Answer to Question #342361 in Statistics and Probability for Vin

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 = 7. (Set the significance level➔ = .05) 

Answer 1

Here we have a two-tailed test.

So,

Alternative hypotesis: Those who take Memory Booster will have a significantly different memory score from that of the general population.

$H_a:\mu_{MB}\not=7$

Null hypothesis: Those who take Memory Booster will have the same memory score as that of the general population.

$H_0: \mu_{MB}=7$

The significance level is $\alpha=0.05$ , so $t_{crit}=\pm2.365$

Now we compute the appropriate statistical test.

So,

$s^2=\frac {\sum (X-\={X})^2} {n-1}=\frac 8 {8-1}=1.143$

$s_{\={X}}=\sqrt{\frac {s_X^2} {n}}=\sqrt{\frac {1.143} 8}=\sqrt{0.143}=0.378$

$t_{obt}=\frac {\=X-\mu}{s_{\=X}}=\frac {8-7}{0.378}=2.65$

Thus, $t(7)=2.65,$ so we can find the p-value using the t-Distribution table with 7 degrees of freedom (2.65 is between 2.365 and 2.998), so our p-value is less than 0.05 ($p<0.05$) :

Also we can calculate p-value using this site: https://www.statology.org/t-score-p-value-calculator/

So, people who took Memory Booster had significantly higher memory scores (M = 8) than the general population  (µ = 7). 

estimated $d=\frac {\=X-\mu} {s_X}=\frac 1 {\sqrt{1.143}}=0.94$

$r^2=\frac {2.65^2} {2.65^2+7}=\frac {7.023} {14.023}=0.5008$

Hence, approximately 50% of the variance in scores on the memory test can be attributed to Memory Booster.