How do u benchmark 8/11/12 and 2/11/20
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Answer:
The test statistic is c. 2.00
The p-value is a. 0.0456
At the 5% level, you b. reject the null hypothesis
Step-by-step explanation:
We want to test to determine whether or not the mean waiting time of all customers is significantly different than 3 minutes.
This means that the mean and the alternate hypothesis are:
Null:
Alternate:
The test-statistic is given by:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
Sample of 100 customers.
This means that
3 tested at the null hypothesis
This means that
The average length of time it took the customers in the sample to check out was 3.1 minutes.
This means that
The population standard deviation is known at 0.5 minutes.
This means that
Value of the test-statistic:
The test statistic is z = 2.
The p-value is
Mean different than 3, so the pvalue is 2 multiplied by 1 subtracted by the pvalue of Z when z = 2.
z = 2 has a pvalue of 0.9772
2*(1 - 0.9772) = 2*0.0228 = 0.0456
At the 5% level
0.0456 < 0.05, which means that the null hypothesis is rejected.
Answer:
Step-by-step explanation:
