BMI for Miss America. A claimed trend of thinner Miss America winners has generated charges that the contest encourages unhealthy diet habits among young women. Listed below are body mass indexes (BMI) for recent Miss America winners. Use a 0.01 significance level to test the claim that recent winners are from a population with a mean BMI less than 20.16, which was the BMI for winners from the 1920s and 1930s. Given that BMI is a measure of the relative amounts of body fat and height, do recent winners appear to be significantly smaller than those from the 1920s and 1930s?

19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8.

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.

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

$$\begin{gathered} H_0:\mu =20.16 \\ {H}_1:\mu <20.16 \end{gathered}$$

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

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.

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Here,

$$\begin{gathered} \overline{x}:\text{sample}\text{mean} \\ s:\text{sample}\text{stadard}\text{deviation} \\ \mu :\text{population}\text{mean} \\ n:\text{sample}\text{size} \end{gathered}$$

The formula for the sample mean is as follows.

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{187.6}{10} \\ =18.76 \end{gathered}$$

The sample mean is$\overline{x}=18.76$.

The standard deviation of the sample is given as follows.

.$$\begin{gathered} s=\sqrt{\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(19.5-18.76\right)}^2+{\left(20.3-18.76\right)}^2+...+{\left(16.8-18.76\right)}^2}{10-1}} \\ =\sqrt{\frac{12.664}{9}} \\ =1.1862 \end{gathered}$$

The sample standard deviation is $s=1.1862$.

The degree of freedom is computed as follows.

$$\begin{gathered} \text{df}=n-1 \\ =10-1 \\ =9 \end{gathered}$$

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

$$\begin{gathered} t_{\text{crit}}=t_{\alpha ,df} \\ =t_{0.01,9} \\ =2.821 \end{gathered}$$

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

Substitute all the computed values in the formula.

$$\begin{gathered} t=\frac{18.76-20.16}{\frac{1.19}{\sqrt{10}}} \\ =\frac{-1.4}{0.376} \\ =-3.723 \end{gathered}$$

The test statistic is $t=-3.723$.

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,

$$\begin{gathered} \left|t\right|=\left|-3.723\right| \\ =3.723 \\ >2.821\left(t_{crit}\right) \end{gathered}$$

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

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.