Question

 A sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years,

does this seem to indicate that the average life span today is greater than 70 years? Use a level of significance:

a. 1%

b. 5%

c. 10%

 

Answer 1

a.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 70$

$H_a:\mu>70$

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 $\alpha = 0.01,$ 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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{71.8-70}{8.9/\sqrt{100}}=2.0225$$

Since it is observed that $z = 2.0225<2.3263=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z>2.0225)=0.021562,$ and since $p=0.021562>0.01=\alpha,$ 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 70, at the $\alpha = 0.01$ significance level.

b.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 70$

$H_a:\mu>70$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{71.8-70}{8.9/\sqrt{100}}=2.0225$$

Since it is observed that $z = 2.0225>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>2.0225)=0.021562,$ and since $p=0.021562<0.05=\alpha,$ 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.

c.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 70$

$H_a:\mu>70$

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 $\alpha = 0.10,$ and the critical value for a right-tailed test is $z_c = 1.2816.$

The rejection region for this right-tailed test is $R = \{z: z > 1.2816\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{71.8-70}{8.9/\sqrt{100}}=2.0225$$

Since it is observed that $z = 2.0225>1.2816=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>2.0225)=0.021562,$ and since $p=0.021562<0.10=\alpha,$ 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 $\alpha = 0.10$ significance level.