Answer to Question #343947 in Statistics and Probability for Feicity Rodrigo

Question

Mapalad Integrated High School determined students’ Body Mass Index (BMI) at the opening of classes. It has been recorded that the average height of female students is 154.2 centimeters with a standard deviation of 9 centimeters. The researcher conducted her own study and she randomly selected 40 female students. In her study, she got an average of 156.7 centimeters. Is there a reason to believe the claims of the school? Use 5% level of significance in testing the hypothesis.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=154.2$

$H_1:\mu\not=154.2$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{156.7-154.2}{9/\sqrt{40}}\approx1.7568$$

Since it is observed that $|z|=1.7568<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(z>1.7568)=0.078952,$ and since $p= 0.078952>0.05=\alpha,$ it is concluded that the null hypothesis is notrejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 154.2, at the $\alpha = 0.05$ significance level.