Answer to Question #92088 in Statistics and Probability for Mark Lester Solis

Question

A magazine survey found that women over the age of 55 consume anaverage of 1660 calories a day. In order to see if the same wastrue for women over 55 in assisted living facilities, a researchersampled 43 women. She found that the mean for the sample was 1446and that the sample standard deviation was 56 calories. Atα=.10 test the claim that the average number of caloriesconsumed by the women is assisted living is the same as the numbercalories consumed by the women magazine survey.

Answer 1

The provided sample mean is $\overline{X} =1446$ and the sample standard deviation is $s=56$, and the sample size is $n=43.$

The following null and alternative hypotheses need to be tested:

$$H_0: \mu=1660$$ $$H_1: \mu\not =1660$$

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

Based on the information provided, the significance level is $\alpha=0.1,$ and the critical value for a two-tailed test is $t_c=1.682.$

The rejection region for this two-tailed test is $R=\{{t: |t|>1.682}\}$  

The t-statistic is computed as follows:

$$t={\overline{X}-\mu \over s/\sqrt{n}}={1446-1660 \over 56/\sqrt{43}}=-25.0588$$

Since it is observed that $|t|=25.0588>t_c=1.682,$  |t| = 25.059 it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=0,$ and since $p=0<0.1,$ it is concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis $H_0$ is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 1660, at the 0.1 significance level.

The 90% confidence interval is 

$$1431.636<\mu<1460.364$$