Question

The reported average cost per workbook at a large college is $27.50. A professor claims that the actual average cost per workbook is higher than $27.50. A sample of 44 different workbooks has an average cost of $28.90. The population standard deviation is known to be $5.00. Can the null hypothesis be rejected at alpha= 0.05?

Answer 1

Answer:

We conclude that the actual average cost per workbook is higher than $27.50.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $27.50

Sample mean, $\bar{x}$ = $28.90

Sample size, n = 44

Alpha, α = 0.05

Population standard deviation, σ = $5.00

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 27.50\text{ dollars}\\H_A: \mu > 27.50\text{ dollars}$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{28.90 - 27.50}{\frac{5.00}{\sqrt{44}} } = 1.8573$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.64$

Since,  

$z_{stat} > z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the actual average cost per workbook is higher than $27.50.