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, = $160
Sample size, n = 40
Alpha, α = 0.025
Population standard deviation, σ = $30.20
First, we design the null and the alternate hypothesis
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:
Putting all the values, we have
Now,
Since,
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
Y = 3
5x^-6y^-2/y