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I am working on a homework problem from the book Applied Linear Statistical Models , specifically problem 2.7b (Page 114 of the linked pdf)

2.7. Refer to Plastic hardness Problem 1.22.

b) The plastic manufacturer has stated that the mean hardness should increase by 2 Brinell units per hour. Conduct a two-sided test to decide whether this standard is being satisfied; use $α = 0.1$. State the alternatives, decision rule, and conclusion. What is the P-value of the test?

I have gone through most of the problem, but am stuck trying to calculate the p-value of the test. I have found that $t^* = 0.332$ and $t(0.95, 14) = 2.145$ in the first half of the problem. I have been following along from an example in the book that is identical to the problem. They "find" the p-value by the logic

Example on page 71 of pdf

In the red box in the image above I can "construct" the same p-value $P\{t(14) >= t^* = 0.332\}$ and also consult table B.2 (on page 1335 of the linked pdf).

Table B2

But I have no idea where the example found their p-value of 0.0005 and thus, how I can find mine. I have consulted a website which purports to calculate p-values from t-test statistics and found mine to be 0.744. Though I would like to properly calculate it.

I have looked around chapter 2 but cannot find any more information on calculating a p-value from t-test statistics. How is this p-value calculated?

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You need to look on the next page, which shows a row looking smoething like this (mine generated in R):

    nu |  .98    .985  .99    .9925  .995    .9975   .9995 
    14 |  2.264  2.415  2.624  2.771  2.977   3.326   4.140 

The t-statistic is 3.57 which lies between the .9975 and .9995 quantiles. The two-tailed p-value is the area below -3.57 and above 3.57. This is smaller than the area below -3.326 and above 3.326 (which is 0.0025 + 0.0025 = 0.005); similarly we can bound it below by 2(1-0.9995) = 0.001.

You can safely say $0.001<p<0.005$.

Via R, the correct p-value is about 0.0031

If you use linear interpolation in the table you get an estimate of about 0.0038 (not so accurate, but it's possible to do better than that if it's necessary)

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