Answer to Question #341208 in Statistics and Probability for Aina

Question

A random sample of ten measurements were obtained from a normally distributed population with mean u=6.5. The sample values are X-4.2 and s 2.

a. Test the null hypothesis that the mean of the population against the alternative hypothesis, μ = 6.5. Use a = 0.05.

b. Test the null hypothesis that the mean of the population against the alternative hypothesis, u 6.5. Use a = 0.05

Answer 1

a.

The following null and alternative hypotheses need to be tested:

$H_0:\mu=6.5$

$H_1:\mu>6.5$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.2-6.5}{2/\sqrt{10}}\approx-3.6366$$

Since it is observed that $z=-3.6366<1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(z>-3.6366)=0.999862,$ and since $p=0.999862>0.05=\alpha,$ 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 90, at the $\alpha = 0.05$ significance level.

b.

The following null and alternative hypotheses need to be tested:

$H_0:\mu=6.5$

$H_1:\mu\not=6.5$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.2-6.5}{2/\sqrt{10}}\approx-3.6366$$

Since it is observed that $z =- 3.6366 <1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-3.6366)=0.000276,$ and since $p=0.000276<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 6.5, at the $\alpha = 0.05$ significance level.