Answer to Question #138978 in Statistics and Probability for Hazel quintos

Question

The new memory enhancer X recently gained a reputation in the market as the most effective memory enhancer ever produced by XY Pharma. The recorded average memory level among people who had taken the enhancer was 95 with a standard deviation of 10. As a neuroscientist, you want to test the claim and conducted a study and measure the memory level of 50 participants. The average memory level of the participants came out to be 80. Is there enough evidence to support the claim about the new memory enhancer?

Answer 1

The provided sample mean is $\bar{X}=80$ and the known population standard deviation is $\sigma=10,$ and the sample size is $n=50.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq95$

$H_1:\mu<95$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.6449$.

The rejection region for this left-tailed test is $R=\{z:z<-1.6449\}$ 

The z-statistic is computed as follows:

$$z=\dfrac{\bar{X}-\mu_0 }{\sigma/\sqrt{n}}=\dfrac{80-95 }{10/\sqrt{50}}\approx-10.6066$$

Since it is observed that $z=-10.6066<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population $\mu$ is less than 95, at the 0.05 significance level.

Using the P-value approach: The p-value is $p<0.00001,$and since $p<0.00001<0.05,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population $\mu$ is less than 95, at the 0.05 significance level.

Therefore, there is enough evidence to reject the claim about the new memory at the 0.05 significance level.