Step-by-step explanation:
n = 52
x bar = 122.31
standard deviation sd = 38.75
1. the hypothesis:
null hypothesis
h0: μ ≤ 35
alternative hypothesis
h1: μ > 35
2.
we have been given the sd of the sample but not that of the population. so what we are supposed to use here is the t test and not the z test. the following conditions have to be met.
- population has to be normal
- sample size has to be more than 30. we have sample size = 52 in this question.
3.
=
= 87.31/5.37
= 16.26
using the T distribution function in excel, the p value was calculated and found to be approximately equal to 0.
TDIST(16.26, 51, 1)
since p value is very small we reject the null and accept the alternate hypothesis.
4. from the result above the answer to this question is yes
