Question

according to nscb statistics, the life expectancy of filipino women is 70.1 years. suppose a random sample of 30 women in town b yields a mean of 72.5 years with a population variance of 4.76 years, would you say that the life expectancy of the women in town b is greater than the average

a. Formulate the null and alternative hypothesis in sentence and in symbol

b. Determine what test statistic to use

c. Compute the test statistic

Answer 1

a.

The null hypothesis $(H_0)$ is: the life expectancy of the women in town b is not greater than the average.

The alternativel hypothesis $(H_1)$ is: the life expectancy of the women in town b is greater than the average.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq 70.1$

$H_1:\mu>70.1$

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 $\alpha=0.05, df=n-1=30-1$

$=29$ degrees of freedom, and the critical value for a right-tailed test is $t_c=1.699127.$

The rejection region for this right-tailed test is $R=\{t:t>1.699127\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{72.5-70.1}{4.76/\sqrt{30}}\approx2.7616$$

Since it is observed that $t=2.7616>1.699127=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than $70.1$, at the $\alpha=0.05$

Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=29, t=2.7616$ is $p=0.004937,$ and since $p=0.004937<0.05=\alpha,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than $70.1$, at the $\alpha=0.05$

Therefore, there is enough evidence to say that the life expectancy of the women in town b is greater than the average, at the $\alpha=0.05$