Answer to Question #300612 in Statistics and Probability for Shaaz

Question

1. At a certain college, new students are weighed when they join the college, new students are weighed when they join the college. The distribution of weights of students at the college when they enroll has a standard deviation of 7.5 kg and a mean of 70kg. A random sample of 90 students from the new entry were weighed and their mean weight was 71.6kg. Assume that the standard deviation has not changed. Test at 5% level, whether there is evidence that the mean of the new entry is more than 70kg.

Perform your hypothesis testing under the 5 steps used when performing a hypothesis test. State your conclusion clearly.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 70$

$H_1:\mu>70$

This corresponds to a right-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 right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z > 1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{71.6-70}{7.5/\sqrt{90}}=2.0239$$

Since it is observed that $z = 2.024 >1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z>2.0239)=0.021492$, and since $p=0.021492<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 70, at the $\alpha = 0.05$ significance level.