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Chapter 13: Q2BSC (page 597)

Identifying Signs For the sign test described in Exercise 1, identify the number of positive signs, the number of negative signs, the number of ties, the sample size n that is used for the sign test, and the value of the test statistic.

Short Answer

Expert verified
  • The number of positive signs is equal to 2.
  • The number of negative signs is equal to 7.
  • The number of ties is equal to 1.
  • The sample size is equal to 10.
  • The value of the test statistic is equal to 2.

Step by step solution

01

Given information

Two samples are given showing the weights of students (kgs) in September and April.

02

Describe the sign test

The sign test is a type of non-parametric test that can is used to test the claim of no difference in the values of the weights between September and April.

The table of weights of the two samples is given below.

September weight (kg)

67

53

64

74

67

70

55

74

62

57

April weight (kg)

66

52

68

77

67

71

60

82

65

58

03

Explain the sign of difference

If the value of the September weight is greater than that of the April weight, the sign of difference is positive.

If the value of the September weight is less than that of the April weight, the sign of difference is negative.

If the value of the September weight is equal to that of the April weight, the difference is zero.

The signs of the differences between September weights and April weights are shown below.

September weight (kg)

67

53

64

74

67

70

55

74

62

57

April weight (kg)

66

52

68

77

67

71

60

82

65

58

Sign of difference

+

+

-

-

0

-

-

-

-

-

The number of positive signs is equal to 2.

The number of negative signs is equal to 7.

The number of ties (equal values of weights) is equal to 1.

04

Find the sample size

The number of observations in each sample is equal to 10.

Therefore, the sample size (n) is equal to 10.

05

Find the test statistic

As n is less than 25, the test statistic is denoted by x, and it is the number of times the less frequent sign occurs.

Here, the less frequent sign is the positive sign.

The number of times the positive sign occurs is equal to 2.

Thus, the value of the test statistic is equal to 2.

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Most popular questions from this chapter

Using the Mann-Whitney U Test The Mann-Whitney U test is equivalent to the Wilcoxon rank-sum test for independent samples in the sense that they both apply to the same situations and always lead to the same conclusions. In the Mann-Whitney U test we calculate

\(z = \frac{{U - \frac{{{n_1}{n_2}}}{2}}}{{\sqrt {\frac{{{n_1}{n_2}\left( {{n_1} + {n_2} + 1} \right)}}{{12}}} }}\)

Where

\(U = {n_1}{n_2} + \frac{{{n_1}\left( {{n_1} + 1} \right)}}{2} - R\)

and R is the sum of the ranks for Sample 1. Use the student course evaluation ratings in Table 13-5 on page 621 to find the z test statistic for the Mann-Whitney U test. Compare this value to the z test statistic found using the Wilcoxon rank-sum test.

Notation What do r, \({r_s}\) , \(\rho \), and \({\rho _s}\) denote? Why is the subscript s used? Does the subscript s represent the same standard deviation s introduced in Section 3-2?

Efficiency of the Wilcoxon Signed-Ranks Test Refer to Table 13-2 on page 600 and identify the efficiency of the Wilcoxon signed-ranks test. What does that value tell us about the test?

Using the Kruskal-Wallis Test. In Exercises 5–8, use the Kruskal-Wallis test.

Car Crash Measurements Use the following listed chest deceleration measurements (in g, where g is the force of gravity) from samples of small, midsize, and large cars. (These values are from Data Set 19 “Car Crash Tests” in Appendix B.) Use a 0.05 significance level to test the claim that the different size categories have the same median chest deceleration in the standard crash test. Do the data suggest that larger cars are safer?

Small

44

39

37

54

39

44

42

Midsize

36

53

43

42

52

49

41

Large

32

45

41

38

37

38

33

Using the Kruskal-Wallis Test. In Exercises 5–8, use the Kruskal-Wallis test.

Correcting the H Test Statistic for Ties In using the Kruskal-Wallis test, there is a correction factor that should be applied whenever there are many ties: Divide H by

\(1 - \frac{{\sum T }}{{{N^3} - N}}\)

First combine all of the sample data into one list, and then, in that combined list, identify the different groups of sample values that are tied. For each individual group of tied observations, identify the number of sample values that are tied and designate that number as t, then calculate\(T = {t^3} - t\). Next, add the T values to get\(\sum T \). The value of N is the total number of observations in all samples combined. Use this procedure to find the corrected value of H for Example 1 in this section on page 628. Does the corrected value of H differ substantially from the value found in Example 1?
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