Question 1041325
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Given Data Set
{35,28,29,33,32,40,27,26,25,29,28,30,36,33,29,30,28,25}

Using a calculator, we find that
sample mean = xbar = 30.16667
sample standard deviation = s = 4.00367



Hypothesis
H0: mu >= 32
H1: mu < 32
This is a one-tailed test. Specifically a left-tailed test. 


Test statistic:
t = (xbar - mu)/(s/sqrt(n))
t = (30.16667 - 32)/(4.00367/sqrt(18))
t = -1.9427 5763 2601 19
t = -1.94


Now use a TI83 or TI84 calculator to compute the area to left of t = -1.94


Type in <font color=blue>tcdf(-99,-1.94,17)</font> to get the result 0.0344


So the p-value is approximately 0.0344


Side note: because the p value is smaller than alpha = 0.05, this means we reject the null hypothesis and conclude that the mean (mu) is smaller than 32
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