Question

A random sample of 100 recorded deaths in the country in 2011 showed an average lifespan of 71.8 years with a standard deviation of 8.9 years. Does this seem to indicate that the average lifespan today is greater than 70 years? Use 0.05 as level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq 70$

$H_1:\mu>70$

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 $\alpha = 0.05,$

$df=n-1=100-1=99$ degrees of freedom,and the critical value for a right-tailed test is $t_c = 1.660391.$

The rejection region for this right-tailed test is $R = \{t: t > 1.660391\}.$

The t-statistic is computed as follows:

$$t=\dfrac{x-\mu}{s/\sqrt{n}}=\dfrac{71.8-70}{8.9/\sqrt{100}}\approx2.022472$$

Since it is observed that $t = 2.022472 >1.660391= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for $df=99$ degrees of freedom, $t=2.022472$, right-tailed from Student T-Value Calculator is the Significance Level $p = 0.022912,$ and since $p = 0.022912<0.05=\alpha,$ 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.