Answer to Question #348563 in Statistics and Probability for ken

Question

In a recent survey, a researcher claimed that the average life expectancy of people in a certain country is 72 years. Is his claim correct if a random sample of 25 deaths from his country showed a mean of 71.2 years with standard deviation of 1.2 years? Use 99% confidence level.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=72$

$H_1:\mu\not=72$

This corresponds to a two-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 $\alpha = 0.01,$ $df=n-1=24$ and the critical value for a two-tailed test is $t_c =2.79694.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.79694\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{71.2-72}{1.2/\sqrt{25}}=-3.33333$$

Since it is observed that $|t|=3.33333>2.79694=t_c,$ it i s then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=24$ degrees of freedom, $t=-3.33333$ is $p=0.002776,$ and since $p= 0.002776<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 72, at the $\alpha = 0.01$ significance level.