Matched Pairs.In Exercises 5–8, use the sign test for the data consisting of matched pairs.

Speed Dating: Attributes Listed below are “attribute” ratings made by couples participating in a speed dating session. Each attribute rating is the sum of the ratings of five attributes (sincerity, intelligence, fun, ambition, shared interests). The listed ratings are from Data Set 18 “Speed Dating” in Appendix B. Use a 0.05 significance level to test the claim that there is a difference between female attribute ratings and male attribute ratings.

Rating of Male by Female

29

38

36

37

30

34

35

23

43

Rating of Female by Male

36

34

34

33

31

17

31

30

42

Short Answer

Expert verified

There is not enough evidence to conclude that there is a difference in the female attribute ratings and the male attribute ratings.

Step by step solution

01

Given information

The two samples of data are given as the ratings of males and females.

02

Define sign test and frame the statistical hypothesis

The sign test is the non-parametric test used to test the difference between male and female attribute ratings.

The null hypothesis is as follows:

There is no difference between male and female attribute ratings.

The alternative hypothesis is as follows:

There is a difference between male and female attribute ratings.

The test is two-tailed.

03

Define sign of difference

If the male rating value is greater than the female rating, the sign of difference is positive.

If the male rating value is less than the female rating, the sign of difference is negative.

If the value of the Male rating is equal to the Female rating, the difference is zero.

The sign of the difference table is as follows:

Rating of Male by Female

29

38

36

37

30

34

35

23

43

Rating of Female by Male

36

34

34

33

31

17

31

30

42

Sign of Difference

-

+

+

+

-

+

+

-

+

The number of positive signs is equal to 6.

The number of negative signs is equal to 3.

The sample size (n) is equal to 9.

Upon observing the number of positive and negative signs, there seems to be a difference in the occurrence of signs, and thus, the observation does not contradict the alternative hypothesis.

04

Calculate test statistic and determine the result of the test

Since n is less than 25, the test statistic value (x) is the number of times the less frequent sign occurs.

The less frequent sign is the negative sign, and it occurs three times.

Thus, x= 3.

Referring to Table A-7, the critical value of x for n = 9 and\(\alpha \)=0.05 for a two-tailed test equals 1.

If the value of the test statistic is greater than the critical value, the null hypothesis is failed to reject.

If the test statistic value is less than or equal to the critical value, the null hypothesis is rejected.

As the calculated value of x equal to 3 is greater than the critical value of 1, the null hypothesis is failed to reject.

05

Determine the conclusion of the test

There is not enough evidence to conclude that there is a difference in the female attribute ratings and the male attribute ratings.

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Radiation in Baby Teeth Listed below are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from Pennsylvania residents and New York residents born after 1979 (based on data from “An Unexpected Rise in Strontium-90 in U.S. Deciduous Teeth in the 1990s,” by Mangano et al., Science of the Total Environment). Use a 0.05 significance level to test the claim that the median amount of strontium-90 from Pennsylvania residents is the same as the median from New York residents.

Pennsylvania

155

142

149

130

151

163

151

142

156

133

138

161

New York

133

140

142

131

134

129

128

140

140

140

137

143

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