Answer to Question #348489 in Statistics and Probability for Ellaine

Question

A History teacher claims that the average height of Filipino males is 163 centimeters. A student taking up Statistics

randomly selects 20 Filipino males and measures their heights. Their heights in cm are shown below.

163 164 165 177

167 163 159 169

159 160 163 163

165 169 163 165

162 174 163 167

Do the collected data present sufficient evidence to indicate that the average heights of Filipino male is different

from 163 cm? Use 0.05 level of significance and assume that the population follows a normal distribution.

(Use critical value method)

Answer 1

Sample mean  

$$\bar{x}=\dfrac{1}{20}(163+164+165+177+167$$

 

$$+163+159+169+159+160+163$$

$$+163+165+169+163+165+162$$

$$+174+163+167)=165$$

Sample variance

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=\dfrac{396}{19}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{396}{19}}\approx4.5653$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=163$

$H_1:\mu\not=163$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=19$ and the critical value for a two-tailed test is $t_c =2.093024.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.093024\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{165-163}{\sqrt{\dfrac{396}{19}}/\sqrt{20}}=1.95918$$

Since it is observed that $|t|=1.95918<2.093024=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=19$ degrees of freedom, $t=1.95918$ is $p=0.064931,$ and since $p=0.064931>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is different than 163, at the $\alpha = 0.05$ significance level.