Question 1196778
<font color=black size=3>
Given info:
xbar = 137
s = 14.2
n = 20
alpha = 0.1


Null Hypothesis: mu = 132
Alternative Hypothesis: mu =/= 132
This is a two-tailed test because of the "not equal" sign in the alternative hypothesis.


Test Statistic:
t = (xbar - mu)/(s/sqrt(n))
t = (137 - 132)/(14.2/sqrt(20))
t = 1.57469575880267
t = 1.57


Use technology to compute the p-value. There are many methods that it would take a while to go over them all, but here's one free online calculator you can use
<a href = "https://www.statology.org/t-score-p-value-calculator/">https://www.statology.org/t-score-p-value-calculator/</a>
Type in 1.57 as the t score and 19 as the degrees of freedom (since df = n-1 = 20-1 = 19)
We're doing a two-tailed test.
Don't worry about the significance level since all we care about is the p-value.


You should get a p-value of roughly 0.1329 from this two-tailed test.
I'm curious to know where your teacher is getting 0.1318 from, but this might be from a reference table of some sort.


Rule: if p-value is smaller than alpha, then reject the null. 


Since the p-value is not smaller than alpha, this means we fail to reject the null.
There isn't enough evidence to overturn it. 
We don't have enough evidence to show the mean is significantly different from 132.
So for now we conclude that mu = 132 is the case.


<font color=red>Answer: Choice B</font>
</font>