Answer to Question #246958 in Statistics and Probability for Keet

Question

A survey found that women over the age of 55 consume an average of 1660 calories a day. In order to see if the number of calories consumed by women over the age 55 living in a certain city is the same, the researcher sampled 30 women over the age of 55 and found the mean number of calories consumed was atleast 1446. The standard deviation of the sample was 446 calories. At a level of significance of 0.05, can it be concluded that there is no difference between the number of calories consumed by the women over the age of 55?

Answer 1

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, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05, df=n-1$

$=30-1=29$ degrees of freedom, and $t_c= 2.04523.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.04523\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{1446-1660}{446/\sqrt{30}}=-2.6281$$

Since it is observed that $|t| = 2.6281 > 2.04523=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $\alpha = 0.05, df=29,t=-2.6281$ is $p= 0.013584,$ and since $p = 0.013584 < 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $1660,$ at the $\alpha = 0.05$ significance level.