examen final de ciberseguridadDescripción completa
Factors contributing to the changing trends of language assessment
jhjFull description
UiPath Final Certification DetailsFull description
UiPath Final Certification DetailsDescripción completa
Tshoot Final Exam CcnpDescripción completa
Full description
Arduino Course Final Exam
Full description
Final Examination in Philippine Government and Politics
ExamFull description
STA 3123 Exam #4 Select the correct answer
The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions.
Clergy Medical Doctor Lawyer Total
Extroverted 62 68
Introverted 45 94
56
81
Total
406
A chi-square for association was done using statistical software and the value obtained for the test statistics was Chi-Sq = 8.649 with a P-Value = 0.013
Under the null hypothesis of no association what would the expected number of people who are Clergy and Introverted be? :
1) a.
45
b.
54
c.
58
How many degrees of freedom are associated with the chisquare test statistic? :
2) a.
6
b.
2
c.
12
Based on the above information is there an association between personality type and profession. Test at a 5% level of significance :
3) a.
Yes
b.
No
c.
Cannot tell
A medical researcher conjectures that heavy smoking can result in wrinkled skin around the eyes. The smoking habits as well as the presence of prominent wrinkles around the eyes are recorded for 500 persons. The following frequency table is obtained:
Prominent Wrinkles Heavy smoker Light or no smoker
95 103
Wrinkles not prominent 55 247
Under the null hypothesis of no association what would be the expected number of Heavy Smokers with Prominent Wrinkles :
4) a.
60
b.
95
c.
150
How many degrees of freedom are associated with the chisquare test statistic? :
5) a.
6
b.
1
c.
4
6)
The value of the chi-square test statistic is a.
50.45
b.
500.00
c.
550.45
7)
:
At a 5% level of significance ,the conclusion is a.
Reject the null hypothesis
b.
Do not reject the null hypothesis
c.
Reject the alternative hypothesis
:
A large scale nationwide poll is conducted to determine the public attitude toward the abolition of capital punishment. The percentages in the various response categories are:
Strongly Favor 20%
Favor
Indifferent
Oppose
30%
20%
20%
Strongly Oppose 10%
From a random sample of 100 law enforcement officers in a metropolitan area, the following frequency distribution is observed Strongly Favor 14
Favor
Indifferent
Oppose
18
18
26
Strongly Oppose 24
Do these data provide evidence that the attitude patterns of law enforcement officers differs significantly from the attitude pattern observed in the large scale national poll? Test at a 5 % level of significance
8)
The appropriate statistical procedure is a a.
Test of independence
b.
Test of goodness of fit
c.
Test of homogeneity
9)
The value of the test statistic is a.
28.2
b.
100.0
c.
128.2
:
:
10)
We reject the null hypothesis if the test statistic is a.
Smaller than 9.49
b.
Greater than 9.49
c.
Greater than 28.2
:
The non-parametric alternative to the independent sample t-test is :
11) a.
Spearman Rank Correlation
b.
Sign Test
c.
Rank Sum Test
The non-parametric alternative to the paired sample t-test is :
12) a.
Sign Test
b.
Rank Sum Test
c.
Kruskall Wallis Test
Kruskall Wallis Test is the non-parametric alternative to :
13) a.
Independent Sample T Test
b.
Rank Sum Test
c.
One Way ANOVA
14)
A positive Spearman correlation indicates a a.
Linear relationship
b.
A monotone increasing relationship
c.
A monotone decreasing relationship
:
A sociologist studying New York City wants to determine if there is a difference in the distribution of income for immigrants from four different countries during their first year in the city. For this purpose he will use a Kruskall Wallis Test. Below is the data collected from a random sample of immigrants from these countries (incomes in thousands of dollars)as well as some partial computer output for this analysis: Country I
Country II
Country III
Country IV
12.7
8.3
20.3
17.2
9.2
17.2
16.6
8.8
10.9
19.1
22.7
14.7
8.9
10.3
25.2
21.3
19.9
19.8
16.4
Kruskal-Wallis Test: Income versus Country Kruskal-Wallis Test on Income Country Country Country Country Country Overall
I II III IV
N 5 4 5 5 19
Median 10.90 13.75 20.30 17.20
Ave Rank 5.8 7.6 15.6 10.5 10.0
Z -1.94 -0.95 2.59 0.23
H = 8.49
15)
The null hypothesis is
:
a.
The four population mean incomes are the same
b.
The four distributions of income are the same
c.
The four distributions of income are not the same
The degrees of freedom associated with the test statistic are :
16) a.
19
b.
3
c.
4
For a 5% level of significance , the chi-square critical value is :
17) a.
7.81
b.
9.49
c.
8.49
18)
The conclusion is
:
a.
Reject the null hypothesis
b.
Do not reject the null hypothesis
c.
Reject the alternative hypothesis
The following table lists the price and number of pages for each of 15 books that were reviewed in the February 1982 issue of the journal Technometrics. Also summary statistics, sample correlation and a scatterplot of the data are shown: Price (in Pages dollars) 302
30
425
24
526
35
532
42
145
25
556
27
426
64
359
59
465
55
246
25
143
15
557
29
372
30
320
25
178
26
Scatterplot of Price (in dollars) vs Pages 70
60
)
50 s r lal o d
40 ni ( e ci r P
30
20
10 100
200
300
400
500
600
Pages
Pearson correlation of Pages and Price (in dollars) = 0.401
A researcher wants to know if there is some statistically significant correlation between Price and number of pages. The test statistic value for this test is :
19) a.
0.401
b.
1.578
c.
13
20)
The alternative hypothesis for the above test is: a.
Population Means are different
b.
Correlation is not equal to zero
c.
There is no association
If the p-value for the above test is 0.139 , at a significance level of 5% our conclusion would be that there is:
21) a.
A positive correlation
b.
A negative correlation
c.
No correlation
Below is a scatterplot and some partial output for a simple linear regression model that relates sepal width (X) to sepal length (Y) for the iris setosa flower, based on data collected by the famous statistician Sir Ronald Fisher. Measurements are in centimeters:
Scatterplot of Y vs X 6.0
5.5
Y
5.0
4.5
2.0
2.5
3.0
3.5
4.0
X
Predictor Constant X
Coef 2.7330 0.66111
S = 0.258174
22)
SE Coef 0.3429 0.09903
T 7.97 6.68
P 0.000 0.000
R-Sq = 0.492
Sepal width and sepal length are: a.
Negatively correlated
b.
Positively Correlated
c.
Uncorrelated
4.5
If the sepal width increases by one centimeter , the sepal length ,on average, increases by :
23) a.
2.73 cms
b.
0.66 cms
c.
6.66 cms
What would be the predicted value for the sepal length of a flower whose sepal width is 3.2 cms :
24) a.
2.73 cms
b.
0.66 cms
c.
4.85 cms
The percentage of variation in sepal length explained by sepal width is :
25) a.
49.2%
b.
70.1%
c.
25.8%
The correlation coefficient between sepal width and sepal length is :
