CHAPTER 10
HYPOTHESIS TESTING
MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 1.
If a rese resear arch cher er tak takes es a larg largee enoug enough h samp sample le,, he/s he/she he wil willl almo almost st alway alwayss obta obtain in:: a. virtua virtually lly signif significa icant nt result resultss b. practically significant results c. consequ consequent ential ially ly sign signifi ifican cantt result resultss d. statis statisti ticall cally y signif significa icant nt resul results ts ANSWER: d
2.
Thee nul Th nulll and and alte altern rnat ativ ivee hyp hypot othe hese sess div divid idee all all poss possib ibil ilit itie iess int into: o: a. two two sets sets that that ove overl rlap ap b. two non-overlapping sets c. two sets sets that that may or may may not not over overlap lap d. as many many sets sets as necessary necessary to to cover cover all possibili possibilities ties ANSWER: b
3.
Whic Which h of of the the foll followi owing ng is true true of the the nul nulll and and alte altern rnat ativ ivee hyp hypoth othes eses es?? a. Exactl Exactly y one one hypoth hypothesi esiss must must be be true true b. both hypotheses must be true c. It is is possib possible le for for both both hypoth hypothese esess to be true true d. It is is possi possible ble for for neit neither her hypo hypothe thesis sis to be true true ANSWER: a
4.
OneOne-ttaile ailed d alt alterna ernattives ives are are phr phras ased ed in ter terms of: of:
α. ≠ b. < or > c. ≈ or = d. ≤ or ≥ ANSWER:
b
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Chapter 10
5.
Thee chiTh chi-sq squa uare re good goodne ness ss-o -off-fi fitt tes testt can can be be used used to to tes testt for for:: a. signif significa icance nce of of samp sample le stat statist istics ics b. difference between population means c. normality d. prob probab abiility ity ANSWER: c
6.
A type II error occurs when: a. the null null hypothesi hypothesiss is incorre incorrectly ctly accepte accepted d when it it is false false b. the null hypothesis is incorrectly rejected when it is true c. the sampl samplee mean diffe differs rs from from the the populati population on mean mean d. the the tes testt is is bia biase sed d ANSWER: a
7.
Of type type I and and typ typee II erro error, r, whi which ch is is tra tradi diti tion onal ally ly rega regard rded ed as as mor moree seri seriou ous? s? a. Type I b. Type II c. Th They ey are are equal equally ly seri seriou ouss d. Neit Neithe herr is is sser erio ious us ANSWER: a
8.
You You cond conduc uctt a hypo hypoth thes esis is tes testt and and you you obser observe ve val value uess for for the the samp sample le mea mean n and sample standard deviation when n = 25 that do not lead to the rejection of H 0 . You calculate a p-value of 0.0667. What will happen to the p-value if you observe the same sample mean and standard deviation for a sample > 25? a. Increase b. Decrease c. Stay Stay the sa same d. May either either increa increase se or or decr decreas easee ANSWER: b
9.
Thee for Th form m of of the the alt alterna ernattive ive hy hypot pothesi hesiss can can be: be: a. one one-tail ailed b. two-tailed c. neith neither er one one nor twotwo-ta tail iled ed d. one one or or two two-t -tai aile led d ANSWER: d
215
Hypothesis Testing
10. 10.
A twowo-tail ailed tes test is one one wher where: e: a. results results in only only one directi direction on can lead lead to rejectio rejection n of the null null hypothesi hypothesiss b. negative sample means lead to rejection of the null hypothesis c. results results in either either of two directi directions ons can lead to to rejection rejection of the the null hypothesi hypothesiss d. no results results lead to to the rejection rejection of the null null hypothesi hypothesiss ANSWER: c
11.
The value set for α is known as: a. the the rej rejec ecti tion on leve levell b. the acceptance level c. the the sign signif ific icanc ancee leve levell d. the error error in in the the hypot hypothes hesis is test test ANSWER: c
12. 12.
A study study in which which rand random omly ly sele select cted ed grou groups ps are are obse observ rved ed and and the the resu result ltss are analyzed without explicitly controlling for other factors is called: a. an obse observ rvat atio iona nall stud study y b. a controlled study c. a field test d. a si simple ple stu study dy ANSWER: a
13. 13.
Thee nul Th nulll hy hypoth pothes esis is usua usuall lly y rep repre rese sent nts: s: a. the theor theory y the rese researc archer her would would like like to to prove. prove. b. the preconceived ideas of the researcher c. the perc percept eption ionss of the the sampl samplee popula populatio tion n d. the the st status atus quo quo ANSWER: d
14. 14.
Thee ANO Th ANOVA VA test test is base based d on on whi which ch assu assump mpti tion ons? s? I. II. II. III. III. IV. IV.
the sample are randomly selected the the pop popul ulat atio ion n var varia ianc nces es are are all all equa equall to to som somee com commo mon n var varia iance nce the the popu popula lati tion onss are are norm normal ally ly dis distr trib ibut uted ed the the pop popul ulat atio ions ns are are stat statis isti tical cally ly signi signifi fica cant nt
a. All All of of the the abov abovee b. II and III only c. I, II, II, aand nd III III onl only y d. I, and and II III onl only y ANSWER: b
216
Chapter 10
15.
In stati statisti stical cal analys analysis, is, the burden burden of proof proof lies lies trad traditi itional onally ly wit with: h: a. the the alte altern rnat ativ ivee hypot hypothes hesis is b. the null hypothesis c. the analyst d. the facts ANSWER: a
16.
When When one refers refers to “how “how sign signifi ificant cant”” the the sample sample evid evidenc encee is, is, he/sh he/shee is referr referring ing to the: a. value of α b. the importance of the sample c. the p-value d. the FF-ratio ANSWER: c
17. 17.
Which hich of the the fol folllowi owing val values ues is is not typically used for α ? a. 0.01 b. 0.05 c. 0.10 d. 0.25 ANSWER:
18.
d
Smaller p -values indicate more evidence in support of: a. the the nul nulll hyp hypot othe hesi siss b. the alternative hypothesis c. the the qual qualit ity y of the the res resea earc rcher her d. furt furthe herr test testin ing g ANSWER: b
19.
Thee chiTh chi-sq squa uare re test test can can be too too sen sensi siti tive ve if if the the sam sampl plee is: is: a. very small b. very large c. hom homogeneo neous d. pred predic icttabl able ANSWER: b
217
Hypothesis Testing
20.
Thee hypot Th hypothe hesi siss that that an an anal analys ystt is tryi trying ng to to prov provee is calle called d the: the: a. elec electi tive ve hypo hypoth thes esis is b. alternative hypothesis c. opti option onal al hyp hypot othe hesi siss d. null null hypo hypoth thes esis is ANSWER: b
21. 21.
A p-v p-val alue ue is is con consi side dere red d “con “convi vinc ncin ing” g” if if it it is: is: a. les less than than 0.01 0.01 b. between 0.01 and 0.05 c. 0.0 0.05 and 0.10 d. grea greate terr tha than n 0.1 0.10 0 ANSWER: a
22. 22.
One-way ANOVA is used when: a. analyzing analyzing the the differenc differencee between between more more than than two population population means b. analyzing the results of a two-tailed test c. analyzi analyzing ng the the result resultss from from a large large sam sample ple d. analyzing analyzing the the differ difference ence between between two population population means means ANSWER: a
23.
A null null hypo hypothes thesis is can can only only be reje rejecte cted d at the 5% sign signifi ificanc cancee level level if and only only if: if: a. a 95% confidence confidence interv interval al includes includes the hypoth hypothesize esized d value of the the parameter parameter b. a 95% confidence interval does not include the hypothesized value of the parameter c. the the null null hyp hypot othes hesis is is is void void d. the null null hypot hypothese hesess include includess sampli sampling ng error error ANSWER: b
24.
Typica Ty pically lly one-way one-way ANOVA ANOVA is used used in in which which of the follow following ing situat situation ions? s? I. II. III. IV.
ther here ar are severa eral dis distinct nct po populations ons ther here are are two two sam sampl plee pop popul ulat atiions ons o ove verr 400 4000 0 randomized exper periments rand random omly ly sel selecte ected d pop popul ulat atio ions ns
a. All All of of the the abov abovee b. II and III only c. I, II, II, aand nd III III onl only y d. I, and and II III onl only y ANSWER: d
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Chapter 10
25.
Thee chiTh chi-sq squa uare re tes testt is is not not very very eff effec ecti tive ve if if the the sam sample ple is: is: a. small b. large c. irregular d. het heterog erogen eneo eous us ANSWER: a
26. 26.
Thee alte Th altern rnat ativ ivee hypo hypoth thes esis is is is also also kno known wn as as the: the: a. elec electi tive ve hypo hypoth thes esis is b. optional hypothesis c. rese resear arch ch hypo hypoth thes esis is d. null null hypo hypoth thes esis is ANSWER: c
27.
An info inform rmal al test test for for norma normali lity ty that that utili utilizes zes a scatt scatter erpl plot ot and look lookss for clus cluste teri ring ng around a 45° line is known as: a. a Lil Lilli lief efor orss test test b. an empirical cdf c. a p-test d. a quant quantil ilee-qua quant ntil ilee plot plot ANSWER: d
28.
Whic Which h of the the foll followi owing ng tes tests ts are are use used d to test test for for norma normali lity ty?? a. A t -test -test and an ANOVA test b. An Empirical CDF test and an F -test -test c. A Chi-Sq Chi-Squar uaree test test and and a Lil Lillie liefor forss test test d. A Quan Quantil tile-Qu e-Quant antil ilee plot plot and a p-value test ANSWER: c
29.
If a teacher teacher is try trying ing to to prove prove that that new new metho method d of teac teachin hing g math math is more more effect effective ive than traditional one, he/she will conduct a: a. oneone-ttail ailed tes testt b. two-tailed test c. point point estim estimate ate of the the popul populati ation on parame parameter ter d. conf confid iden ence ce inte interv rval al ANSWER: a
219
Hypothesis Testing
30. 30.
A type I error occurs urs when: a. the null null hypothesi hypothesiss is incorre incorrectly ctly accepte accepted d when it it is false false b. the null hypothesis is incorrectly rejected when it is true c. the sampl samplee mean diffe differs rs from from the the populati population on mean mean d. the the tes testt is is bia biase sed d ANSWER: b
220
Chapter 10
TEST QUESTIONS 31.
A sport sport pre prefe fere renc ncee poll poll yield yielded ed the the follo followi wing ng data data for for men men and wome women. n. Use Use the 5% significance level and test to determine is sport preference and gender are independent.
Sport Preference Basketbal ball
Football
Socc occer
Men
20
25
30
75
Women
18 38
12 37
15 45
45 120
Gender
ANSWER:
We reject the null hypothesis of independence at the 5% significance level (since p-value = 0.019 < 0.05). We may conclude that sport preference and gender are not independent; that is, there is evidence that sport preference of men is different from that of women. 32. 32.
Supp Suppos osee that that we we obs observe erve a ran rando dom m sam sample ple of si size n from a normally distributed population. If we are able to reject H 0 : µ = µ 0 in favor of H a : µ ≠ µ 0 at the 5% significance level, is it true that we can definitely reject H 0 in favor of the appropriat appropriatee one-tailed one-tailed alternative alternative at the 2.5% significan significance ce level? Why or why not?
221
Hypothesis Testing ANSWER: This is not true for certain. Suppose µ 0
= 50
and the sample mean we observe observe is X = 55. If the alternative for the one-tailed test is H a : µ < 50, then we obviously can’t can’t reject reject the null null becaus becausee the obs observ erved ed sample sample mean mean X is in the wrong direction. But if the alternative alternative is H 1 : µ > 50, we can reject the null at the 2.5% level. The reason is that we know the p-value for the two-tailed test was less than 0.05. The p-value for a one-tailed test is half of this, or less than 0.025, which implies rejection at the 2.5% level. 33.
An invest investor or want wantss to to comp compare are the risks risks associ associate ated d with with two differ different ent stocks stocks.. One way to measure the risk of a given stock is to measure the variation in the stock’s daily daily price changes. changes. The investo investorr obtain obtainss a random sample sample of 20 daily price price changes for stock 1 and 20 daily price changes for stock 2. These data are shown shown in the table below. Show how this investor can compare compare the risks associated with the two stocks by testing the null hypothesis that the variances of the stocks are equal. Use α = 0.10 and interpret the results of the statistical test.
Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
ANSWER: n1 = 20, s1 = 0.8487, n2 2
2
H 0 : σ 1 / σ 2
H a : σ 12 / σ 22
Price Change for stock 1 1.86 1.80 1.03 0.16 -0.73 0.90 0.09 0.19 -0.42 0.56 1.24 -1.16 0.37 -0.52 -0.09 1.07 -0.88 0.44 -0.21 0.84
= 20, s = 0.5291 2
=1 ≠1
Test statistic: F
=s
2
1
/ s22
= 2.573
222
Price Change for stock 2 0.87 1.33 -0.27 -0.20 0.25 0.00 0.09 -0.71 -0.33 0.12 0.43 -0.23 0.70 -0.24 -0.59 0.24 0.66 -0.54 0.55 0.08
Chapter 10 -value=0.023 P -value=0.023 Since the P -values -values is less than 0.10, we reject the null hypothesis of equal variances and conclude that the variances of the stocks are not equal at the 10% level. QUESTIONS 34 THROUGH 37 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
BatCo (The Battery Battery Company) Company) produces your typical typical consumer battery. battery. The company claims claims that their batteries batteries last at least least 100 hours, on average. Your experience experience with the BatCo battery has been somewhat different, so you decide to conduct a test to see if the companies companies claim is true. You believe that the the mean life is actually actually less than the 100 hours BatCo claims. You decide to collect data on the average battery life (in (in hours) of a random sample and the information related to the hypothesis test is presented below. Test of µ ≥ 100 versus one-tailed alternative Hypothesized mean 1 0 0 .0 Sample mean 9 8 .5 Std error of mean 0.777 Degrees of freedom 19 t-test statistic -1.932 p-value 0.034
34.
Can Can the the samp sample le size size be deter determi mined ned from from the the inf inform ormat atio ion n abov above? e? Yes Yes or or no? no? If yes, what is the sample size in this case? ANSWER: Yes. 19 + 1 = 20.
35.
You believ believee that that the the mean mean life life is actual actually ly less less than than 100 100 hours hours,, shoul should d you you conduc conductt a one-tailed or a two-tailed hypothesis hypothesis test? Explain your answer. ANSWER: One-tailed test. You are interested in the mean mean being less than 100.
36.
What What is is the the samp sample le mea mean n of thi thiss data data?? If you you use use a 5% sig signi nifi fica canc ncee leve level, l, woul would d you conclude that the mean life of the batteries is typically more than 100 hours? Explain your answer. ANSWER: 98.5 hours. No. You would reject the null hypothesis in favor of the alternative, which is less than 100 hours (0.034 < 0.05).
37.
If you you were were to to use a 1% sig signi nifi fica canc ncee level level in thi thiss case, case, woul would d you you concl conclude ude tha thatt the mean life life of the batteries batteries is typically typically more more than 100 hours? Explain Explain your answer. 223
Hypothesis Testing
ANSWER: Yes. You cannot reject the null hypothesis hypothesis at a 1% level of significance (0.034 > 0.01).
QUESTIONS 38 AND 39 ARE BASED ON THE FOLLOWING FOLLOWING INFORMATION: INFORMATION:
Two Two team teamss of work worker erss asse assemb mble le auto automo mobi bile le engin engines es at a manuf manufac actu turi ring ng plan plantt in Michig Michigan. an. A random random sample sample of 145 assemb assembli lies es from from team team 1 sho shows ws 15 unaccep unacceptab table le asse assemb mbli lies es.. A simi simila larr rand random om samp sample le of 125 125 asse assemb mbli lies es from from team team 2 show showss 8 unacceptable assemblies. 38.
Constr Construct uct a 90% 90% confi confidenc dencee inter interval val for for the the diffe differen rence ce betwe between en the the prop proport ortion ionss of unacceptable assemblies generated by the two teams. ANSWER: n = 145, Pˆ = 0.1034, n 1
1
2
= 125, Pˆ = 0.0640, z − multipl e = 1.645 2
ˆ) Pˆ1 (1 − Pˆ1 ) Pˆ2 (1 − P 2 SE ( Pˆ1 − Pˆ2 ) = + n1 n2
= 0.0334
ˆ ) = 0.0394 ± 0.0549 ( Pˆ1 − Pˆ2 ) ± Z ⋅ SE( Pˆ1 − P 2 Lower limit = -0.0155, and Upper limit = 0.0943 39.
Base Based d on the conf confid iden ence ce inte interv rval al cons constr truc ucte ted d in Quest Questio ion n 38, is ther theree suffi suffici cient ent evidence o conclude, at the 10% significance level, that the two teams differ with respect to their proportions of unacceptable assemblies? ANSWER: Because the 90% confidence interval includes the value 0, we cannot reject the null hypothesis of equal proportions.
40.
Stapl Staples es,, a chain chain of large large offi office ce supp supply ly stor stores es,, sells sells a line line of desk deskto top p and lapt laptop op comput computers ers.. Company Company execut executive ivess want to know whether whether the demands demands for these these two types of computers are related in any way. Each day's demand for each type of computers is categorized categorized as Low, Medium-Low, Medium-Low, Medium-High, Medium-High, or High. The data shown in the table below is based on 200 days of operation. Based on these data, can Staples conclude that demands for these two types of computers are independent? Test at the 5% level of significance.
Desktops Med-Low Med-High
Low
224
High
Chapter 10
Laptops
Low Med-Low Med-High High
3 6 13 8 30
14 18 16 14 62
14 17 11 15 57
4 22 16 9 51
35 63 56 46 200
ANSWER:
We fail to reject the null hypothesis of independence at the 5% significance level (since p-value = 0.083 > 0.05). We may conclude that demands for these two types of computers are independent 41. 41.
Suppo ppose th that you you ar are as asked to to te test H 0 : µ = 100 versus H a : µ < 100 at the α = 0.05 significan significance ce level. Furthermor Furthermore, e, suppose that you observe observe values of the sample mean and sample standard deviation when n = 50 that lead to the rejection of H 0 . Is it true true that you might might fail to reject H 0 if you were to observe the same values of the sample mean and standard deviation from a sample with n > 50? Why or why not? ANSWER: No. When n increases and the standard deviation of the sample mean stays the same, same, the standard standard error will decrease. Therefore, Therefore, the test statistic statistic will become
225
Hypothesis Testing more significant. If you rejected H 0 with n = 50, you will continue to reject with n > 50. QUESTIONS 42 THROUGH 44 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Do graduates of undergraduate business programs with different majors tend to earn disparate disparate starting starting salaries? salaries? Below you will will find the StatPro output output for 32 randomly randomly selected graduate with majors in accounting (Acct), marketing (Mktg), finance (Fin), and information systems (IS).
Summary statistics statistics for for samples
Sample sizes Sample means Sample standard deviations Sample variances Weights for pooled variance
Acct.
Mktg.
9 32711.67 2957.438 8746437.5 0.286
6 27837.5 754.982 5 69 997 .5 0.179
Number of samples Total sample size Grand mean Pooled variance Pooled standard deviation
4 32 3 1 0 3 9 .2 2 5308612.5 2304.043
One Way ANOVA table Source Between variation Within variation Total variation
SS 1 176 0 9 807 1 48 64 11 49 2 662 5 0 95 5
df 3 28 31
Confidence Intervals for Differences Difference Mean diff Acct. - Mktg. 4874.167 Acct. – Fin. 2537.667 Acct. - IS -157.619 Mktg. – Fin. -2336.500 Mktg. - IS -5031.786 Fin. - IS -2695.286
MS 392 03 26 9 530 8 612
Lower limit 1263.672 -609.890 -3609.912 -5874.048 -8843.014 -6071.216
226
Fin. 10 30174 1354.613 1834976.7 0.321
F 7 .3 85
IS 7 32869.3 3 143 .906 9884145.2 0.214
p-value 0.0009
Upper limit 8484.661 5685.223 3294.674 1201.048 -1220.557 680.644
Chapter 10 42.
Assuming Assumi ng that that the the vari varianc ances es of of the the four four under underly lying ing popul populati ations ons are equal, equal, can can you you reject at a 5% significance level that the mean starting salary for each of the given business majors? Explain why or why not? ANSWER: Yes. Because of the the F-test and the p-value is less than 0.05 ( p-value = 0.0009)
43.
Is the there re any any reaso reason n to doubt doubt the the equal equal-v -var aria ianc ncee assum assumpti ption on made made in Ques Questi tion on 42? 42? Support your answer. ANSWER: Yes, there is some cause for concern. The F-test is rather robust, however, is this case, the sample sizes are rather small and of different sizes.
44.
Use the inform informati ation on above above relate related d to the 95% confid confidenc encee interv intervals als for each pair pair of differences to explain which ones are statistically significant at α = 0.05. ANSWER: These confidence intervals show that the accounting majors stating salaries, on average, are larger than the marketing majors. There is not a significant difference for the other pairs using a 95% confidence interval.
QUESTIONS 45 THROUGH 47 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Do graduates of undergraduate business programs with different majors tend to earn disparate average starting salaries? salaries? Consider the data given in the table below.
Accounting $37,220 $30,950 $32,630 $31,350 $29,410 $37,330 $35,700
Marketing $28,620 $27,750 $27,650 $27,640 $28,340
Finance $29,870 $31,700 $31,740 $32,750 $30,550 $29,250 $28,890 $30,150
Management $28,600 $27,450 $26,410 $27,340 $27,300
227
Hypothesis Testing 45.
Is the there re any any reaso reason n to doubt doubt the the equa equall-va vari rian ance ce assu assump mpti tion on made made in the the oneone- way ANOVA model in this particular case? Explain. ANSWER: Summary measures measures table table Accounting Sample sizes 7 Sample means 33512.857 Sample standard deviations 3213.413 Sample variances 10326023.810 Weights for pooled variance 0.286
Marketing 5 28000.000 451.276 203650.000 0.190
Finance 8 30612.500 1342.458 1802192.857 0.333
Management 5 27420.000 780.096 608550.000 0.190
There There certain certainly ly is reason reason to doubt equal varian variances ces.. The ratio ratio of the larges largestt standard deviation to the smallest is about 7.12, so the ratio of corresponding variances is about 51. 46.
Assumi Assu ming ng that that the vari varian ance cess of the four four underl underlyi ying ng popul populat atio ions ns are are indee indeed d equal, equal, can you reject at the 10% significance level that the mean starting salary is the same for each of the given given business majors? Explain why or why not. ANSWER: One Way ANOVA table Source of variation Between groups Within gr groups Total variation
H 0 : µ1 = µ2
SS 140927283.143 77820292.857 218747576.000
=µ =µ 3
4
df 3 21 24
MS 46975761.048 3705728.231
F 12.677
p-value 0.0001
H α : At least two population means are unequal. unequ al. The ANOVA table indicates indicates definite mean difference, difference, even at the 1% level (since the p-value is is less than than .01). Even if the the test is is not perfectly perfectly valid valid (because (because of unequal variances), we can still be pretty confident that the means are not all equal.
47.
Gene Genera rate te 90% conf confid iden ence ce inter interva vals ls for for all pair pairss of diffe differe renc nces es betwe between en means means.. Whic Which h of the the diff differ eren ences ces,, if any, any, are are stat statis isti tica call lly y signi signifi fica cant nt at the the 10% significance level?
228
Chapter 10 ANSWER: Simultaneous Simultaneous confidence confidence intervals for mean differences differences with confidence confidence level of 90% Difference Accounting - Marketing Marketing Accounting - Finance Finance Accounting - Management Management Marketing - Finance Marketing - Management Finance - Management
Mean difference 5512.857 2900.357 6092.857 -2612.500 580.000 3192.500
Lower limit 2510.523 246.644 3090.523 -5535.603 -2662.892 269.397
Upper limit 8515.191 5554.071 9095.191 310.603 3822.892 6115.603
Significant? Yes Yes Yes No No Yes
The a Accounting mean is significantly different (larger) than each of the others. Also, the Finance mean is significantly different (larger) than the Management mean. The other means are not significantly different from each other. QUESTIONS 48 THROUGH 52 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Q-Mart is interested interested in comparing its male and female customers. Q-Mart would like like to know if its female charge customers spend more money, on average, than its male charge customers. customers. They have collected collected random samples samples of 25 female customers customers and 22 male custom customers ers.. On average, average, women charge charge customers customers spend spend $10 $102.23 2.23 and men charge charge customers spend $86.46. Additional information information are shown below: Summary statistics statistics for for two samples Sales (Female) Sample sizes 25 Sample means 102.23 Sample standard deviations 93.393
Test of difference=0 Sample mean difference Pooled standard deviation Std error of difference t-test statistic p-value
48. 48.
Sales (Male) 22 86.460 59.695
15.77 79.466 23.23 0.679 0.501
Given the the inf info ormation abo above, wha what is H 0 and H a for this comparison comparison?? Also, does this represent a one-tailed one-tailed or a two-tailed test? test? Explain your answer.
229
Hypothesis Testing ANSWER: H 0 : µ F ≤ µ M , H a : µ F
49. 49.
> µ M .
This represents a one-tail test. What are are the the deg degrees of fr freedom for for the the t -statisti -statisticc in this calculati calculation? on? Explain Explain how you would calculate the degrees of freedom in this case.
ANSWER: d.f = 25 + 22 – 2 = 45
50. 50.
What What is the the assum assumpt ptio ion n in this this case case that that allo allows ws you you to use use the the poole pooled d stand standar ard d deviation for this confidence interval? ANSWER: The assumption is that the populations’ standard deviations are equal (σ F
51. 51.
= σ M ).
Using Usin g a 10% 10% leve levell of sign signif ific ican ance ce,, is ther theree suff suffic icie ient nt evide evidenc ncee for for Q-Ma Q-Mart rt to conclude that women charge customers on average spend more than men charge customers? Explain your answer. ANSWER: No. There is not a statistical difference between women and men spending at QMart, since p-value = 0.501 > 0.10.
52. 52.
Using Usin g a 1% leve levell of sign signif ific ican ance ce,, is ther theree suff suffic icie ient nt evide evidenc ncee for for Q-Mar Q-Martt to conclude that women charge customers on average spend more than men charge customers? Explain your answer. ANSWER: No. There is not a statistical difference between women and men spending at QMart, since p-value = 0.501 > 0.01.
53. 53.
The CEO of a softw The softwar aree compa company ny is comm commit itte ted d to expa expand ndin ing g the prop propor orti tion on of highly qualified women in the the organization’s staff of salespersons. He claims that the proportion of women in similar sales positions across the country in 1999 is less than 45%. Hoping to find support support for his claim, he directs directs his assistant assistant to collect a random sample of salespersons employed by his company, which is thought to be representative of sales staffs of competing organizations in the indust industry ry.. The collect collected ed random random sample sample of size 50 showed showed that that only 18 were were α women. Test this this CEO’s claim at the =.05 significance level and report the pvalue. Do you find statistical statistical support support for his hypothesis hypothesis that the proportion proportion of women in similar sales positions across the country is less than 40%? ANSWER: H 0 : P ≥ 0.45
H a : P < 0.45
Test statistic: Z =-1.279
230
Chapter 10 -value = 0.10 P -value There is not enough evidence to support this claim. The P -value -value is large (0.10). QUESTIONS 54 THROUGH 56 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Joe owns a sandwich sandwich shop near a large university university.. He wants to know if he is servings servings approximately the same number of customers customers as his competition. His closest competitors competitors are Bob and Ted. Joe decides to use a couple of college students to collect collect some data for him on the number of lunch customers served by each sandwich shop during a weekday. The data for two weeks (10 days) and additional information are shown below (the tables have been generated using StatPro). Summary stats stats for samples samples
Sample sizes Sample means Sample standard deviations Sample variances Weights for pooled variance
Joe’s 10 50.700 4.244 18.011 0.333
Number of samples Total sample size Grand mean Pooled variance Pooled standard deviation
3 30 4 6 .8 0 0 17 .044 4.128
One-way ANOVA Table Source SS Between variation 264.60 Within variation 460.20 Total variation 724.80
df 2 27 29
Bob’s 10 46.200 4.492 20.178 0.333
MS 132.30 1 7 .04 4
Ted’s 10 43.500 3.598 12.944 0.333
F 7.762
p-value 0.0022
Confidence Intervals for mean difference using 95% confidence level Difference Mean diff Lower Upper Joe’s – Bob’s 4.500 -0.282 9.282 Joe’s – Ted’s 7.200 2.418 11.982 Bob’s – Ted’s 2.700 -2.082 7.482
54.
Are Are all thr three ee sand sandwi wich ch shop shopss servi serving ng the the same same numbe numberr of custo custome mers rs,, on avera average ge,, for lunch each weekday? Explain how you arrived arrived at your answer. ANSWER:
231
Hypothesis Testing No. You should reject H o at a 5% significance level ( p p-value = 0.0022). Means are not all equal. 55. 55.
Explai Expl ain n why the the weigh weights ts for for the pool pooled ed vari varian ance ce are are the same same for for each each of the samples. ANSWER: The weights for the pooled variance are the same for each of the samples, because sample sizes are equal (sample of 10 customers cu stomers from each sandwich shop).
56.
Use the inf Use infor orma mati tion on rela relate ted d to the the 95% conf confid idenc encee inter interva vall to expla explain in how how the number of customers Joe has each weekday compares to his competition. ANSWER: These intervals intervals show that there is not a significan significantt difference difference between Joe’s and Bob’s. However, there is a significant significant difference between Joe’s Joe’s and Ted’s using using a 95% confidence interval.
QUESTIONS 57 AND 58 ARE BASED ON THE FOLLOWING FOLLOWING INFORMATION: INFORMATION:
The manag The manager er of a cons consul ulti ting ng firm firm in Lans Lansin ing, g, Michi Michiga gan, n, is tryi trying ng to asse assess ss the the effectiveness of computer skills skills training given to to all new entry-level entry-level professionals. In an effort to make such an assessment, he administers a computer skills test immediately before and after the training program to each of 20 randomly chosen employees. The pre-training and post-training scores of these 20 individuals are shown in the table below.
Empl Employ oyee ee Scor Score e befor before e Scor Score e after after 1 62 77 2 63 77 3 74 83 4 64 88 5 84 80 6 81 80 7 54 83 8 61 88 9 81 80 10 86 88 11 75 93 12 71 78 13 86 82 14 74 84 15 65 86 16 90 89 17 72 81 18 71 90 19 85 86 20 66 92
232
Chapter 10
57. 57.
Using Usin g a 10% 10% level level of sign signif ific ican ance ce,, do the the given given samp sample le data data supp suppor ortt that that the firm’s training programs is effective in increasing the new employee’s working knowledge of computing? ANSWER: H 0 : µ D ≥ 0
where D = Before score ore-After score H 1 : µ D < 0, whe Test statistic: t = - 4.471 (paired t -test) -test) -value = 0.00013 P -value The test scores scores have improved improved by an average of 11 points. points. Since the P -value -value is virtually 0, there is enough evidence to conclude that the given sample data suppor sup portt that that the firm’ firm’ss train training ing progra program m is increa increasin sing g the new employ employee’ ee’ss knowledge of computing at the 10% significance level. 58.
Re-d Re-do o Que Quest stio ion n 57 57 usi using ng a 1% 1% leve levell of of sig signi nifi fican cance ce.. ANSWER: Again, since the P -value -value is virtually zero, there is plenty of evidence to support the effectiveness of the program at the 1% level of significance.
QUESTIONS 59 THROUGH 62 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Suppose a firm that produces light bulbs wants to know whether it can claim that it light bulbs typically last more than 1500 hours. Hoping to find support for their claim, the firm collects collects a random sample sample and records records the lifetime lifetime (in hours) hours) of each bulb. The information related to the hypothesis test is presented below. Test of µ ≤ 1500 versus one-tailed alternative Hypothesized mean 1500.0 Sample mean 1509.5 Std error of mean 4.854 Degrees of freedom 24 t-test statistic 1.953 p-value 0.031
59.
Can Can the the samp sample le size size be deter determi mined ned from from the the inf inform ormat atio ion n abov above? e? Yes Yes or or no? no? If yes, what is the sample size in this case? ANSWER: Yes. 24 + 1 = 25.
233
Hypothesis Testing
60.
The firm firm beli believes eves that that the the mean mean lif lifee is actual actually ly grea greater ter than than 1500 1500 hour hours, s, shou should ld you conduct a one-tailed or a two-tailed two-tailed hypothesis test? Explain your answer. ANSWER: One-tailed, since the firm is interested in finding whether the mean is actually greater than 1500.
61.
What What is is the the samp sample le mea mean n of thi thiss data data?? If you you use use a 5% sig signi nifi fica canc ncee leve level, l, woul would d you conclude that the mean life of the light bulbs is typically more than 1500 hours? Explain your answer. ANSWER: 1509.5 hours. hours. Yes, you would reject reject the null hypothesi hypothesiss in favor of the mean being greater than 1500 hours (0.031 < 0.05).
62.
If you you were were to to use a 1% sig signi nifi fica canc ncee level level in thi thiss case, case, woul would d you you concl conclude ude tha thatt the mean life of the light light bulbs is typically more more than 1500 hours? Explain your answer. ANSWER: No. You cannot cann ot reject the null nu ll hypothesis at a 1% level of significance (0.031 > 0.01).
QUESTIONS 63 AND 64 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
A study is performed in San Antonio to determine whether the average weekly grocery bill per five-person family in the town is significantly different from the national average. A random sample of 50 five-person families in San Antonio showed a mean of $133.474 and a standard deviation of $11.193. 63.
Assume Assu me that that the the natio nationa nall average average week weekly ly groc grocery ery bill bill for for a fivefive-per perso son n famil family y is $131. Is the sample evidence statistically significant? If so, so, at what significance significance levels can you reject the null hypothesis? ANSWER: H 0 : µ = 131
H α : µ ≠ 131
Test statistic: t = 1.563 p-value: 0.124 The sample mean is not significantly different from 131 at even the 10% level because the p-value is greater than 0.10
234
Chapter 10
64.
For For which which valu values es of the the samp sample le mean mean (i.e. (i.e.,, avera average ge week weekly ly groc grocer ery y bill) bill) wou would ld you decide to reject the null hypothesis at the α = 0.01 significance level? For which values of the sample mean would you decide to reject the null hypothesis at the α 0.10 significance level? =
ANSWER: For either p-value (0.01 or 0.10), we find the t -value -value that would lead to the rejection of the null hypothesis, and then solve the equation 1.583 for X on eithe t = ( X − 131) / 1. eitherr side side of 131. 131. Th This is leads leads to the the foll follow owin ing g results: α -value
0.01 0.10
t -value 2.680 1.677
Lower limit 1 26 .758 1 28 .346
For example, at the 10% level, if X the null hypothesis.
Upper limit 135.242 133.654
< 128.346 or X > 133.654, we would reject
QUESTIONS 65 THROUGH 68 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Do undergraduate business students who major in information systems (IS) earn, on average, higher annual starting salaries than their peers who major in marketing (Mktg)? Before addressing this question with a statistical hypothesis test, a comparison should be done to determine whether the variances of annual starting salaries of the two types of majors majors are equal. Below you will find find the StatPro output for for 20 randomly selected selected IS majors and 20 randomly selected Mktg Mk tg majors. Summary statistics statistics for for two samples
IS Salary 20 30401.35 1937.52
Sample sizes Sample means Sample standard deviations Test of difference ≤ 0 Sample mean difference Pooled standard deviation Std error of difference Degrees of freedom t-test statistic p-value
2685.5 2515.41 795.44 38 3.376 0.0009
Test of equality of variances
235
Mktg Salary 20 2 7 7 1 5 .8 5 2 983 .39
NA 7 9 5 .4 4 33 3.376 0.0009
Hypothesis Testing
65.
Ratio of sample variances 2.371 p-value 0.034 Use the inform informati ation on above above to to perfor perform m the the test test of equal equal varian variance. ce. Explai Explain n how how the the ratio ratio of sample variances is calculate calculated. d. What type of distributi distribution on is used to test for equal variances variances?? Also, would would you conclude that the variances variances are equal or not? Explain your answer. ANSWER: (2983.39)2 / (1937.52)2 = 2.371. Since the p-value is 0.034, you can conclude that there is a significant significant difference between the sample variance. variance. They are not equal.
66.
Base Based d on your your conc conclu lusi sion on in Que Quest stio ion n 65, whi which ch test test stat statis isti ticc shoul should d be used used in performing a test for the existence of a difference between population means? ANSWER: Conduct the t-test with individual sample variances (do not n ot use pooled variance).
67.
Using Usin g a 5% level level of signi signifi fican cance ce,, is there there suff suffic icie ient nt evid eviden ence ce to concl conclud udee that IS IS majors earn, on average, a higher annual starting salaries than their peers who major in Mktg? Explain your answer. ANSWER: Yes. The average starting starting salary for IS majors majors is significantl significantly y larger than the starting salary for MKT majors, since p-value = 0.0009 < 0.05.
68.
Using Usin g a 1% level level of signi signifi fican cance ce,, is there there suff suffic icie ient nt evid eviden ence ce to concl conclud udee that IS IS majors earn, on average, a higher annual starting salaries than their peers who major in Mktg? Explain your answer. ANSWER: Yes. The average starting salary for IS majors is significantly larger than the starting salary for MKT majors even at a 1% significance level, since p-value = 0.0009 < 0.01.
69. 69.
A rece recent nt stud study y of educ educat atio iona nall leve levels ls of 1000 1000 vote voters rs and and thei theirr poli politi tica call part party y affiliations in a Midwestern state showed the results given in the table below. Use the 5% significance level and test to determine if party affiliation is independent of the educational level of the voters. Party Affiliation Democrat Republican Independent Didn't Complete High School Educational Level Has High School Diploma Has College Degree
ANSWER:
236
95 135 160 390
80 85 105 270
115 105 120 340
290 325 385 1000
Chapter 10
We fail to reject the null hypothesis of independence at the 5% significance level (sin (since ce p-val p-value ue = 0.08 0.087 7 > .05). .05). We may may concl conclude ude that that part party y affi affili liat atio ion n is independent of the educational level of the voters.
QUESTIONS 70 THROUGH 73 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
A marketing research consultant hired by Coca-Cola is interested in determining if the proportion of customers who prefer Coke to other brands is over 50%. A random sample of 200 consumers was selected from the market under investigation, 55% favored CocaCola over other brands. Additional information information is presented below. Sample proportion Standard error of sample proportion Z test statistic p-value 70. 70.
0.55 0.03518 1 .421 3 0.07761
If you you were were to cond conduc uctt a hypo hypoth thes esis is test test to to deter determi mine ne if grea greate terr than than 50% of customers prefer Coca-Cola to other brands, would you conduct a one-tail or a two-tail hypothesis hypothesis test? Explain your answer. ANSWER: One-tailed, since the consultant is interested in finding whether the proportion is actually greater than 50%.
237
Hypothesis Testing 71. 71.
How How many many cust custom omer erss out of the the 200 200 sampl sampled ed must must have have favo favore red d Coke Coke in this this case? ANSWER: (200)(0.55) = 110
72. 72.
Using Usin g a 5% sign signif ific ican ance ce level level,, can can the the mark market etin ing g cons consul ulta tant nt concl conclud udee that that the proportion of customers who prefer Coca-Cola exceeds 50%? Explain your answer. ANSWER: No. You cannot reject the null hypothesis at a 5% level of significance, since pvalue = 0.07761 > 0.05.
73. 73.
If you you were were to to use a 1% sign signif ific ican ance ce leve level, l, woul would d the conc conclu lusi sion on from from part part c change? Explain your answer. ANSWER: No. You still cannot reject the null hypothesis at a 1% level of significance, since p-value = 0.07761 > 0.01.
QUESTIONS 74 THROUGH 77 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
The owner of a popular Internet-based auction site believes that more than half of the people who sell items on her site are women. To test this hypothesis, h ypothesis, the owner sampled 1000 customers who sale items on her site and she found that 53% of the customers sampled were women. Some calculations are shown in the table table below Sample proportion Standard error of sample proportion Z test statistic p-value 74. 74.
0.53 0.01578 1 .900 8 0.0287
If you you were were to cond conduc uctt a hypo hypoth thes esis is test test to to deter determi mine ne if grea greate terr than than 50% of customers who use this Internet-based site are women, would you conduct a onetail or a two-tail hypothesis test? Explain your answer. ANSWER: One-tailed, since the owner is interested in finding whether the proportion is actually greater than 50%.
75. 75.
How How many many cust custom omer erss out of the the 1000 1000 samp sample led d must must have have been been women women in thi thiss case? ANSWER: (1000)(0.53) = 530
238
Chapter 10
76.
Using Usin g a 5% sign signif ific ican ance ce leve level, l, can can the own owner er of thi thiss site site concl conclud udee that that women women make up more than than 50% of her customers? Explain your answer. ANSWER: Yes. You can reject the null null hypothesis hypothesis at a 5% level of significance, significance, since since pvalue = 0.0287 < 0.05.
77.
If you you were were to use use a 1% sign signif ific icanc ancee level level,, would would the the concl conclus usio ion n from from Quest Questio ion n 76 change? Explain your answer. ANSWER: Yes. Your answer would now change. You cannot reject reject the null hypothesis at a 1% level of significance, since p-value = 0.0287 > 0.01.
QUESTIONS 78 THROUGH 82 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Q-Mart is interested in comparing customer who used it own charge card with those who use other types of credit cards. cards. Q-Mart would like to to know if customers who use the the QMart card spend more money per visit, on average, than customers who use some other type of credit card. card. They have collected collected informati information on on a random sample sample of 38 charge customers and the data is presented below. On average, the person using using a Q-Mart card spends $192.81 per visit and customers using another type of card spend $104.47 per visit. Summary statistics statistics for two two samples
Sample sizes Sample means Sample standard deviations
Q-Mart 13 192.81 115.243
Test of difference=0 Sample mean difference Pooled st standard deviation Std error of difference t-test statistic p-value
8 8 .34 88.323 30.201 2.925 0.006
78. 78.
Other Charges 25 104.47 71.139
Given the the inf info ormation abo above, wha what is H 0 and H a for this comparison comparison?? Also, does this represent a one-tailed one-tailed or a two-tailed test? test? Explain your answer. ANSWER: H 0 : µQ Mart ≤ µOthers , H a : µQ −
−
Mart
> µ Others . This represents a one-tail test. 239
Hypothesis Testing
79. 79.
What are are the the deg degrees of fr freedom for for the the t -statisti -statisticc in this calculati calculation? on? Explain Explain how you would calculate the degrees of freedom in this case. ANSWER: d.f = 13 + 25 – 2 = 36
80. 80.
What What is the the assum assumpt ptio ion n in this this case case that that allo allows ws you you to use use the the poole pooled d stand standar ard d deviation for this confidence interval? ANSWER: The assumption is that the two populations standard deviations are equal; that is σQ
81. 81.
−
Mart
= σ Others
Using Usin g a 5% leve levell of sign signif ific ican ance ce,, is ther theree suff suffic icie ient nt evide evidenc ncee for for Q-Mar Q-Martt to conclude that customers who use the Q-Mart card charge, on average, more than those who use another charge card? Explain your answer. ANSWER: Yes. There is a statistica statisticall difference difference between those using using the Q-Mart card and those who use other types of charge cards, since p-value = 0.006 < 0.05.
82. 82.
Using Usin g a 1% leve levell of sign signif ific ican ance ce,, is ther theree suff suffic icie ient nt evide evidenc ncee for for Q-Mar Q-Martt to conclude that customers who use the Q-Mart card charge, on average, more than those who use another charge card? Explain your answer. ANSWER: Yes. There is still still a statisti statistical cal difference difference between those using using the Q-Mart card and those who use other types of charge cards, since p-value = 0.006 < 0.01.
83.
The num number ber of cars cars ssold old by thre threee sales salesper person sonss over over a 6-mon 6-month th peri period od are are shown shown in the table below. Use the 5% level of significance to test for independence of salespersons and type of car sold. Insurance Preference Ali Salesperson Bill Chad
Chevrolet
Ford
Toyota
15 20 13 48
9 8 4 21
5 15 11 31
240
29 43 28 100
Chapter 10
ANSWER:
We fail to reject the null hypothesis of independence at the 5% significance level (since p-value = 0.305 > 0.05). We may conclude that salespersons and type of car sold are independent.
QUESTIONS 84 AND 85 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
An automobile manufacturer needs to buy aluminum sheets with an average thickness of 0.05 inch. The manufacturer manufacturer collect collectss a random sample of 40 sheets sheets from a potential potential supplier. The thickness of each sheet in this sample is measured measured (in inches) and recorded. The information below are pertaining to the Chi-square goodness-of-fit test. Upper limit 0.03 0.04 0.05 0.06
Category ≤ 0 .03 0.03 but ≤ 0.04 0.04 but ≤ 0.05 0.05 but ≤ 0.06 >0.06
Frequency 1 10 13 12 4
241
Normal 1.920 8.074 14.947 11.218 3.842
Distance measure 0 .4 4 1 0 .4 5 9 0 .25 4 0 .05 5 0 .00 7
Hypothesis Testing
Test of normal fit Chi-square statistic p-value
84.
1.214 0.545
Are these these measur measureme ements nts normal normally ly distri distribut buted? ed? Summar Summarize ize you yourr result results. s. ANSWER: Yes. Based on the Chi-square test, with a p-value of 0.545, you can conclude that the values are normally normally distributed. distributed. The frequency frequency distribution distribution also shows that the values are fairly close to the expected values.
85.
Are Are there there any any weakne weakness sses es or conc concer erns ns abou aboutt your your concl conclus usio ions ns in in Quest Questio ion n 84? Explain your answer. ANSWER: Yes. There are are a couple of concerns concerns.. The sample sample size is is rather rather small (n (n = 40), you should should use a larger sample sample size for this test test to be more effective. effective. Also, the the test depends depends on which and how many categorie categoriess are used for the histogram histogram.. A different choice could result in a different answer.
QUESTIONS 86 THROUGH 88 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
Do undergraduate business students who major is computer info inform rmat atio ion n syst system emss (CIS) earn, on average, higher annual starting starting salaries than thei heir peers who major in internation international al business business (IB)?. (IB)?. To address address this que question through a stati statisti stical cal hyp hypoth othesi esiss test test,, the the tabl tablee show shown n below contains the start starting ing salari salaries es of 25 randomly randomly selected selected CIS majors and 25 random randomly ly select selected ed IB majors.
Graduate 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Finance 29,522 31,444 29,275 26,803 28,727 32,531 33,373 31,755 31,393 26,124 30,653 30,795 30,319 31,654 27,214 30,579 30,249 31,024 31,940 31,387 29,479 30,735 29,271 30,215 31,587
242
Marketing 28,201 29,009 29,604 26,661 26,094 22,900 24,939 23,071 29,852 27,213 23,935 25,794 28,897 27,890 27,400 26,818 27,603 26,880 28,791 24,000 25,877 24,825 28,423 28,956 29,758
Chapter 10
86. 86.
Is it it appr approp opri riat atee to per perfo form rm a pai paire redd-co com mpari pariso son n t -test -test in this case? case? Explain why or why not. ANSWER: A two-sample, not paired-sample, procedure should be used because there is no evidence of pairing.
87.
Perfor Perform m an an appr appropr opriat iatee hypot hypothes hesis is test test with with a 1% signif significa icance nce level. level. Ass Assume ume that that the population variances are equal. ANSWER: H 0 : µ1 − µ 2 ≤ 0 , H a : µ1 − µ 2
> 0 , Test statistic t = 6.22, P -value=0. -value=0. Since P -value -value
is virtually 0, we can conclude at the 1% level that the mean salary for CIS majors is indeed larger. 88.
How How large large woul would d the diff differ eren ence ce betw between een the the mean mean star starti ting ng sala salari ries es of CIS CIS and and IB majors have to be before you could conclude that CIS majors earn more on average? Employ a 1% significance significance level in answering this question. ANSWER: -value=0.01, t =2.41, and Standard error of difference = P -value=0.01,
SE ( X 1
−
X 2 )
=
544.439 .
Then X 1 − X 2 = t ⋅ SE ( X1 − X 2 ) = 1312.20 A mean difference of 1312.20 is all that would be required to get the conclusion in Question 87 at the 1% level.
243
Hypothesis Testing
89. 89.
A stat statis isti tics cs profe profess ssor or has has just just give given n a fina finall exam examin inat atio ion n in his his line linear ar mode models ls course. He is particularly interested interested in determining whether the distribution distribution of 50 exam scores scores is normally normally distrib distribute uted. d. The data are shown in the table table below. below. Perform the Lilliefors Lilliefors test. Report and interpret the results results of the test. 77 73 79 91 80
71 89 62 70 91
78 74 73 76 74
83 75 88 74 69
84 93 76 68 88
71 74 76 80 84
81 88 76 87 83
82 83 80 92 87
79 90 84 84 82
71 82 84 79 72
ANSWER: Thee maxi Th maximu mum m dist distan ance ce betw betwee een n the the empi empiri rica call and and norm normal al cumu cumula lati tive ve distributions is 0.0802. This is less than 0.1247, the maximum allowed with a sample size of 50. Therefore, the normal hypothesis cannot be rejected at the 5% level.
Normal (smooth) and empirical cumulative distributions 1.0
0.8
0.6
0.4
0.2
0.0 3 1 9 7 5 3 1 9 7 5 3 1 9 7 5 6 8 0 2 4 6 8 0 2 4 6 8 0 2 9 7 4 2 0 8 6 3 1 9 7 5 2 0 1 3 5 8 0 2 4 6 9 1 3 5 7 0 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 2 2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 2 2 2 2 3 3 -
Standardized values of Score
244
Chapter 10
90.
An ins insur uran ance ce firm firm int inter ervi view ewed ed a random random sam sampl plee of 600 col colle lege ge stude student ntss to find find out the type of life insurance preferred, if any. The results are shown in the table below. Is there evidence that life insurance preference of male students is different than that of female students. Test at the 5% significance level.
Insurance Preference Term erm
Whol Whole e Life Life
No Insu Insura ranc nce e
Male
80
30
240
350
Female
50 130
40 70
160 400
250 600
Gender
ANSWER:
We reject the null hypothesis of independence at the 5% significance level (since p-value = 0.019 < 0.05). We may conclude that there is no evidence that life insurance preference of male students is different than that of female students.
QUESTIONS 91 THROUGH 93 ARE BASED ON THE FOLLOWING INFORMATION: INFORMATION:
245
Hypothesis Testing
The retailing manager of Meijer supermarket chain in Michigan wants to determine whether product location has any effect on the sale of children toys. Three different aisle locations are considered: front, middle, and rear. A random sample of 18 stores is selected, with 6 stores randomly assigned to each aisle location. The size of the display area and price of the product are constant for all the stores. At the end of onemonth trial period, the sales volumes (in thousands of dollars) of the product in each store were as shown below: Front Aisle 10.0 8.6 6.8 7.6 6.4 5.4
91.
Middle Aisle 4.6 3.8 3.4 2.8 3.2 3.0
Rear Aisle 6.0 7.4 5.4 4.2 3.6 4.2
At the the 0.05 0.05 leve levell of signif significa icance nce,, is ther theree eviden evidence ce of a signi signific ficant ant diff differe erence nce in in average sales among the various va rious aisle locations? ASNWER: StatPro’s one-way ANOVA produces the following results:
246
Chapter 10 To test at the 0.05 level of significance whether the average sales volumes in thousands of dollars are different across the three store aisle locations, we conduct an F test: H 0: µ Front µ Middle µ Rear H 1: At least one mean is different. =
=
Since p-value = 0.0004 < α = 0.05, we reject H 0. There is enough evidence to conclude that the average sales volumes in thousands of dollars are different across the three store aisle locations. 92. 92.
If appro appropr pria iate te,, whic which h aisl aislee loca locati tion onss appe appear ar to diffe differr sign signif ific ican antl tly y in aver averag agee sales? (Use α = 0.05) ANSWER:
It appears that the front and middle aisles and an d also the front and rear aisles differ significantly in average sales at α = 0.05. 93. 93.
What hat shoul hould d the the ret retail ailing manag anager er conc conclu lude de?? Ful Fully des descri cribe the retai etaili ling ng manager’s options with respect to aisle locations? ANSWER: The front aisle is best for the sale of this product. The manager should evaluate the tradeoff in switching the location of this product and the product that is currently intended for the front location.
QUESTIONS 94THROUGH 97 ARE BASED ON THE FOLLOWING INFORMATION:
A real estate agency wants to compare the appraised values of single-family homes in two cities in Michigan. A sample of 60 listings in Lansing and 99 listings in Grand Rapids yields the following results (in thousands of dollars):
X S n
Lansing 1 9 1 .3 3 32.60 60
Big Rapids 172.34 16.92 99
247
Hypothesis Testing 94.
Is ther theree evide evidenc ncee of a signi signifi fica cant nt diff differ eren ence ce in the the avera average ge appr apprai aise sed d value valuess for single-family homes in the two Michigan cities? Use 0.05 level of significance. ANSWER: Populations: 1 = Lansing, 2 = Grand Rapids
H 0: µ 1
= µ 2
H 1: µ 1
≠ µ 2
(The average appraised values for single-family homes are the same in Lansing and Grand Rapids) (The average appraised values for single-family homes are not the same in Lansing and Grand Rapids)
Decision rule: df = 157. If t < – 1.9752 or t > 1.9752, reject H 0.
S p
=
2
(n1 – 1) ⋅ S 1
+ (n 2 – 1) ⋅ S 2 2 (59) ⋅ 32.60 2 + (98) ⋅ 16.92 2 = = 578.0822 (n1 – 1) + (n2 – 1) 59 + 98 2
Test statistic:
t =
( X 1 – X 2 ) – ( µ 1 – µ 2 ) S p
2
1 1 + n1 n2
=
(191.33 – 172.34) – 0
1 + 1 = 4.8275 60 99
578.0822
Decision: Since t calc calc = 4.8275 is above the upper critical bound of 1.9752, reject H 0. There is enough evidence to conclude that there is a difference in the average appraised values for single-family homes in the two Michigan cities. The p value is 3.25E-06 using Excel. 95.
Do you you thin think k any of of the the assum assumpt ptio ions ns neede needed d in Ques Questi tion on 94 have have been been viol violat ated ed?? Explain. ANSWER: The assumption of equal variances may be violated because the sample variance in Lansing is nearly four times the size of the sample variance in Grand Rapids and the two sample sizes are not small. small. Nevertheless, the results of the the test for the differences in the two means were overwhelming (i.e., the p value is nearly 0).
96. 96.
Cons Constr truc uctt a 95% 95% conf confid iden ence ce inter interva vall esti estima mate te of the the diff differ eren ence ce betwe between en the population means of Lansing and Grand Rapids. ANSWER:
( X − X ) + t 1
2
1 1 + = ( 191.33 − 172.34) + 1.9752 n n
S p2
1
11.2202< µ1 − µ 2
2
< 26.7598
248
1 + 1 60 99
578.08
Chapter 10 97.
Explai Explain n how how to use the confid confidence ence interv interval al in in Quest Question ion 96 to to answe answerr Quest Question ion 94. ANSWER: Since the 95% confidence interval in Question 96 does not include 0, we reject the null hypothesis at the 5% level of significance that the average appraised values for single-family homes are the same in Lansing and Grand Rapids.
QUESTIONS 98THROUGH 100 ARE BASED ON THE FOLLOWING INFORMATION:
In a survey of 1,500 customers who did holiday shopping on line during the 2000 holiday season, 270 indicated that they were not satisfied with their experience. Of the customers that were not satisfied, 143 indicated that they did not receive the products in time for the holidays, while 1,197 of the customers that were satisfied with their experience indicated that they did receive the products in time for the holidays. The following complete summary of results were reported:
Received Products in Time for Holidays Satisfied with their Experience Yes No Total
98.
Y es
No
Total
1,197 127 1,324
33 143 176
1,230 270 1,500
Is ther theree a sign signif ific icant ant diff differ erenc encee in satisf satisfact actio ion n betw betwee een n those those who recei receive ved d thei their r products in time for the holidays, and those who did not n ot receive their products in time for the holidays? Test at the 0.01 0.0 1 level of significance. ANSWER: Populations: 1 = received product in time, 2 = did not receive product in time
= P 2 H1 : P1 ≠ P 2 H 0 : P1
Decision rule: If Z Z < -2.5758 or Z Z > 2.5758, reject H 0.
249
Hypothesis Testing Test statistic:
Z =
( pˆ1 – pˆ 2 )
1 1 + n n
pˆ c ⋅ (1 – pˆ c )
1
2
=
0.9041 – 0.1875
1 + 1 0.82 ⋅ ( 1 − 0.82) 1324 176
= 23.248
Decision: Since Z calc calc = 23.248 is well above the upper critical bound of Z = 2.5758, reject reject H 0. There There is suffic sufficien ientt eviden evidence ce to conclu conclude de that that a signif significa icant nt difference in satisfaction exists between those who received their products in time for the holidays and those who did not receive their products in time for the holidays. 99.
Find the p-value in Question 98 and interpret its meaning. ANSWER: The p-value is virtually 0. The probability of obtaining a difference in two sample proportions as large as 0.7166 or more is virtually 0 when H 0 is true.
100.
Based Based on the the results results of of Questio Questions ns 98 and and 99, if you you were the the market marketing ing direc director tor of a company selling products online, what would you do to improve the satisfaction of the customers? ANSWER: Ensuring that the customers receive their products in time for the holidays will be one effective way to improve the satisfaction of the customers.
TRUE / FALSE QUESTIONS 101.
The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed given that the null hypothesis is true. ANSWER:
102.
The p-value is usually 0.01 0r 0.05. ANSWER:
103.
F
A null hypo hypothe thesis sis is is a stateme statement nt about about the value value of a popula populatio tion n paramete parameter. r. It is usually the current thinking, or “status quo”. ANSWER:
104.
T
T
An altern alternati ative ve or researc research h hypothes hypothesis is is usuall usually y the hypoth hypothesi esiss a resear researche cherr wants wants to prove. ANSWER:
T
250
Chapter 10
105.
A two-tai two-tailed led altern alternative ative is one one that that is supported supported by evidence evidence in in a single single directio direction. n. ANSWER:
106.
A one-tai one-tailed led alter alternative native is one that that is supported supported by evidence evidence in either either directi direction. on. ANSWER:
107.. 107
which the null
T
F
In order to determine the p-val -value ue,, it is unnec unneces essa sary ry to know know the the level level of significance. ANSWER:
113.
α at
If a null null hypo hypoth thes esis is abou aboutt a popu popula lati tion on mean mean µ is rejected at the 0.025 level of significance, it must be rejected at the 0.01 0 .01 level. ANSWER:
112.
T
The p-value of a test is the smallest level of significance hypothesis can be rejected. ANSWER:
111. 111.
F
The probab The probabil ilit ity y of making making a Ty Type pe I erro errorr and the leve levell of signif signific ican ance ce are the the same. ANSWER:
110.
T
A Type Type II error error is commi committe tted d when we incor incorrec rectly tly accep acceptt an altern alternati ative ve hypoth hypothesi esiss that is false. ANSWER:
109.. 109
F
A Type Type I err error or pro proba babi bili lity ty is is repr repres esent ented ed by α ; it is the probability of incorrectly rejecting a null hypothesis that is true. ANSWER:
108.
F
T
If we we reject reject a null null hypot hypothesi hesiss about about a popul populati ation on propor proportio tion n p at the 0.025 level of significance, then we must also reject it at the 0.05 level. ANSWER:
T
251
Hypothesis Testing 114.
Using the confid confidence ence interv interval al when when conducti conducting ng a two-tailed two-tailed test for the population population mean µ , we do not reject the null hypothesis if the hypothesized value for µ falls between the lower and upper confidence limits. ANSWER:
115.. 115
T
A prof profes esso sorr of stati statist stic icss refu refute tess the the clai claim m that that the propor proporti tion on of inde indepen pende dent nt vote voters rs in Minn Minnes esot otaa is at most ost 40%. 40%. To test test the the clai claim, m, the the hypo hypoth thes eses es:: H : p = 0.40 , H : p ≠ 0.40 , should be used. 0 a
ANSWER:
116.
Using the confid confidence ence interv interval al when when conducti conducting ng a two-tailed two-tailed test for the population population proportion p, we reject the null hypothesis if the hypothesized value for p falls inside the confidence interval. ANSWER:
117.
F.
In testi testing ng the diff differ eren ence ce betwee between n two two pop popul ulat atio ion n mean meanss usin using g two two inde indepen pende dent nt samples samples,, we use the pooled pooled varian variance ce in estima estimatin ting g the standard standard error error of the sampling distribution of the sample mean difference x1 − x 2 if the populations are normal with equal variances. ANSWER:
121.
T
The poole oled-va -variances t -tes - testt requi require ress that that the the two two pop popul ulat atio ion n vari varian ance cess are are different. ANSWER:
120.. 120
F
Tests Tests in whic which h sample sampless are not not indepe independe ndent nt are refe referre rred d to as matc matched hed pairs pairs.. ANSWER:
119 119.
F
When When testing testing the equal equality ity of two popul populati ation on variance variances, s, the test test statist statistic ic is the ratio ratio 2 2 of the population variances; namely σ 1 / σ 2 . ANSWER:
118.
F
T
In conduct conducting ing hypot hypothesi hesiss testing testing for for differ difference ence betwe between en two means means when when samples samples are dependent, the variable under consideration is D ; the sample mean difference between n pairs.
252
Chapter 10 ANSWER:
122.. 122
The number The number of of degre degrees es of fre freed edom om asso associ ciat ated ed with with the the t test, when the data are gathered from a matched pairs experiment with 12 pairs, is 22. ANSWER:
123. 123.
T
F
2 2 Thee test Th est stat statis isti ticc empl employ oyed ed to test est H 0 : σ 1 / σ 2
=1
is F
= s12 / s22 ,
which is F
distributed with n1 − 1 and n2 − 1 degrees of freedom. ANSWER:
124.
T
When When the necess necessary ary condi conditi tions ons are are met, met, a two-tai two-taill test test is being being conduct conducted ed to test test the difference between between two population proportions. proportions. The two sample sample proportions are p = 0.35 and p = 0.42 , and the standard error of the sampling distribution of 1
2
p1 − p2 is 0.054. The calculated value of the test statistic will be 1.2963.
ANSWER:
125.. 125
The equa The equall-var varia ianc nces es test test stat statis isti ticc of µ1 − µ 2 is Student t distributed with n1 + n2 -2 degrees of freedom, provided that the two populations are normally distributed. ANSWER:
126.
F
T
When When the necess necessary ary condi conditi tions ons are are met, met, a two-tai two-taill test test is being being conduct conducted ed at α = 2 2 0.05 to test H 0 : σ 1 / σ 2 = 1 . The two sample sample varian variances ces are s12 = 700 and s22 = 875 , and the sample sizes are n1 = 40 and n2 statistic will be F = 0.80. ANSWER:
127. 127.
T
T
Give Given n the the sign signif ific ican ance ce leve levell 0.01, 0.01, the the F -value -value for the degrees of freedom, d.f. = (6,9) is 7.98. ANSWER:
129.
The calculated value of the test
Stat Statis isti tics cs prac practi titi tion oner erss use use the the anal analy ysis sis of vari varian ance ce (ANO (ANOVA VA)) tech techni niqu quee to compare more than two population means. ANSWER:
128. 128.
= 40 .
F
The analy analysis sis of vari varianc ancee (ANOVA) (ANOVA) techn techniqu iquee analyze analyzess the varian variance ce of the the data to to determine whether differences exist between the population means. ANSWER:
T
253
Hypothesis Testing
130.
The F-tes F-testt of the the analysis analysis of vari varianc ancee requires requires that that the the populati populations ons be norm normall ally y distributed with equal variances. ANSWER:
131.
One-way One-way ANOVA ANOVA is appl applied ied to to four indepen independen dentt sample sampless having having means means 13, 13, 15, 18 and 20, respectively. respectively. If each observation in the forth sample were increased by 30, the value of the F -statistics -statistics would increase by 30. ANSWER:
132.. 132
T
The numbe The numberr of degre degrees es of fre freed edom om for for a contin contingen gency cy tabl tablee with with r rows and c columns is rc rc - 1 , provided that both r and c are greater than or equal to 2. ANSWER:
135.
F
A test test for inde indepe pend nden ence ce is appl applie ied d to a cont contin inge genc ncy y tabl tablee with with 4 rows rows and 4 columns. The degrees of freedom for this chi-square test must equal 9. ANSWER:
134.. 134
F
The degree The degreess of freed freedom om for for the denom denomin inat ator or of a one-way one-way ANOV ANOVA A test test for 4 population means with 10 observations sampled from each population are 40. ANSWER:
133. 133.
T
F
The Lillie Lilliefor forss test test is is used used to to test test for for norm normali ality ty.. ANSWER:
T
254
