Question

A manager of a grocery store wants to determine if consumers are spending more than the national average. The national average is $150.00 with a standard deviation of $30.20. The manager collects 40 random receipts and finds that the average is $160. Complete a hypothesis test with a significance level of 2.5% to determine if the average customer spends more in his store than the national average. Which of the following is a valid conclusion for the manager based on this test?

Answer 1

We conclude that consumers are spending more than the national average.

Explanation:

We are given the following in the question:

Population mean, μ = $150.00

Sample mean,
`\bar{x}` = $160

Sample size, n = 40

Alpha, α = 0.025

Population standard deviation, σ = $30.20

First, we design the null and the alternate hypothesis

`H_(0): \mu = 150\text{ dollars}\\H_A: \mu > 150\text{ dollars}`

We use One-tailed 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(160 - 150)/((30.20)/(√(40)) ) = 2.0942`

Now,
`z_(critical) \text{ at 0.025 level of significance for one tailed test } = 1.96`

Since,

`z_(stat) > z_(critical)`

We reject the null hypothesis and accept the alternate hypothesis. We conclude that consumers are spending more than the national average.