Answer to Question #129935 in Statistics and Probability for Ntokozo Kabini

Question

A college professor randomly selects 25 freshmen to study the mathematical background of the  incoming freshman class. The average SAT score of these 25 students is 565 and the standard deviation is estimated to be 40. Using this information, can the professor reject the null hypothesis that the average test score is 550 or less? 

Use a 5 % level of significance.

Answer 1

The provided sample mean is $\bar{x}=565$ and the sample standard deviation is $s=40,$

and the sample size is $n=25.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq550$

$H_1:\mu>550$

This corresponds to a right-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 $\alpha=0.05,$ $df=25-1=24,$ and the critical value for a right-tailed test is $t_c=1.711.$

The rejection region for this right-tailed test is $R=t:t>1.711$  

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}=\dfrac{565-550}{40/\sqrt{25}}=1.875$$

Since it is observed that $t=1.875>1.711=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 550, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.036506,$ and since $p=0.036506<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 550, at the 0.05 significance level.