The management of Quick and Easy Health Club claims that its members lose a mean
ID: 3243844 • Letter: T
Question
The management of Quick and Easy Health Club claims that its members lose a mean of more than 10 pounds in their first month with them. (the population standard deviation is known to be 2.4 pounds in the first month) You are working for the Consumer Protection Agency in your city and your boss assigns you the task to determine if their claim is correct. You take a sample of 36 members and find the mean weight loss of 10.24 pounds in the first month. Test the Health Club claim at the .01 significance level. (be sure to show the hypothesis and the claim) What do you conclude about null? What do you conclude about the claim? Would your answer change if you tested at the .10 significance level? Why or Why not?
Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: <= 10 (that its members lose a mean of less than or equal to 10 pounds in their first month with them)
Alternative hypothesis: > 10 (claim, that its members lose a mean of more than 10 pounds in their first month with them)
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too large.
Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 2.4 / sqrt(36) = 0.4
DF = n - 1 = 36 - 1 = 35
t = (x - ) / SE = (10.24 - 10) / 0.4 = 0.6
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Here is the logic of the analysis: Given the alternative hypothesis ( > 10), we want to know whether the observed sample mean is large enough to cause us to reject the null hypothesis.
The observed sample mean produced a t statistic test statistic of 0.6. We use the t Distribution Calculator to find P(t < 0.6) = 0.19.
The P-Value is 0.276186.
The result is not significant at p < 0.01.
Interpret results. Since the P-value (0.276) is greater than the significance level (0.01), we cannot reject the null hypothesis.
Conclusion. Fail to reject the null hypothesis. We have insufficient evidence to prove the claim that its members lose a mean of more than 10 pounds in their first month with them.
If we test the claim at 0.10 significance level, still our answer won't change because P-value (0.276) is still greater than the significance level (0.10), thus we cannot reject the null hypothesis.
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