Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 2500
For the alternative hypothesis,
µ < 2500
This is a left tailed test.
Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 2500
x = 2470
σ = 75
n = 25
z = (2470 - 2500)/(75/√25) = - 2
Since it is a left tailed test, we would determine the area to the left of z = - 2 from the normal distribution table. the probability corresponding to the z score is 0.023
Since alpha, 0.01 < the p value, 0.023, then we would fail to reject the null hypothesis. Therefore, at a 1% level of significance, there is not enough evidence that the mean cost is lower than $2,500
