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

Speed Dating: Attractiveness Listed below are “attractiveness” ratings (1 = not attractive; 10 = extremely attractive) made by couples participating in a speed dating session. The listed ratings are from Data Set 18 “Speed Dating”. Use a 0.05 significance level to test the claim that there is a difference between female attractiveness ratings and male attractiveness ratings.

Rating of Male by Female

4

8

7

7

6

8

6

4

2

5

9.5

7

Rating of Female by Male

6

8

7

9

5

7

5

4

6

8

6

5

Short Answer

Expert verified

There is not enough evidence to conclude that there is a difference between female and male attractiveness ratings.

Step by step solution

01

Given information

The two samples of data are given ratings of male attractiveness and female attractiveness.

02

Define sign test and frame the statistical hypothesis

The sign test belongs to the non-parametric (or distribution-free) category used to test the claim of difference between male and female attractiveness ratings.

Null hypothesis: There is no difference between male and female attractiveness ratings.

Alternative hypothesis: There is a difference between male and female attractiveness 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

4

8

7

7

6

8

6

4

2

5

9.5

7

Rating of Female by Male

6

8

7

9

5

7

5

4

6

8

6

5

Sign of Difference

-

0

0

-

+

+

+

0

-

-

+

+

The number of positive signs is equal to 5.

The number of negative signs is equal to 4.

The number of ties is equal to 3.

The sample size (n) is equal to 12.

04

Calculate the test statistic

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 four times.

Thus, x = 4.

05

Determine the result and conclusion of the test

From Table A-7, the critical value of x for n=12 and\(\alpha \)=0.05 for a two-tailed test equals 2.

Since the calculated value of x equal to 2 is greater than the critical value of 1, the null hypothesis is failed to reject.

There is not enough evidence to conclude that there is a difference between female and male attractiveness ratings.

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Most popular questions from this chapter

Blanking Out on Tests In a study of students blanking out on tests, the arrangement of test items was studied for its effect on anxiety. The following scores are measures of “debilitating test anxiety” (based on data from “Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance,” by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two samples are from populations with different medians? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Use a 0.01 significance level.

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7.1

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21.06

28.89

28.71

31.73

30.02

21.96

25.49

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30.29

30.72

Questions Arranged From Difficult To Easy

33.62

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35.91

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In Exercises 1–4, use the following sequence of political party affiliations of recent presidents of the United States, where R represents Republican and D represents Democrat.

R R RR D R D R RR D R RR D D R D D R R D R R D R D

Testing for Bias Can the runs test be used to show the proportion of Republicans is significantly greater than the proportion of Democrats?

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

Arsenic in Rice Listed below are amounts of arsenic in samples of brown rice from three different states. The amounts are in micrograms of arsenic and all samples have the same serving size. The data are from the Food and Drug Administration. Use a 0.01 significance level to test the claim that the three samples are from populations with the same median.

Arkansas

4.8

4.9

5

5.4

5.4

5.4

5.6

5.6

5.6

5.9

6

6.1

California

1.5

3.7

4

4.5

4.9

5.1

5.3

5.4

5.4

5.5

5.6

5.6

Texas

5.6

5.8

6.6

6.9

6.9

6.9

7.1

7.3

7.5

7.6

7.7

7.7

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Height 180.3 175.3 184.8 177.8 182.3 185.4 180.3 175.3 177.8 185.4

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