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

Answer:

A manager of a grocery store wants to determine if consumers are spending more than the national average.

Yes is the answer

Step-by-step explanation:

Given that 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 Manager's contention is right.

Answer 2

Answer:

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

Step-by-step 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:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{160 - 150}{\frac{30.20}{\sqrt{40}} } = 2.0942$

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.