Answer to Question #338017 in Statistics and Probability for KLTino

Question

A rural health unit conducted a survey on the heights of the male aged 18 to 24 years old. It was found out that the mean height of male aged 18 to 24 years old was 70 inches. Test the hypothesis that the mean height of the male aged 18 to 24 years old is not 70 inches if a random sample of 20 male aged 18 to 24 years old had a mean height of 65 inches with a standard deviation of 3. Use 1% level of significance.

Parameter: _______________________________________

Statistic: ________________________________

Null Hypothesis: ____________________________________________________

Alternative hypothesis: _____________________________________________

Answer 1

Parameter: $\mu,$ the mean height of male 

Statistic: $t-$ statistic, two-tailed test

Null Hypothesis: $H_0:\mu=70$

Alternative hypothesis: $H_a:\mu\not=70$

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu=70$

$H_a:\mu\not=70$

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.

2. Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=19$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.860935.$

3. The rejection region for this two-tailed test is $R = \{t: |t| > 2.860935\}.$

4. The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{65-70}{3/\sqrt{20}}\approx-7.45356$$

5. Since it is observed that $|t| =7.45356>2.860935=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=19$ degrees of freedom, $t=-7.45356$ is $p=0,$ and since $p= 0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

6.Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 70, at the $\alpha = 0.01$ significance level.