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?

a.The customers spend more than the national average in his store.b.The manager should decrease prices in his store.c.The customers do not spend more than the national average in his store. d.The customers in his store just come from a rich neighborhood.

Answer 1

Option A) The customers spend more than the national average in his store.

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}`

The null hypothesis states that the consumers are spending equal to the national average. The alternative hypothesis states that consumers are spending more than the national average.

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)) ) = 2.094`

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

Since,

`z_(stat) > z_(critical)`

We reject the null hypothesis and accept the alternate hypothesis. Thus, the customers spend more than the national average in his store.

Thus, option A) is a valid conclusion for the manager

Answer 2

The correct answer is:

The customers spend more than the national average in his store

Explanation:

The national average is $150.00 with a standard deviation of $30.20.

Sample size n =40

H0: x bar = mu

Ha: x bar >mu

(one tailed test for a single mean)

Sample average x bar = 160

Mean difference = 160-150 =10

std error = 30.20/sqrt 40

=4.775

Test statistic = 2.094

Z critical for 2.5% = 1.96 (one tailed)

Since test statistic > z critical we reject null hypothesis.

Hence the correct answer is:

The customers spend more than the national average in his store.