Question

Short answer

There is insufficient evidence to support the claim that the variation of the recent winners is different from the winners of the 1920s and 1930s at a 0.01 level of significance.

Step-by-step solution

Given information

Refer to Exercise 6 for the BMI indexes of the recent winners. The claim states that the variation of the recent winners’ BMI indexes is different from that of the ones in the 1920s and 1930s, which is 1.34.

The sample size is \(n = 10\) with the level of significance \(\alpha = 0.01\).

Hypotheses for the problem

The claim is about the variability of the BMI of the recent years’ Miss America.

The test that is used for this type of problem is the chi-square test of variance.

In the given problem, the test is two-tailed with an equality statement.

\(\begin{array}{l}{H_0}:\sigma = 1.34\\{H_1}:\;\sigma \ne 1.34\end{array}\)

Here, \(\sigma \) is the standard deviation of the population of winners.

State the test statistic

This is the chi-square test for the variance or standard deviation.

The test statistic for this test is

\({\chi ^2}\; = \frac{{\left( {n - 1} \right) \times \;{s^2}}}{{{\sigma ^2}}}\).

Here,

\(n\): sample size,

\({s^2}\): sample variance, and

\({\sigma ^2}\): Hypothesised variance.

Calculate the sample mean and the sample standard deviation

Refer to the data in Exercise 6 to compute the mean and the standard deviation.

Therefore, the sample mean is as follows.

\(\begin{array}{c}\bar x\; = \;\frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\;\\ = \frac{{187.6}}{{10}}\;\\ = 18.76\end{array}\).

The sample mean is\(\bar x = 18.76\).

The standard deviation of the sample is given as follows.

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}} \;\\ = \sqrt {\frac{{{{\left( {19.5 - 18.76} \right)}^2} + {{\left( {20.3 - 18.76} \right)}^2} + ... + {{\left( {16.8 - 18.76} \right)}^2}}}{{10 - 1}}} \;\\ = 1.1862\end{array}\).

The sample standard deviation is \(s = 1.1862\).

Compute the critical value

The degree of freedom for computing the critical value is

\(\begin{array}{c}{\rm{df}} = n - 1\\ = 10 - 1\\ = 9\end{array}\).

It is a two-tailed test.

The left- and right-tailed critical values are obtained as\({\chi ^2}_{\frac{\alpha }{2}},{\chi ^2}_{1 - \frac{\alpha }{2}}\).

The critical values with \(\alpha = 0.01\) and 9 degrees of freedom are obtained from the chi-square table as follows.

\(\begin{array}{c}\chi _R^2 = {\chi ^2}_{\frac{\alpha }{2}}\\ = {\chi ^2}_{\frac{{0.01}}{2}}\\ = {\chi ^2}_{0.005}\\ = 23.589\end{array}\),

and

\(\begin{array}{c}\chi _L^2 = {\chi ^2}_{1 - \frac{\alpha }{2}}\\ = {\chi ^2}_{1 - \frac{{0.01}}{2}}\\ = {\chi ^2}_{0.995}\\ = 1.735\end{array}\).

The critical values are obtained from the chi-square distribution.

This is a two-tailed test. So, the two critical values are\(\chi _L^2 = 1.735\;and\;\chi _U^2 = 23.589\). \(\)

Compute the test statistic

The test statistic for the chi-square test for variance is

\(\begin{array}{c}{\chi ^2}\; = \frac{{\left( {n - 1} \right) \times \;{s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {10 - 1} \right) \times {{\left( {1.186} \right)}^2}}}{{{{\left( {1.34} \right)}^2}}}\\ = 7.053\end{array}\).

The test statistic is \({\chi ^2} = 7.053\).

Test criteria using the critical value

The test criteria for a two-tailed test of the chi-square test for variance are given below.

If\(\chi _L^2 < {\chi _{stat}} < \chi _U^2\) , fail to reject the null hypothesis at an\(\alpha \)level of significance.

If the value does not belong in the range, reject the null hypothesis at an \(\alpha \)level of significance.

Here, the test statistic lies between the critical values.

\(\chi _L^2\; = \;1.753\; < \chi _{stat}^2 = 7.071\; < \chi _U^2 = 23.589\).

Therefore, the null hypothesis is failed to be rejected.

Thus, there is insufficient evidence to support the claim that the recent winners have BMI values different from those of the winners of the 1920s and 1930s.