Answer to Question #333842 in Statistics and Probability for Claire

Question

Test the hypothesis. Show the step-by-step process in testing hypothesis.

1. A researcher claims that the average salary of a private school teacher is greater than P35,000 with a standard deviation of P7,000. A sample of 35 teachers has a mean salary of P37,000. Test the claim of the researcher. At 0.05 level of significance, test the claim of the researcher.

2. A researcher reports that the average salary of a college dean is more than P65,000. A sample of 35 college deans has a mean salary of P67,000. At 0.01 level of significance, test the claim that the college deans earn more than P65,000 a month. The standard deviation of the population is P5,250.

Answer 1

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

$H_0:\mu\le35000$

$H_1:\mu>35000$

This corresponds to a right-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 $\alpha=0.05,$ $df=n-1=35-1=34$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 1.690924.$

The rejection region for this right-tailed test is $R = \{t: t> 1.690924\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{37000-35000}{7000/\sqrt{35}}\approx1.6903$$

Since it is observed that $t =1.6903 <1.690924=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=34$ degrees of freedom, $t=1.6903$ is $p=0.05006,$ and since $p=0.05006>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is greater than 35000, at the $\alpha = 0.05$ significance level.

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

$H_0:\mu\le65000$

$H_1:\mu>65000$

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

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a right-tailed test is $z_c = 2.3263.$

The rejection region for this right-tailed test is $R = \{z: z> 2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{67000-65000}{5250/\sqrt{35}}\approx2.2537$$

Since it is observed that $z =2.2537<2.3263=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $z=2.2537$ is $p=P(Z>2.2537)=0.012108,$ and since $p=0.012108>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is greater than 65000, at the $\alpha = 0.01$ significance level.