Question

One​ year, the mean age of an inmate on death row was 40.3 years. A sociologist wondered whether the mean age of a​ death-row inmate has changed since then. She randomly selects 32 ​death-row inmates and finds that their mean age is 38.5 with a standard deviation of 9.6 Construct a​ 95% confidence interval about the mean age. What does the interval​ imply?

Determine the nature of the hypothesis test. Recall that the mean age of death row inmates is being tested.

Though technology or the​ t-distribution table can be used to find the critical​ value, for the purpose of this​ exercise, use the​ t-distribution table.

Answer 1

Solution :

Given that,

= 38.5

s = 9.6

n = 32

Degrees of freedom = df = n - 1 = 32 - 1 = 31

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,31 = 2.039

Margin of error = E = t/2,df * (s /n)

= 2.039 * (9.6 / 25)

= 3.46

Margin of error = 3.46

The 95% confidence interval estimate of the population mean is,

- E < < + E

38.5 - 3.46 < < 38.5 + 3.46

30.04 < < 41.96

(30.04 , 41.96 )

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :    = 40.3

Ha :     40.3

The information provided, the significance level is α=0.05, and the critical value for a two-tailed test is tc​= 2.04

Test statistic = t

= ( - ) / / n

= (38.5 - 40.3 ) / 9.6 / 32

= −1.061

Test statistic = t = −1.061

P-value =0.2970

= 0.05  

P-value ≥

0.2970 ≥ 0.05

Fail to reject the null hypothesis .

There is insufficient evidence to suggest that