Question

The manager of a pizza restaurant claims his employees make an average of $8.00 per hour with a standard deviation of $0.75. Jack does not believe the manager's claim, so he takes a sample of 40 employees and finds their average salary to be $7.80. Jack tests the manager's claim at α = 0.05.

Answer 1

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: Employees make an average of $8.00 per hour.

Alternative hypothesis: H_a: Employees make an average different from $8.00 per hour

H₀: µ = 8 versus H_a: µ ≠ 8

This is a two tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 8

Xbar = 7.80

σ = 0.75               

n = 40

α = 0.05

Critical value = -1.96 and 1.96

(by using z-table or excel)

Z = (7.80 - 8)/[0.75/sqrt(40)]

Z = -1.6865

P-value = 0.0917

(by using Z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that employees make an average of $8.00 per hour.