Hypothesis testing (parameters)

Question

I am new to statistics, but familiar with basic concepts. The lecture that confuses me is Hypothesis testing. Although the idea behind it seems ok, I couldn't understand the first exercise I found.

Exercise:

Average time of execution of computer program is 45s. We want to buy a better PC so we tested that program on 30 other PCs and got average time of 44.5s and standard deviation 2. Based on these assumptions, should we buy a new PC?

Result from the book:

$$H_0:m=45$$ $$H_1:m<45$$ $$n=30$$ $$\overline{X_n}=44.5$$ $$\overline{S_n}=2$$ $$T_n=\frac{\overline{X_n}-m_0}{\sqrt{\overline{S_n^2}}}=\frac{44.5-45}{2}=\sqrt{30}=-1.3639$$ $$W=\{T_n<c\}$$ $$p=P\{T_n<\overline{T_n}\}=0.0907$$ p is greater then alpha so we accept hypothesis null hypothesis

My understanding: We want to prove that we shouldn't buy new PC because it runs as long on the others average, and alternative hypothesis says that we should buy a new one. So I don't understand what the last three rows, from Tn to p. Can you please tell me what is the idea behind this?

Answer 1

Comments:

(1) I believe there is something wrong with the following equation in your Question.

$$T_n=\frac{\overline{X_n}-m_0}{\sqrt{\overline{S_n^2}}}=\frac{44.5-45}{2}=\sqrt{30}=-1.3639.$$

You say that the 'standard deviation' of the sample is 2. Using the usual definition and notation, this means $$S = \sqrt{\frac{\sum_{i=1}^n (X_i - \bar X)^2}{n-1}}.$$ Then the $T$-statistic for your problem would be $$T = \frac{\bar X - 45}{S/\sqrt{n}} = \frac{44.5 - 45}{2/\sqrt{30}} = -1.369306.$$

I believe you may be using the notation $\sqrt{\overline{S_n^2}}$ for the 'standard error' $S/\sqrt{n}$ of $\bar X$, not for the 'standard deviation' of the sample. Also, the part with "$= \sqrt{30}$" may be the result of a mistake in typesetting--easy enough to make with such a complicated formula. Finally, many statistics texts use $\mu_0$ instead of $m_0$ to denote the hypothetical value of the population mean.

Anyhow, the result $T = -1.369306$ is OK. And then the p-value is 0.0907 (not significant), as in the useful comment by @GregoryGrant. In R statistical software, this computation is as follows:

 pt(-1.369306, 29)
 ## 0.09070778

Below is output from Minitab statistical software for the problem.

 MTB > Onet 30 44.5 2;
 SUBC>   Test 45;
 SUBC>   Alternative -1.  # indicates alternative is 'less than'

 One-Sample T 

 Test of mu = 45 vs < 45

                             95% Upper
  N    Mean  StDev  SE Mean      Bound      T      P
 30  44.500  2.000    0.365     45.120  -1.37  0.091

The 'upper bound' is for a one-sided confidence interval. Roughly speaking, it indicates that the population mean is likely below 45.12.

(2) There is a difference between 'statistical significance' and 'practical importance'. Even if you had rejected the null hypothesis that new computers take 45s to run your program in favor of the alternative that they are faster, what does that mean to you? Your best estimate of the speed of new computers is 44.5s. Are you willing to buy a new computer for half a second difference in running time for such a program? That is not really a direct statistical question, but a personal one that depends on your needs and finances.