In: Statistics and Probability
The CEO of the company claims that the mean work day of the company’s mechanical engineers is less than 8.5 hours. A random sample of 25 of the company’s mechanical engineers has a mean work of 8.2 hours. Assume the population standard deviation is 0.5 hour and the population is normally distributed. At α = 0.01, test the CEO’s claim.
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: the mean work day of the company’s mechanical engineers is 8.5 hours.
Alternative hypothesis: Ha: the mean work day of the company’s mechanical engineers is less than 8.5 hours.
H0: µ = 8.5 versus Ha: µ < 8.5
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 = 8.5
Xbar = 8.2
σ = 0.5
n = 25
α = 0.01
Critical value = -2.3263
(by using z-table or excel)
Z = (8.2 – 8.5)/[0.5/sqrt(25)]
Z = -3.0000
P-value = 0.0013
(by using Z-table)
P-value < α = 0.01
So, we reject the null hypothesis
There is sufficient evidence to conclude that the mean work day of the company’s mechanical engineers is less than 8.5 hours.