Question

A business students claims that on average an MBA students is required to prepare more than five cases per week. To examine the claim, a statistics professor ask a random sample of ten MBA students to report the number of cases they prepare weekly. The results are given below. Can the professor conclude that the claim is true, at the .05 level of significance, assuming the number of cases is normally distributed with a standard deviation of 1.5?

Answer 1

Answer:

The proffesor can conclude that the claim is true.

There is  enough evidence to support the claim that on average an MBA students is required to prepare more than five cases per week.

Step-by-step explanation:

The question is incomplete:

The sample of cases per week for students is [2​, 7, ​4, ​8, ​9, ​5, ​11, ​3, ​7, ​4].

This is a hypothesis test for the population mean.

The claim is that on average an MBA students is required to prepare more than five cases per week.

Then, the null and alternative hypothesis are:

$H_0: \mu=5\\\\H_a:\mu> 5$

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=6, calculated as:

$M=\dfrac{1}{10}\sum_{i=1}^{10} x_i=\dfrac{1}{10}(2+7+4+8+9+5+11+3+7+4)=\dfrac{60}{10}=6$

The standard deviation of the population is known and has a value of σ=1.5.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1.5}{\sqrt{10}}=0.474$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{6-5}{0.474}=\dfrac{1}{0.474}=2.108$

This test is a right-tailed test, so the P-value for this test is calculated as:

$P-value=P(z>2.108)=0.018$

As the P-value (0.018) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that on average an MBA students is required to prepare more than five cases per week.