Question 1190015
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Ho: mu >= 100
Ha: mu < 100
This is a left-tailed test because of the inequality sign in the alternative hypothesis.
The rejection region is to the left of the critical value.


n = 40 = sample size
n > 30 so we can use the Z distribution here


Given info:
mu = 100 is what we assume is the case
xbar = sample mean = 95
s = sample standard deviation = 20


Test Statistic:
z = (xbar - mu)/(s/sqrt(n))
z = (95 - 100)/(20/sqrt(40))
z = -1.58
The value is approximate


Use a Z table or calculator to find that
P(Z < -1.58) = 0.05705
This is the approximate p-value


The p-value is not less than alpha = 0.05, so we fail to reject the null. 
In other words, we don't have enough info to overturn the null. 
The null was that mu >= 100, so either mu = 100 or mu > 100. 
We can interpret this as "The average textbook is $100 or more", or something along those lines.


If you were to use a table or calculator, you should find that the z critical value is roughly -1.645 when considering a left-tailed test.
Effectively,
P(Z < -1.645) = 0.05 approximately


Compare the test statistic (-1.58) and the critical value (-1.645)
The test statistic is not to the left of the critical value, so the test statistic is not in the rejection region. 


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Summary:


Test Statistic: z = -1.58
critical value: z = -1.645
P-value = 0.05705
Decision: Fail to reject the null
Interpretation: The average textbook is $100 or more

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