Answer to Question #279812 in Statistics and Probability for arnold

Question

A nutrition store in the mall is selling “Memory Booster,” which is a concoction of herbs and minerals that is intended to improve memory performance, but there is no good reason to think it couldn't possibly do the opposite. To test the effectiveness of the herbal mix, a researcher obtains a sample of 8 participants and asks each person to take the suggested dosage each day for 4 weeks. At the end of the 4-week period, each individual takes a standardized memory test. In the general population, the standardized test is known to have a mean of μ = 8 (Set the significance level➔ = .06

Answer 1

$$12, 11, 13, 15, 8, 11, 13, 15$$

$$mean=\bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nx_i=$$

$$=\dfrac{12+11+13+15+8+11+13+15}{8}$$

$$=\dfrac{98}{8}=12.25$$

$$s^2=\dfrac{1}{n-1}\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2=$$

$$=\dfrac{1}{8-1}((12-12.25)^2+(11-12.25)^2+$$

$$+(13-12.25)^2+(15-12.25)^2+(8-12.25)^2$$

$$+(11-12.25)^2+(13-12.25)^2+(15-12.25)^2)$$

$$=\dfrac{37.5}{7}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{37.5}{7}}\approx2.31455$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=8$

$H_1:\mu\not=8$

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 $\alpha = 0.06,$ $df=n-1=8-1=7$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.240875.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.240875\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{12.25-8}{2.31455/\sqrt{8}}\approx5.1936$$

Since it is observed that $|t| = 5.1936 > 2.240875=t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=7$ degrees of freedom, $t=5.1936$ is $p = 0.001262,$ and since $p=0.001262<0.06=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than $8,$ at the $\alpha = 0.06$ significance level.