Critical value: with n-1 degrees of freedom= 15.This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not вЂњtoo highвЂќ a number. The null and alternative hypothesis are for the parameter Вµ because the number of dollars of the contracts is a continuous random variable. Both methods reach the same conclusion that we cannot accept the null hypothesis. The p-value has been put on the graph as the shaded area beyond -2.08 and it shows that it is smaller than the hatched area which is the alpha level of 0.05. 05 we see that we cannot accept the null. If we wished to use the p-value system of reaching a conclusion we would calculate the statistic and take the additional step to find the probability of being 2.08 standard deviations from the mean on a t-distribution. A formal conclusion would be stated as: вЂњWith a 95% level of significance we cannot accept the null hypothesis that the swimming time with goggles comes from a distribution with a population mean time of 16.43 minutes.вЂќ Less formally, вЂњWith 95% significance we believe that the goggles improves swimming speedвЂќ Step 5 has us state our conclusions first formally and then less formally. The probability that an average time of 16 minutes could come from a distribution with a population mean of 16.43 minutes is too unlikely for us to accept the null hypothesis. We see that the test statistic is in the tail and thus we move to step 4 and reach a conclusion. Step 4 has us compare the test statistic and the critical value and mark these on the graph. We find that the calculated test statistic is 2.08, meaning that the sample mean is 2.08 standard deviations away from the hypothesized mean of 16.43. Step 3 is the calculation of the test statistic using the formula we have selected. This is the critical value and we can put this on our graph. Looking up 14 degrees of freedom at the 0.05 column of the t-table we find 1.761. For this problem the degrees of freedom are n-1, or 14. We can now draw the graph of the t-distribution and mark the critical value. We have concluded that this is a t-test on the basis of the sample size and that we are interested in a population mean. To find the critical value we need to select the appropriate test statistic. (Reject the null hypothesis when the null hypothesis is true.) For this case the only concern with a Type I error would seem to be that JefferyвЂ™s dad may fail to bet on his sonвЂ™s victory because he does not have appropriate confidence in the effect of the goggles. The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds.

It is worth thinking about the meaning of this choice.

Our step 2, setting the level of significance, has already been determined by the problem. Оњ 0 = 16.43 comes from H 0 and not the data. The sample size is less than 30 and we do not know the population standard deviation so this is a t-test. Random variable: = the mean time to swim the 25-yard freestyle. The null and alternative hypotheses are thus:ДЇor Jeffrey to swim faster, his time will be less than 16.43 seconds. Remember that the status quo must be defeated with a high degree of confidence, in this case 95 % confidence. The claim will always be in the alternative hypothesis because the burden of proof always lies with the alternative. The effect of this is to set the hypothesis as a one-tailed test. This is that the goggles will reduce the swimming time. In this case there is an implied challenge or claim. Since the problem is about a mean, this is a test of a single population mean.

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