Answer to Question #338810 in Statistics and Probability for ron

Question

A random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years, with a standard deviation of 8.9 years. Does this seem to indicate that the average life span today is greater than 70 years? Use a 0.05 level of significance. The computed value is

Answer 1

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{71.8-70}{8.9/\sqrt{100}}\approx2.0225$$

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.