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The right-hand strength in pounds was measured and data is as follows

29.4  30.8  30.6  31.5  32.1  31.7  30.3  30.8

a. What is P-value for rejecting the null hypotheses that the population mean is equal to 32?

b. Find acceptance region in testing hypotheses in (a). with a significance level (alpha) equal to 0.1. (Assume normality in both)

BruceET
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Pinalen
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    Hi Pinalen. I am worried that you have not made any research effort. For example, questions about meaning of p-value have been asked before. https://math.stackexchange.com/questions/582945 , https://math.stackexchange.com/questions/2738017 – 311411 Jun 13 '21 at 12:35
  • Yes, Sir I am aware of these concepts presented in the link, but this I am not able to figure out how to apply it in this particular question. – Pinalen Jun 13 '21 at 12:40

1 Answers1

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Find a discussion of one sample t tests in your text or class notes. And I suppose you are using some kind of software in your course.

How about looking to see what happens when these (claimed normal) observations are used in a t test in R (or some other statistical program or using a statistical calculator).

x = c(29.4, 30.8, 30.6, 31.5, 32.1, 31.7, 30.3, 30.8)
t.test(x, mu = 32)

    One Sample t-test




data:  x
t = -3.6238, df = 7, p-value = 0.008468
alternative hypothesis: true mean is not equal to 32
95 percent confidence interval:
 30.18222 31.61778
sample estimates:
mean of x 
     30.9

The P-value is given in the output. When the t statistic has DF = 7, the probability of a smaller value than 30.9 is $P(T < -3.6238) = 0.00423.$ But this is a 2-sided test (Why do I say that?), so you have to consider the probability of a value at least as far into the upper tail. So the P-value is $0.00847$ as in the output.

pt(-3.6238, 7)
[1] 0.004233656
2 * pt(-3.6238, 7)
[1] 0.008467311

The critical value $c$ for a test at the $10\%$ level is the probability that $|T| \ge c,$ again with DF = 7. You should be able to use use a printed table of percentage point or quantiles of t distributions to find the value that cut 5% from each tail of that t distribution. Look on row DF = 7 of such a table and see if you can find something close to $c = 1.895.$

qt(.95, 7) 
[1] 1.894579

Notice that you need the actual data (or at least $\bar X$ and $S_X$ in order to find the P-value, but to find $c.$ you need only know that you have $n = 8$ normally distributed observations.

Now match this whole answer with examples in your text or notes.

BruceET
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    Thank you so much Sir !!, I'll take care while posting questions next time in the view of downvotes :) – Pinalen Jun 14 '21 at 04:51
  • Once you get used to how the site works you can learn a lot here, and as time goes on maybe others can learn from you. // Maybe you can find something worthwhile adding to your question, based on what you know now. – BruceET Jun 14 '21 at 05:03