In: Statistics and Probability
Ten years ago, the mean age of death row inmates was 50.4 according to the US Dept. of Justice. A sociologist wants to test whether that data is still accurate. He randomly selects 9 death row inmates and finds that their mean age is 34.2 with a standard deviation of 19.1. Use a 0.01 significance level to test the claim that the mean age of death row inmates is now equal to 50.4, which was the mean age ten years ago.
The test statistic is: (to 3 decimals)
The Critical Value is: ± (to 3 decimals)
Solution :
= 50.4
=34.2
S = 19.1
n = 9
This is the two tailed test .
The null and alternative hypothesis is ,
H0 : = 50.4
Ha : 50.4
Test statistic = t
= ( - ) / S / n
= (34.2 -50.4) / 19.1 / 9
= −2.545
Test statistic = t = −2.545
The critical value = 3.355
P-value =0.0345
= 0.01
P-value >
0.0345 > 0.01
Fail to reject the null hypothesis .
There is not sufficient evidence to claim that the population mean μ is different than 50.4, at the 0.01 significance level.