Question

Short answer

There is sufficient evidence to support the claim that the BMI of the recent winners is significantly smaller than the models in the 1920s and 1930s at a 0.01 level of significance.

Step-by-step solution

Given information

The claim states that the population mean BMI is less than\(20.16\), which was the BMI for the winners from the 1920s and 1930s.

The level of significance to test this claim is 0.01.

The sample size is \(n = 10\)with the observations of the recent winners.

State the hypotheses

The claim does not include the equality statement. Therefore, the claim of the problem becomes the alternate hypothesis.

\(\begin{array}{l}{H_{0\;}}:\;\mu = 20.16\\{H_1}\;:\mu \; < 20.16\end{array}\)

Here, \(\mu \) is the population mean BMI of the recent winners of Miss America.

State the formula for the test statistic

Assume that the samples are randomly selected, and the population follows the normal distribution.

As the population standard deviation is unknown, the t-test applies.

The formula for the test statistic is given below.

\(t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\)

Here,

\(\begin{array}{l}\bar x\;:\;{\rm{sample}}\;{\rm{mean}}\\s\;:\;{\rm{sample}}\;{\rm{stadard}}\;{\rm{deviation}}\\\mu \;:\;{\rm{population}}\;{\rm{mean}}\\n\;:{\rm{sample}}\;{\rm{size}}\end{array}\)

Find the sample mean and the sample standard deviation

The formula for 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.

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}} \;\\ = \sqrt {\frac{{{{\left( {19.5 - 18.76} \right)}^2} + {{\left( {20.3 - 18.76} \right)}^2} + ... + {{\left( {16.8 - 18.76} \right)}^2}}}{{10 - 1}}} \;\\ = \sqrt {\frac{{12.664}}{9}\;\;} \\ = 1.1862\end{array}\).

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

Find the degree of freedom and the critical value

The degree of freedom is computed as follows.

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

For the one-tailed test , at a \(0.01\) level of significance, that is, \(\alpha = 0.01\) with \(9\)degrees of freedom, the critical value will be

\(\begin{array}{c}{t_{{\rm{crit}}}}\; = {t_{\alpha ,df}}\;\\ = {t_{0.01,9}}\;\\ = 2.821\;\end{array}\)

This critical value is observed from the t-table corresponding to row 9 and the level of significance 0.01 (one-tailed).

Compute the test statistic

Substitute all the computed values in the formula.

\(\begin{array}{c}t = \frac{{18.76 - 20.16}}{{\frac{{1.19}}{{\sqrt {10} }}}}\;\\ = \frac{{ - 1.4}}{{0.376}}\;\\ = - 3.723\end{array}\)

The test statistic is \(t = - 3.723\).

State the decision using the critical value

The test criteria for the critical value are given below.

If \(\left| t \right|\; > \;{t_{crit}}\), reject the null hypothesis at the given level of significance; otherwise, fail to reject the null hypothesis.

Here,

\(\begin{array}{c}\left| t \right|\; = \left| { - 3.723} \right|\\ = 3.723\\ > 2.821\left( {{t_{crit}}} \right)\end{array}\)

Thus, the null hypothesis is rejected at the 0.01 level of significance.

Conclusion 

As the null hypothesis is rejected, it can be concluded that there is sufficient evidence to support the claim that the mean body mass index of the recent Miss America is significantly smaller than those in the 1920s and 1930s, which is 20.16.