Answer to Question #337711 in Statistics and Probability for JohnMark

Question

A fitness center claims that its members lose an average of 12 pounds or more the first month after joining the center. An independent agency that wanted to check this claim took a sample of 45 members and found that they lost an average of 10 pounds within the first month with standard deviation of 3 pounds. Find the p-value for this test. What will your decision be if 𝛼 = 0.05?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge12$

$H_a:\mu<12$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=44$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.68023.$

The rejection region for this left-tailed test is $R = \{t: t <-1.68023\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{10-12}{3/\sqrt{45}}\approx-4.472$$

Since it is observed that $t =-4.472 <-1.68023=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=44$ degres of freedom, $t=-4.472$ is $p= 0.000027,$ and since $p= 0.000027<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is less than 12, at the $\alpha = 0.05$ significance level.