Question

The dean of a business school claims that the average starting salary of its graduates is more than 85 (in $000’s). It is known that the population standard deviation is 10 (in $000’s). Sample data on the starting salaries of 64 randomly selected recent graduates yielded a mean of 88(in $000s). What is the critical value for the rejection region if the level of significance is 5%?

Answer 1

Critical value for the rejection region if the level of significance is 5% = 1.645

We conclude that the average starting salary of graduates is more than 85,000$.

Explanation:

We are given the following in the question:

Population mean, μ = 85,000$

Sample mean,
`\bar{x}` = 88,000$

Sample size, n = 64

Alpha, α = 0.05

Population standard deviation, σ = 10,000

First, we design the null and the alternate hypothesis

`H_(0): \mu = 85,000\text{ dollars}\\H_A: \mu > 85,000\text{ dollars}`

We use One-tailed(right) z test to perform this hypothesis.

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(88000 - 85000)/((10000)/(√(64)) ) = 2.4`

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 average starting salary of graduates is more than 85,000$.