# Statistical Analysis II

##### Question 1 of 20
5.0 Points
The __________ test is useful for before/after experiments.

A. goodness-of-fit
B. sign
C. median
D. chi-square
##### Question 2 of 20
5.0 Points
The __________ test is useful for drawing conclusions about data using nominal level of measurement.

A. goodness-of-fit
B. sign
C. median
D. chi-square
##### Question 3 of 20
5.0 Points
In an experiment, a sample size of 10 is drawn, and a hypothesis test is set up to determine: H
0 : p = 0.50; H
1:p < or = 0.50; for a significance level of .10, the decision rule is as follows:

A. Reject H0 if the number of successes is 2 or less.
B. Reject H0 if the number of successes is 8 or more.
C. Reject H0 if the number of successes is three or less.
D. Reject H0 if the number of successes is less than 2 or more than 8.
##### Question 4 of 20
5.0 Points
For a “before and after” test, 16 of a sample of 25 people improved their scores on a test after receiving computer-based instruction. For H
0 : p = 0.50; H
1:p is not equal to 0.50; and a significance level of .05:

A. z = 1.2, fail to reject the null hypothesis.
B. z = 1.4, reject the null hypothesis.
C. z = 1.4, fail to reject the null hypothesis.
D. z = 1.64, reject the null hypothesis.
##### Question 5 of 20
5.0 Points
A sample group was surveyed to determine which of two brands of soap was preferred. H
0 :p = 0.50; H
1: p is not equal to 0.50. Thirty-eight of 60 people indicated a preference. At the .05 level of significance, we can conclude that:

A. z = 0.75, fail to reject H0.
B. z = 1.94, fail to reject H0.
C. z = 1.94, reject H0.
D. z = 2.19, reject H0.
##### Question 6 of 20
5.0 Points
The performance of students on a test resulted in a mean score of 25. A new test is instituted and the instructor believes the mean score is now lower. A random sample of 64 students resulted in 40 scores below 25. At a significance level of Î± = .05:

A. H0 : p = 0.50; H1:p < 0.50.
B. H0 : p = 0.50; H1:p > 0.50.
C. H0 : p = 25; H1:p > 25.
D. H0 : p = 25; H1:p < 25.
##### Question 7 of 20
5.0 Points
From the information presented in question #6:

A. z = 3.75, we can reject the null hypothesis.
B. z = 1.875, we fail to reject the null hypothesis.
C. z = -1.625, we fail to reject the null hypothesis.
D. z = -1.875, we can reject the null hypothesis.
##### Question 8 of 20
5.0 Points
A golf club manufacturer claims that the median length of a drive using its driver is 250 yards. A consumer group disputes the claim, indicating that the median will be considerably less. A sample of 500 drives is measured; of these 220 were above 250 yards, and none was exactly 250 yards. The null and alternate hypotheses are:

A. Ho: 0 = 250; H1: 0 < 250.
B. Ho: median = 250; H1: median > 250.
C. Ho: 0 > 250; H1: 0 < 250.
D. Ho: median = 250; H1: median < 250.
##### Question 9 of 20
5.0 Points
From the information presented in question #8, using a level of significance = .05:

A. z = -1.74; we should fail to reject the null hypothesis.
B. z = 2.64; we should fail to reject the null hypothesis.
C. z = -2.72; we should fail to reject the null hypothesis.
D. z = -3.17; we should reject the null hypothesis.
##### Question 10 of 20
5.0 Points
The Wilcoxon rank-sum test:

A. is a nonparametric test for which the assumption of normality is not required.
B. is used to determine if two independent samples came from equal populations.
C. requires that the two populations under consideration have equal variances.
D. Both A and B
##### Question 11 of 20
5.0 Points
A nonparametric test which can evaluate ordinal-scale data of a non-normal population is called the:

A. Wilcoxon signed rank test.
B. Kruskal-Wallis test.
C. sign test.
D. median test.
##### Question 12 of 20
5.0 Points
A researcher wishes to test the differences between pairs of observations with a non-normal distribution. She should apply the:

A. Wilcoxon signed rank test.
B. Kruskal-Wallis test.
C. Wilcoxon rank-sum test.
D. t test.
##### Question 13 of 20
5.0 Points
The data below indicate the rankings of a set of employees according to class theory and on-the-job practice evaluations:

 Theory 1 7 2 10 4 8 5 3 6 9 Practice 2 8 1 7 3 9 6 5 4 10

What is the Spearman correlation of coefficient for the data?

A. -0.0606

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