Question

An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the mean daily revenue was $675 with a population standard deviation of $75. A sample of 30 days reveals a daily mean revenue of $640. If you were to test the null hypothesis that the daily mean revenue was $675 and decide not to reject the null hypothesis, what can you conclude

Answer 1

Answer:

$z=\frac{640-675}{\frac{75}{\sqrt{30}}}=-2.556$    

For this case we don't have the significance level provided $\alpha$ but we know the final conclusion and on this case is not reject the null hypothesis, so that means :

$p_v >\alpha$

Where $p_v = 2*P(z$ represent the p value for this test.

So then we can't conclude that we have a significant difference from the specified value of 675 for the true mean.

Step-by-step explanation:

Data given and notation  

$\bar X=640$ represent the sample mean

$\sigma=75$ represent the population standard deviation for the sample

$n=30$ sample size  

$\mu_o =675$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 675 or no, the system of hypothesis would be:  

Null hypothesis:$\mu = 675$  

Alternative hypothesis:$\mu \neq 675$  

If we analyze the size for the sample is = 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{640-675}{\frac{75}{\sqrt{30}}}=-2.556$    

For this case we don't have the significance level provided $\alpha$ but we know the final conclusion and on this case is not reject the null hypothesis, so that means :

$p_v >\alpha$

Where $p_v = 2*P(z$ represent the p value for this test.

So then we can't conclude that we have a significant difference from the specified value of 675 for the true mean.