Answer:
b. 0.1216
Step-by-step explanation:
Given that a sample of 15 from a normal population yields a sample mean of 43 and a sample standard deviation of 4.7.
We have to check the p value for the claim that mean <45
[tex]H_0: \mu =45\\H_a: \mu <45[/tex]
(Left tailed test for population mean)
Sample size n = 15
Sample mean = 45
Sample std dev s = 4.7
Since sample std deviation is being used, we use t test only
Std error of mean = [tex]\frac{s}{\sqrt{n} } \\=1.214[/tex]
Mean difference = 43-45 = -2
t statistic = mean difference/std error
= -1.176
df = n-1 = 14
p value = 0.1216
