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Step 5 has us state our conclusions first formally and then less formally. A formal conclusion would be stated as: “With a 95% level of confidence 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.” Less formally, “With 95% confidence we believe that the goggles improves swimming speed”
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. This value is .0187. Comparing this to the α-level of .05 we see that we cannot accept the null. 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. Both methods reach the same conclusion that we cannot accept the null hypothesis.
The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.
First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.
Since the problem is about a mean, this is a test of a single population mean.
H 0 : μ = 40
H a : μ >40
p = 0.0062
Because p < α , we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.
Mat is an incredible singer. In fact, people even faint when he sings. When we sample 16 of his amazing performances, we find the sample mean number of fainters to be 108 people with a sample standard deviation to be 12 people. Test at 5% significance that the population mean is at least 108 people fainting against the alternative that it is less than 108 people. Mat has claimed that during this set of performances he could get 108 people to faint.
The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 fainters is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Mat's claim.
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