Question

3. A fitness center claims that its members lose an average of 15 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 13

pounds within the first month with standard deviation of 3 pounds. What will your

decision be if a = 0.05?

Answer 1

Hypothesized Population Mean $\mu=15$

Sample Standard Deviation $s=3$

Sample Size $n=45$

Sample Mean $\bar{x}=13$

Significance Level $\alpha=0.05$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq15$

$H_1: \mu<15$

This corresponds to 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 fredom, and the critical value for two-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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{13-15}{3/\sqrt{45}}\approx-4.47214$$

Since it is observed that $t=-4.47214<-1.68023=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than $15,$ at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value for left-tailed, the significance level $\alpha=0.05, df=44, t=-4.47214,$ is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than $15,$ at the $\alpha=0.05$ significance level.

Therefore, there is enough evidence that claim of fitness center is false at the $\alpha=0.05$ significance level.