Answer to Question #345572 in Statistics and Probability for Chelle

Question

HYPOTHESIS TESTINGABOUT POPULATION MEAN 𝝁

USING THE CRITICAL VALUE APPROACH

The records of SCA Registrar show that the average final grade in Mathematics for STEM students is 91 with a standard deviation of 3. A group of student-researchers found out that the average final grade of 37 randomly selected STEM students in Mathematics is no longer 91. Use 0.05 level of significance to test the hypothesis and a sample mean within the range of 88 to 94 only.

A. State the hypotheses.

B. Determine the test statistic to use.

C. Determine the level of significance, critical value, and the decision rule.

D. Compute the value of the test statistic.

E. Make a decision.

F. Draw a conclusion.

Answer 1

A. The following null and alternative hypotheses need to be tested:

$H_0:\mu=91$

$H_1:\mu\not=91$

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

C. Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

D. The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{90-91}{3/\sqrt{37}}=-2.0276$$

E. Since it is observed that $|z|=2.0276>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-2.0276)= 0.042601,$ and since $p= 0.042601<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 91, at the $\alpha = 0.05$ significance level.