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Nominal Data. In Exercises 9鈥12, use the sign test for the claim involving nominal data.

Births A random sample of 860 births in New York State included 426 boys and 434 girls. Use a 0.05 significance level to test the claim that when babies are born, boys and girls are equally likely.

Short Answer

Expert verified

There is not enough evidence to conclude that boys and girls are not equally likely to be born.

Step by step solution

01

Given information

The number of male births is equal to 426, and the number of female births is equal to 434.

The researcher wants to test that the boys and girls are equally likely to be born at the significance level of 0.05.

02

Frame the statistical hypothesis

The sign test is the non-parametric test used to test the claim of difference between the proportions of male and female births.

Let p be the proportion of male births.

Considering that the proportions of male and female births should be the same, the null hypothesis is as follows:

\({H_0}:p = 0.5\)

Boys and girls are equally likely to be born.

The alternative hypothesis is as follows:

\({H_0}:p \ne 0.5\)

Boys and girls are not equally likely to be born.

The test is two-tailed.

03

Define the sign of the two categories

A negative sign denotes the male births.

A positive sign denotes the female births.

The number of negative signs =426.

The number of positive signs =434.

The sample size (n) is equal to 860.

04

Define test statistic

Let x be the number of times the less frequent sign occurs.

The less frequent sign is the negative sign corresponding to the number of male births.

The value of xis equal to 426.

As the sample size n is greater than 25, the value of z is calculated.


The test statistic z is calculated as shown:

\(\begin{array}{c}z = \frac{{\left( {x + 0.5} \right) - \frac{n}{2}}}{{\frac{{\sqrt n }}{2}}}\\ = \frac{{\left( {426 + 0.5} \right) - \frac{{860}}{2}}}{{\frac{{\sqrt {860} }}{2}}}\\ = - 0.24\end{array}\)

05

Determine the result and the conclusion of the test

Critical value:

The critical value of z from the standard normal table for a two-tailed test with a value of\(\alpha \)= 0.05 is equal to\( \pm 1.96\).

Moreover, the absolute value of z equal to 0.24 is less than the critical value; the null hypothesis is failed to reject.

P-value:

The corresponding p-value for z-score equal to -0.24 and\(\alpha \)equal to 0.05 from the table is equal to 0.4052.

Since it is a two-tailed test, the p-value becomes as follows:

\(\begin{array}{c}p{\rm{ - value}} = 2 \times 0.4052\\ = 0.8104\end{array}\)

As the p-value equal to 0.8104 is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that boys and girls are equally likely to be born.

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Sign Test vs. Wilcoxon Signed-Ranks Test Using the data in Exercise 1, we can test for no difference between body temperatures at 8 AM and 12 AM by using the sign test or the Wilcoxon signed-ranks test. In what sense does the Wilcoxon signed-ranks test incorporate and use more information than the sign test?

Using the Kruskal-Wallis Test. In Exercises 5鈥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?

Wilcoxon Signed-Ranks Test for Body Temperatures The table below lists body temperatures of seven subjects at 8 AM and at 12 AM (from Data Set 3 鈥淏ody Temperatures in Appendix B). The data are matched pairs because each pair of temperatures is measured from the same person. Assume that we plan to use the Wilcoxon signed-ranks test to test the claim of no difference between body temperatures at 8 AM and 12 AM.

a. What requirements must be satisfied for this test?

b. Is there any requirement that the samples must be from populations having a normal distribution or any other specific distribution?

c. In what sense is this sign test a 鈥渄istribution-free test鈥?

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