Answer to Question #341255 in Statistics and Probability for Errol

Question

The Guidance Counselor of your school claims that the Grade 11 students spend an average of 11.28 hours in a week doing performance tasks with standard deviation of 1.64. Your adviser thinks that students spend more time in doing performance tasks, so he decided to conduct his own research. He used a sample of 46 Grade 11 students and obtained a mean of 11.83. Is there enough evidence at 0.05 level of significance that the students spend 11.28 hours in a week doing performance tasks?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=11.28$

$H_1:\mu>11.28$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{11.83-11.28}{1.64/\sqrt{46}}\approx2.27456$$

Since it is observed that $z=2.27456 > 1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>2.27456)=0.011466,$ and since $p=0.011466<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 11.28, at the $\alpha = 0.05$ significance level.