Respuesta :
Answer:
We would use Z-test of a population mean.
Step-by-step explanation:
We are given that an entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625.
And we have to test the hypothesis that the daily average revenue was $675.
Firstly, as we know that testing is always done on the population parameter.
So, let [tex]\mu[/tex] = daily average revenue over the pat 5 years
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = $675
Alternate Hypothesis, [tex]H_a[/tex] : [tex]\mu \neq[/tex] $675
Here, null hypothesis states that the owner's claim of average daily revenue was $675 over the past 5 years is correct.
And alternate hypothesis states that the owner's claim of average daily revenue was $675 over the past 5 years is not correct.
The test statistics that will be used here is Z-test of a population mean because here we have knowledge of population standard deviation of $75.
Test statistics = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample average revenue = $625
[tex]\sigma[/tex] = population standard deviation = $75
n = sample of days = 30
Therefore, to test the null hypothesis that the daily average revenue was $675, we should use Z-test of a population mean.
The Z-test of a population mean is used here because it is given that the population standard deviation is $75. The correct option is a).
Given :
- The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75.
- A sample of 30 days reveals a daily average revenue of $625.
Let [tex]\mu[/tex] is the daily average revenue. So:
Null Hypothesis is, [tex]\rm H_0:\mu=$675[/tex]
The Alternate Hypothesis is, [tex]\rm H_a : \mu\neq $675[/tex]
Here, the Z-test of a population mean is used because it is given that the population standard deviation is $75.
[tex]\rm Test \;Statistics = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n} }}[/tex]
where [tex]\rm \bar {X}[/tex] is the sample average revenue, [tex]\sigma[/tex] is the standard deviation and n is the sample size.
Therefore, the correct option is a).
For more information, refer to the link given below:
https://brainly.com/question/13245779
