Question

According to the Centers for Disease Control and Prevention, average life expectancy has risen to 77.6 years. It has been suggested has rural areas in the Southwest have a shorter life span than the general population. In a random sample of 18 recently deceased residents of the rural Southwest, the average life span was 75.9, with a sample standard deviation of 9.8 years.

          a.       State the null and alternative hypotheses to be used to test whether residents in the Southwest have a

          shorter life span than the general population.

b.       If the significance level is a = 5%, can the null hypothesis be rejected?

c.       What is the p-value of the appropriate test statistic (you will need Excel’s =T.DIST(x,df,True) function)?

Answer 1

Given

n = 18 ,   , S =9.8

Level of significance = 0.05

a)

Null and alternative hypothesis is

b)

Here population standard deviation is not known so we use t-test statistic.

Test statistic is

Degrees of freedom = n - 1 = 18-1=17

P-value = 0.235 ( using t table)

P-value , Failed to Reject H0

No,The null hypothesis Failed to Reject

c)

By using EXCEL the p-value= 0.235