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As the title says, this is a review question. However, I'm having trouble solving it, particularly at the final part when I reject/accept the null hypothesis.

The question is: Twenty years ago, entering male high school students of Central High could do an average of 24 pushups in 60 seconds. To see whether this remains true today, a random sample of 36 freshmen was chosen. If their average was 22.5 with a sample standard deviation of 3.1, can we conclude that the mean is no longer equal to 24? Use the 5 percent level of significance.

Ho: mu-not = 24

H1: mu != 24

n = 36 students

sample std dev = 3.1

sample mean = 22.5

95% Confidence Interval

Because this question involves a sample standard deviation, I need to use the t-distribution.

(24 - 22.5 ) / ( 3.1 / sqrt(6) ) = 2.9

Degrees of freedom = 35

It's at this point that I'm not sure how to proceed. Looking up the value on the t-dist table, since the interval is double-sided, gives me a value of roughly 1.96. However, I'm not sure if this is correct and what to do with it compared to the value 2.9.

I was given the answer to this question:

2*P(T-35 > 2.903) = .0064

However, I don't know where .0064 is from and the answer sheet doesn't go into detail. Can someone please give me an explanation?

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First, you want $|T| = \frac{|22.4 - 24|}{3.1/\sqrt{36}} = 2.903.$ (You seem to have taken the square root twice. Also, the usual expression in the numerator of $T$ is $\bar X - \mu_0$, and you have the sample mean and the hypothetical population mean in the reverse order. And you need to take the absolute value for a two-sided test.)

Then the computation you show is to find the probability P-value, which is the sum of the areas under the density of T(35) to the left of -2.903 and to the right of 2.903: the probability of a 'more extreme result' than you observed.

You can't find exact P-values from printed tables of the t distribution. You'll need a statistical calculator or a statistical computer package. I used R as follows:

 pt(-2.903,35)*2   # 'pt' is the CDF
 ##  0.006358315

Because the P-value is smaller than 5% = 0.05, you reject the null hypothesis at the 5% level and conclude that the population mean of the students you tested is not 24.

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  • $\begingroup$ Sorry about that, made a typo. I did 24 - 22.5 because I realized that the value was going to be positive anyway. Thanks for the answer, I understand where that value came from (the printed t-dist tables didn't have much for it). $\endgroup$ Jun 8 '15 at 6:43

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