Answer to Question #339498 in Statistics and Probability for rai

Question

A random sample of 11 observations was taken from normal population. The sample mean and

standard deviation are 74.5 and 9 accordingly. Can we infer at 5% significance level that the

population mean is greater than 70?

5. Repeat number 4 with assuming the population standard deviation = 9

Answer 1

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

$H_0:\mu\le70$

$H_a:\mu>70$

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 $\alpha = 0.05,$ $df=n-1=10$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 1.812461.$

The rejection region for this right-tailed test is $R = \{t: t> 1.812461\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{74.5-70}{9/\sqrt{11}}\approx1.6583$$

Since it is observed that $t=1.6583< 1.812461=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=10$ degrees of freedom, $t=1.6583$ is $p=0.064124,$ and since $p=0.064124>0.05=\alpha,$ 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 70, at the $\alpha = 0.05$ significance level.

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

$H_0:\mu\le70$

$H_a:\mu>70$

This corresponds to a right-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 right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z> 1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{74.5-70}{9/\sqrt{11}}\approx1.6583$$

Since it is observed that $z=1.6583> 1.6449=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed is $p=P(Z>1.6583)=0.048628,$ and since $p=0.048628<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 70, at the $\alpha = 0.05$ significance level.