a null hypothesis that the difference between a given pair of means is zero does not imply that the means are equal, because power is not necessarily high, especially when the differences are small. If the means of the “Medium” and “Isolated” treatments differ from one another, then one or both of them must differ from the means from the other two groups, but we don’t know which. (d)
(e) The critical value for a t -test has a Type 1 error rate of 0.05 when comparing two means. The critical value for the Tukey comparison of all pairs of means is larger so that the probability of making at least one Type 1 error in all the comparisons is only 0.05. (Note that degrees of freedom are 12 in either case because we are using the error mean square to calculate the standard error of the difference between means).
4. (a) Transformations or a nonparametric Kruskal-Wallis test. (b) The transformation should be attempted first, because this would yield the more powerful test. 5. (a) H0: The mean number of shoots is equal in the three treatments (µ 1 = µ 2 = µ 3). HA: At least one of the treatment means µ i is different. (b) Source of variation Sum of squares df Mean squares F -ratio P Groups (treatments) 2952.808 2 1476.40 5.32 0.011 Error 8049.067 29 277.55 Total 11001.875 31 The critical value is F 0.05(1),2,29 = 3.33. Since F > 3.33, P < 0.05, reject H0. Conclude that there are differences between treatments in mean shoot number. (c) The variable is normally distributed with equal variance in the three treatment populations. (d) Fixed-effects. The levels chosen were set by the researcher and are repeatable, they were not randomly sampled from a population of treatments.
" Y = "0.004 " "0.195 = 0.191. MS error = 0.0345, df = 42, SE = 0.0679, t 0.05(2),42 = 2.01, 0.054 < µ 1 " µ 2 < 0.328. (b) Yes, because the study was designed mainly to compare PLP1 gene
6. (a)
Y 1
2
expression in persons with schizophrenia to that of control individuals. It was a single focused comparison not a broad search for differences between groups. (c) The expression measurements are normally distributed in the populations with equal variances. (We also assume random samples.) 7. (a) H0: Mean PLP1 gene expression is equal in the three groups (µ 1 = µ 2 = µ 3). HA: At least one of the group means µ i is different. Source of variation Sum of squares df Mean squares F -ratio Groups 0.5403 2 0.2701 7.82 Error 1.4502 42 0.0345 Total 1.9905 44
P 0.0013
The critical value is F 0.05(1),2,42 = 3.22. Since F > 3.22, P < 0.05, reject H0. Conclude that mean PLP1 expression differs among groups. (b) The expression measurements are normally distributed in the populations with equal standard de viations. (c) Fixed-effects ANOVA: we are comparing predetermined and repeatable treatment groups, not a random selection of groups in a 2 2 population. (d) Use R to describe the fraction of the variance explained by group differences: R = 0.27. (e) Use the Tukey-Kramer method. 8. (a) Random effect. The sites were randomly chosen; the goal was to determine whether sites varied in general in Spain, not just whether these four sites were different. (b) Because this is a random effect ANOVA, the hypotheses should b e stated in terms of the “population” of sites, from which we have a sample. H0: Mean carotenoid plasma concentration in vultures is the same among sites in Spain. HA: Mean carotenoid plasma concentration in vultures differs among sites in Spain. Source of variation Groups (Sites) Error Total
Sum of squares 712.25 1388.26 2100.52
df 3 178 181
Mean squares 237.42 7.80
F -ratio 30.44
P < 0.0001
The critical value is F 0.05(1),3,178 = 2.66. Since F > 2.66, P < 0.05, reject H0. Conclude that sites vary in the mean carotenoid plasma concentration. (c) We assume that carotenoid plasma concentration has a normal distribution in every population with equal variances. We also assume that sites were randomly chosen and that site means have a normal distribution. (We also assume random samples.) 9. (a) H0: Mean tested weight is the same in the four treatments ( µ 1 = µ 2 = µ 3 = µ 4). HA: At least one of the treatment means µ i is different. Source of variation Groups (treatment) Error Total
Sum of squares 19057.5 130050.0 149107.5
df 3 68 71
Mean squares 6352.5 1912.5
F -ratio 3.32
P 0.025
The critical value is F 0.05(1),3,68 = 2.74. Since F > 2.74, P < 0.05, reject H0. Conclude that mean testes weight varies among treatments. (b) Testes weight is normally distributed in the four populations with equal variance. (We also assume random samples). (c) Yes, this was an experimental study because the treatments were assigned (randomly) to the subjects (mice) by 2 the researchers. (d) R = 19057.5/149107.5 = 0.13. 10. Need to use a & contingency test (Chapter 9). Observed and expected frequencies (in parentheses) are shown in the contingency table below. H0: The proportion of impregnated females is the same in the four treatments HA: The proportion of impregnated females is not the same in the four treatments 2
Treatment oil THC cannabinol cannabidiol Column sum
Number of females impregnated not impregnated 14 (12.75) 4 (5.75) 13 (12.75) 5 (5.75) 11 (12.75) 7 (5.75) 13 (12.75) 5 (5.75) 51 21
Row sum 18 18 18 18 72
&2 = 1.28, df = 3. The critical value is ! 02.05(1),3 = 7.81. Since &2 not greater than 7.81, do not reject H0. ( P = 0.73). We cannot reject the null hypothesis that the proportion o f females impregnated is equal between treatments. 11. Random-effects: the males were chosen at random from the population. A given male is not a specific, repeatable treatment. The goal is to generalize to the population of males (the hypothesis statements should reflect this). (b) H0: Mean condition of offspring is the same for all males in the population. HA: Mean condition of offspring differs between males in the population. Source of variation Sum of squares df Mean squares F -ratio Groups (males) 9.9401 11 0.9036 4.63 Error 4.682 24 0.1951 Total 14.622 35
P 0.0008
The critical value F 0.05(1),11,24 = 2.21 (between 2.18 and 2.25). Since F > 2.21, P < 0.05, reject H0. Conclude that mean offspring conditions differs among males in the beetle population. 2 (c) s A = (0.904 " 0.195) / 3 = 0.236. Repeatability = 0.236 / (0.236 + 0.195) = 0.548.
Chapter 16 1. r = "0.5; r = "0.8; r = 0.5; r = 0. 2. (a) 95%: 0.22 <* < 0.63. 99%: 0.14 <* < 0.68. (b) 95%: "0.95 <* < "0.76. 99%: "0.96 <* < "0.70. (c) 95%: "0.17 <* < 0.25. 99%: "0.24 <* < 0.31 3. (a) Scatter plot:
(b) The relationship is linear, positive, and strong. (c) r = 0.93. SE = 0.13. (d) The standard error is the standard deviation of the sampling distribution of r . (e) 0.72 < * < 0.98
4 (a) No change to the correlation coefficient (r = 0.93). Adding a constant to one of the variables does not alter the correlation coefficient. (b) No change to the correlation coefficient (r = 0.93). Dividing one of the variables by a constant does not alter the correlation coefficient. 5. We can use a paired t -test (Chapter 12). H0: Mean arrival date of male and female partners is the same (µ d = 0) HA: Mean arrival date of male and female partners is different (µ d $ 0) = 0.3 days (males are slightly earlier on average), SE = 1.667, t = 0.18, df = 9, P = 0.86. t 0.05(2),9 = 2.26; since observed t is less than t 0.05(2),9 , P > 0.05, do not reject H0. Conclude that we cannot reject null hypothesis of equal mean arrival times of males and females. Assume a normal distribution of differences between arrival rates of males and females. d
6. When there is measurement error in one or both of the variables X and Y . 7. A narrower range of values for inbreeding coefficient should lower the correlation with the number of surviving pups compared with a wider range of inbreeding coefficients. 8. Use the Spearman rank correlation: H0: The population rank correlation is zero (* S = 0) HA: The population rank correlation is not zero (* S $ 0) r S = 0.66. P = 0.17. r S (0.05(2),7) = 0.821. Since r S is not greater than or equal to r S (0.05(2),7) , P > 0.05, do not reject H0. Conclude that we cannot reject the null hypothesis of zero correlation. (b) Assume random sample and a linear relationship between the ranks of the two variables. 9. Sampling error in the estimates of earwig d ensity and proportion males with forceps means that true density and proportion on an island are measured with error. Measurement error will tend to decrease the estimated correlation. Therefore, the actual correlation is expected to be higher on average than the estimated correlation. 10. (a) There is a negative linear relationship between telomere length and c hronicity, but it is not strong. (b) "0.43. (c) "0.66 < * < "0.13. (d) It is the range of most plausible values for the parameter * . If you were to repeatedly and randomly sample individuals from the same population and compute the 95% confidence interval each time, 19 out of 20 of the intervals are expected to include the population correlation * . (e) Assume random sampling, and that the two variables have a bivariate normal distribution in the population. (f) (Answers may vary) The scatter plot suggests that the relationship between telomere length and chronicity might be mildly non-linear, which would violate the assumption of bivariate normality. 11. (a)
(b) r = 0.82 (c) H0: There is no correlation between second language proficiency and grey matter density (* = 0). HA: There is a correlation between second language proficiency and grey matter density (* $ 0). r = 0.82, SE = 0.13, t = 6.37, df = 20, P = 0.000003 t 0.05(2),20 = 2.09. Since t is greater than t 0.05(2),20 , P > 0.05. Reject H0. Conclude that second language proficiency and grey matter density are correlated. (d) Random sampling and a bivariate distribution of gray matter density and language proficiency in the population. (e) No, because there appears to be two outlying observations, which violates the assumption of bivariate normality. (f) No, correlation alone does not imply causation. Perhaps individuals with high grey matter densities are able to achieve a high proficiency in a second language. An experiment would be necessary to test whether proficiency affects grey matter.
Chapter 17 1. (a) y = 0.5x + 1 (b) y = x – 1 (c) y = "0.5 x + 2 (d) y = x – 5 2. (a)
The percent infant mortality increases approximately linearly with the log of the home range size. (b) Mortality = 16.37 + 10.26(log home range) (c) H0: Home range size does not predict infant mortality (+ = 0). HA: Home range size predicts infant mortality (+ $ 0). b = 10.26, SE = 2.69, t = 3.81, df = 18, P = 0.0013, t 0.05(2),18 = 2.10. Since t > 2.10, P < 0.05. Reject H0. Conclude that home range size predicts infant mortality. (d) Mortality = 17.51 + 6.60(log home range). The slope is much lower with the polar bear removed (6.60 instead of 10.26, a reduction of more than a third).
3. (a) The “least squares” regression line is the one the minimizes the sum of squared differences between the predicted Y values on the regression line for each X and the observed Y values. (b) Residuals are the differences between the predicted Y value on the estimated regression line and the observed Y values. (c) The most conspicuous problem is that the variance of the residuals increases with increasing progesterone concentration, violating the a ssumption that the variance of Y is the same at all values of X . There are no conspicuous departures from the other two assumptions, linearity and a normal distribution of Y values at every X . 4. (a)
(b) H0: Rate if heat loss does not change with body leanness (+ = 0). HA: Rate if heat loss changes with body leanness (+ $ 0). b = 0.0190, SE = 0.0023, t = 8.29, df = 10, P = 0.000009, t 0.05(2),18 = 2.29. Since t > 2.29, P < 0.05. Reject H0. Conclude that rate of heat loss increases with body leanness. (c) 0.0139 < + < 0.0241 (d) For every X there is a normal distribution of Y values, of which we have a random sample; the relationship between leanness and h eat loss is linear; the variance of Y is the same for all values of X .
5. Caution is warranted because the prediction is based on extrapolation, which is risky because the relationship between winning time and year might not be linear beyond the range of the existing data. 6. (d) 7. (a) The arcsine square root transformation is a good bet for data that are proportions. (b) Y = 0.416 + 7.10 X , where X is genetic distance and Y is arcsine square root transformed proportion sterile.
(c) 5.67 < + < 8.52
8. (a) Such a complicated curve is unwarranted by the data. It would do a poor job of predicting new observations because it does not describe a general trend. A curve fit should be as simple as possible. (b) First try to transform the data to make the relationship linear. If that fails, consider non-linear regression. 9. (a) b = 11.68, SE = 4.85. (b) "4.08 < + < 27.43 (c) The range of most plausible values for the parameter. In 99% of random samples, the confidence interval will bracket the population value for the slope. (d) Measurement error in the X variable (bite force) will tend to lead to an underestimation of the population slope. (e) Measurement error in the Y variable (territory area) will not bias the estimate of slope (though it will increase its uncertainty). 10. (a) Source of variation Regression Residual Total
Sum of squares 194.3739 301.5516 495.9255
df 1 9 10
Mean squares 194.3739 33.5057
F -ratio 5.8012
(b) H0: The slope of the regression of territory area o n bite force is zero (+ = 0). HA: The slope of the regression of territory area on bite force is not zero (+ $ 0). F = 5.8012, df = 1,9, P = 0.039. F0.05(1),1,9 = 5.12. Since F > 5.12, P < 0.05. Reject H0. Conclude that the slope is not zero (it is positive). (c) That the relationship between X and Y is linear; for each X there is a normal distribution of Y values in the population, of which we have a random sample; the variance of Y is the same at all 2 values of X . (d) The variance of the residuals. (e) R = 0.392. It measures the fraction of the variation in Y that is explained by X .
11. (a) Perform the analysis with and without the outlier included in the data set to determine whether it has an influence on the outcome. If it has a big influence, then it is probably wise to leave it out and limit predictions to the range of X values between 0 and about 200 (and urge them to obtain more data at the higher X values.) (b) Confidence bands give the confidence interval for the predicted mean time since death for a given hypoxanthine concentration. (c) Confidence bands. (d) The prediction interval, because it measures uncertainty when predicting the time of death of a single individual.
12. (a)
(b) The assumptions of equal variance of residuals, and of a normal distribution of Y values at each X , are not met because of the presence of an outlier. (c) A transformation of the data might improve matters, but success is doubtful. Alternatively, use nonparametric correlation to measure association between the two variables.
Chapter 18 1. (a) HOURS = CONSTANT + MUTANT. HOURS is hours of resting. CONSTANT is the grand mean. MUTANT is the effect of each lines of mutant flies. (b) HOURS = CONSTANT (c) Long horizontal line indicates the grand mean, which is the predicted value for the mean of each group under the null hypothesis. The short horizontal lines give the group means, which are the predicted values under the “full” general linear model including the MUTANT term.
(d) F .
2. (a) Plasma corticosterone concentration (b) Chick age group and disturbance regime (c) That the two explanatory variables interact to affect corticosterone concentrations. (d) Observational. The penguins were not assigned by the researcher to age groups or disturbance regimes. (e) Yes. Every combination of the two factors (age group and disturbance regime) is included in the design.
3. (a) CORTICOSTERONE = CONSTANT + AGE + DISTURBANCE + AGE*DISTURBANCE. CORTICOSTERONE is the corticosterone concentration, CONSTANT is the grand mean corticosterone concentration, AGE is the age group of the penguin, DISTURBANCE indicates whether the penguin lived in the undisturbed or tourist-disturbed area, and AGE*DISTURBANCE is the interaction between age and disturbance. (b) H0: Penguin age group has no effect on mean corticosterone concentration. H0: Disturbance regime has no effect on mean corticosterone concentration. H0: There is no interaction between penguin age group and disturbance regime. (c) Penguins were randomly samples from each area and each age class. Corticosterone concentrations are normally distributed within age class and disturbance regime. The variance of the corticosterone concentration is the same for each combination of age and disturbance. 4. Three of the following: a) To investigate the effect of more than one variable with the same experimental design; b) To include the effects of blocking in an experimental design; c) To investigate interactions between the effects of different factors; d) To control for the effects of a confounding variable by including it as a covariate. 5. In these plots, different symbols refer to different groups of factor B.
6. (a) MEMORY = CONSTANT + LESION. MEMORY is the spatial memory score of a rat, CONSTANT is a constant indicating the value of MEMORY when LESION is zero (Y intercept), and LESION is the extent of the lesion. (b) MEMORY = CONSTANT (c) The line having negative slope in the following graph represents the predicted values from the “full” general linear model. The flat (horizontal) line represents the predicted values under the null model.
7. (a)
(b) EXPRESSION = CONSTANT + WORKERTYPE + BLOCK. EXPRESSION is the for gene expression, CONSTANT is the grand mean of for gene expression, WORKERTYPE is the worker type, and BLOCK identifies which colony the bee comes from. (c) EXPRESSION = CONSTANT + BLOCK (d) Fixed effect, because the types are repeatable and of direct interest. The worker types in the analysis are n ot randomly sampled from a population of worker types. (e) Blocking variables allow extraneous variation caused by the block variable to be accounted for in the analysis and eliminated. When a block variable is included, the error variance is smaller, making it easier to detect real effects of the factor.
Chapter 19 1. (a) For example: Singleton: 3.5, 3.5, 2.6, 4.4 Twin: 3.4, 4.2, 3.4 (b) Singleton: 3.5, 4.2, 4.4, 3.4 Twin: 2.7, 2.6, 1.7 (c) In a bootstrap replicate, the data for each group is a sample of the data from that same group. Sampling is with replacement (some points might not occur an d
others might occur more than once). In a randomization replicate, the data for each group is sampled from all the data without regard to original group. Sampling is without replacement (every data point occurs exactly once). (d) No. 2. (a) Yes " by chance the randomization might reassign data to correct groups. (b) Yes. (c) No " Sample sizes of groups must be the same in each randomization as in the data. (d) No " Each data point can occur only once in a randomization. (e) No " 3.8 is not in the data. (f) No " Each data point can occur only once in a randomization. 3. (a) Yes " by chance bootstrap samples might contain the same observations as the data. (b) No " The bootstrap sample for a group can contain only data from that group. (c) No " Sample sizes of groups must be the same in each randomization as in the data. (d) Yes (e) No " 3.8 is not in the data (f) Yes. 4. 10.20 < µ < 10.40 5. B is the real data. In A, there is little or no relationship between the two variables, whereas B shows a strong relationship. Since randomization tends to break up associations, it is more likely that B is the data and A is a randomization. 6. (a) A comes from randomization and B comes from the bootstrap. (b) Approximately 10 units. The range from X =10 to X =50 should span about 4 standard deviations of the distribution, so 1 standard deviation should be about 10 units. 7. (a) The null distribution for the ratio. (b) H0: Mean ratio of range size equals that expected of a randomly broken stick. HA: Mean ratio of range size differs from that expected of a randomly broken stick. P is approximately 2 # (42/10,000) = 0.0084. Since P < 0.05, reject H0. Conclude that mean ratio in birds exceeds that expected from a randomly broken stick. 8. (a) 98. (b) 100. (c) “Giant pandas are most closely related to bears” 9. (a) For example: Group A 7.8 4.5 2.1 7.8 7.8 4.5 4.5 2.1 7.8 7.8 2.1 4.5 7.8 2.1 4.5 7.8 4.5 7.8 4.5 4.5 7.8 2.1 7.8 4.5 4.5 7.8 4.5 2.1 7.8 7.8
Group B 12.4 8.9 8.9 12.4 12.4 12.4 8.9 8.9 12.4 10.8 12.4 10.8 12.4 8.9 10.8 8.9 10.8 10.8 10.8 12.4 8.9 8.9 8.9 10.8 10.8 10.8 12.4 8.9 12.4 12.4
(b) Difference in median (B minus A) for these 10 bootstrap replicates: 3.0, 4.4, 4.4, 4.4, 4.6, 4.6, 6.3, 6.3, 6.3, 6.3 4.4 < difference between population medians < 6.3
10. (a) For example: Group A
Group B
8.9 4.5 4.5 4.5 10.8 4.5 8.9 12.4 2.1 7.8
2.1 4.5 8.9 2.1 12.4 2.1 8.9 2.1 4.5 2.1 10.8 2.1 12.4 4.5 10.8 2.1 10.8 7.8 4.5 10.8
10.8 7.8 7.8 10.8 8.9 12.4 2.1 7.8 8.9 12.4
7.8 10.8 8.9 12.4 12.4 8.9 7.8 4.5 12.4 2.1
12.4 12.4 10.8 7.8 7.8 7.8 10.8 8.9 4.5 8.9
(b) H0: The two groups have the same median. HA: The two groups do not have the same median Test statistic: observed difference between medians (B minus A) = 6.3 Null distribution: "3.0, "1.1, 1.1, 1.1, 4.4, 4.4, 4.4, 6.3, 6.3, 6.3 3 out of 10 outcomes is greater than or equal to the observed value, 6.3 P is approximately 2 # 3/10 = 0.6. Since P > 0.20, do not reject H0. Conclude that the null hypothesis of no difference between medians is not rejected.
11. Yes. 12. Yes. 13. 10.4 seconds. 14. Welch’s approximate t -test. 15. (a) A randomization test. The variable “shell volume” has been kept fixed and values of the second variable are all included but in random order. (b) A bootstrap estimate. Each row of the data set is a different individual sampled from the original sample of individuals. Sampling is with replacement because the same individual sometimes occurs more than once. (c) The linear correlation coefficient, r .
Chapter 20 ' 47 $ 12 35 " p (1 ! p ) . It measures " & 12 #
1. (a) Binomial distribution with n = 47. (b) L[ p | 12 heterozygotes ]= %%
the probability of getting 12 heterozygotes (the data) given that the true value of the proportion '- 47 * $ of heterozygotes is p. (c) ln L[ p ]= ln %++ ( ( " + 12 ln[ p] + 35 ln[1 ! p] . (d) –7.90 12 &, ) # 2. (a) 0.26 (b) 0.14 < p < 0.40 3. (a) The log of the probability of the DNA data given different possible times since the split between Neanderthals and modern humans. (b) 710 thousand years (c) 470 < time since split < 1000 (in thousands of years) (d) This interval is like a 95% confidence interval. It describes the most plausible values of the time since the split between human and Neanderthals.
2
4. (a) –4400 (b) –4396.5 (c) G = 2(–4396.5 – (–4400)) = 7. (d) & with df = 1. (e) ! 02.05,1
=
3.84 .
Since G is greater than 3.84, P < 0.05 (exact P = 0.008). Reject H0. Conclude that heterogeneity is not zero. 5. (a) Pr[Yes] = 1/2 + s/2 = (1 + s)/2
& 185 # & 1 + s # 113 & 1 ' s # 72 ! (b) L[ s | 113 yeses]= $$ ! $ ! ! $ 113 % " % 2 " % 2 " '- 185 * $ ( (c) ln L[ s | 113 yeses]= ln %++ ( " + 113 ln[(1 + s ) / 2] + 72 ln[(1 ! s) / 2] 113 , ) # & (d) "7.39
6. (a) The maximum likelihood estimate is s = 0.22. (b) 0.07 < s < 0.36. (c) H0: The fraction of thieves s is zero. HA: The fraction of thieves s is not zero. 2 G = 2 (- 2.81 - (- 7.39)) = 9.16 . ! 0.05,1 3.84 . Since G > 3.84, P < 0.05 (exact P = 0.002). Reject ˆ
=
H0. Conclude that there are thieves among us. 7. (a)
(b) 0.60 (c) –121.653 (d) The data seem to fit a geometric distribution very well.
(e) & Goodness-of-fit test 2
8. (a)
(b) 0.58. (c) 0.36 < , < 0.86
Chapter 21 1. Results are seen as more interesting or exciting; results are more likely to be accepted for publication; results are more likely to be accepted in a better journal. 2. (a) Yes: average heritability declines with increasing sample size. (b) Small-sample studies yielding low heritability estimates are unlikely to be submitted for publication (researchers) or accepted for publication (editors). 3. The value of including this type of study is low because we need to assume that success in tennis is a measure of aggression. Including it might b e an act of desperation resulting from a shortage of human studies that directly measure aggression. 4.
