Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Heights of Supermodels Listed below are the heights (cm) for the simple random sample of female supermodels Lima, Bundchen, Ambrosio, Ebanks, Iman, Rubik, Kurkova, Kerr,Kroes, Swanepoel, Prinsloo, Hosk, Kloss, Robinson, Heatherton, and Refaeli. Use a 0.01 significance level to test the claim that supermodels have heights with a mean that is greater than the mean height of 162 cm for women in the general population. Given that there are only 16 heights represented, can we really conclude that supermodels are taller than the typical woman?

178 177 176 174 175 178 175 178 178 177 180 176 180 178 180 176

Heights of the supermodels are listed below:

178 177 176 174 175 178 175 178 178 177 180 176 180 178 180 176

Claim of the study is to test the mean height of supermodels.

Assume that the distribution is normal and the samples are randomly selected. In this case, the population standard deviation is unknown. Thus, t-distribution would be used.

Sample size (n) of heights of supermodels is 16.

Null hypothesis $\left(H_0\right)$: The mean height of supermodels is equal to than the mean height of 162 cm for women in the general population.

Alternate hypothesis$\left(H_1\right)$: The mean height of supermodels is greater than the mean height of 162 cm for women in the general population.

Mathematically, it can be expressed as,

$$\begin{gathered} H_0:\mu =162 \\ {H}_1:\mu >162 \end{gathered}$$

The test is right-tailed.

Formula for test statistic is given by,

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

Where , $\overline{x}$is the sample mean and s is the standard deviation of observations.

The mean is computed as,

$$\begin{gathered} \overline{x}=\frac{\sum x_i}{n} \\ =\frac{178+177+176+...+176}{16} \\ =177.25 \end{gathered}$$

The sample standard deviation is computed as,

$$\begin{gathered} s=\sqrt{\frac{{\sum \left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(178-177.25\right)}^2+{\left(177-177.25\right)}^2+...+{\left(176-177.25\right)}^2}{16-1}} \\ =1.844 \end{gathered}$$

Substituting values in the test statistic,

$$\begin{gathered} t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}} \\ =\frac{177.25-162}{\frac{1.844}{\sqrt{16}}} \\ =33.082 \end{gathered}$$

Significance level is 0.01.

Sample size is 16 (n).

The degree of freedom is,

$$\begin{gathered} \text{df}=n-1 \\ =16-1 \\ =15 \end{gathered}$$

In the t-distribution table, find the value corresponding to the row value of degree of freedom 15 and column value of area in one tail 0.01 is 2.602 which is critical value${\text{t}}_{\text{0.01}}$.

Thus, the critical value ${\text{t}}_{\text{0.01}}$is 2.602.

The rejection region is $\left(t:t>2.602\right)$.

Test statistic is 33.152 and the critical value is 2.602.

According to this, we can conclude that the test statistic33.152 will fall in rejection region.

Therefore, we reject the null hypothesis.

Thus, it can be concluded that there is sufficient evidence to support the claim that the mean height of supermodels is greater than the mean height of women in general population.