Answer:
The correct answer is:
The customers spend more than the national average in his store
Step-by-step 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.
Answer:
Option A) The customers spend more than the national average in his store.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $150.00
Sample mean, [tex]\bar{x}[/tex] = $160
Sample size, n = 40
Alpha, α = 0.025
Population standard deviation, σ = $30.20
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 150\text{ dollars}\\H_A: \mu > 150\text{ dollars}[/tex]
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:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{160 - 150}{\frac{30.20}{\sqrt{40}} } = 2.094[/tex]
Now, [tex]z_{critical} \text{ at 0.025 level of significance } = 1.96[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
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
