Question

at a local university a samaple of 49 evening students was selected in order to determine whether the average age of the evening students is significantly differrent from 21. the average age of the students in the sample was 23 with a standard deviation of 3.5 test this hypotheses and make a related business decision. let significance level is 95%

Answer 1

Solution :

Given that,

= 21

= 23

= 3.5

n = 49

The null and alternative hypothesis is ,

H₀ :   = 21

H_a :    21

This is the two tailed test .

Test statistic = z

= ( - ) / / n

= (23 - 21) / 3.5 / 49

= 4.00

Test statistic = 4.00

P-value = 2 * P (Z < 4) = 2 * 1.00 = 2.00

P-value = 2.00

= 1 - 95% = 1 - 0.95 = 0.05

2.00 > 0.05

P-value >

Fail to reject the null hypothesis.

There is not sufficient evidence to test the claim.