Answer to Question #139279 in Statistics and Probability for Jeanina Muñoz

Question

Using the sample T-Test formula, work on the following problem.
Suppose the average GPA of college students who use memory enhancer is 1.75. You surveyed seven college students who do not use memory enhancers. Their GPAs are as follow: 1.65,1.78, 1.74, 1.6, 1.79, 1.81, and 1.85. Using the .05 alpha level, do students who use memory enhancers greatly differ from those who don't use memory enhancers in terms of GPA?

Answer 1

$$\bar{x}=\dfrac{1.65+1.78+1.74+1.6+1.79+1.81+1.85}{7}\approx1.7457$$

$$\sum_{i=1}^7(x_i-1.7457)^2=$$

$$=(1.65-1.7457)^2+(1.78-1.7457)^2+(1.74-1.7457)^2+$$

$$+(1.6-1.7457)^2+(1.78-1.7457)^2+(1.81-1.7457)^2+$$

$$+(1.85-1.7457)^2\approx0.048571$$

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

$$=\sqrt{\dfrac{0.048571}{7-1}\displaystyle}\approx0.09$$

The provided sample mean is $\bar{x}=1.7457$ and the sample standard deviation is $s=0.09,$ and the sample size is $n=7.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=1.75$

$H_1:\mu\not=1.75$

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  $\alpha=0.05,$ and the critical value for a two-tailed test is $t_c=2.446899.$  

The rejection region for this two-tailed test is $R=\{t:|t|>2.446899\}$  

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{1.7457-1.75}{0.09/\sqrt{7}}\approx-0.1264$$

Since it is observed that $|t|=0.1264<2.446899,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 1.75, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.903545,$ and since $p=0.903545\geq0.05,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 1.75, at the 0.05 significance level.