Answer :
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: [tex]H_{0} = 3[/tex]
Alternate: [tex]H_{a} = 3[/tex]
The test-statistic is given by:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
Sample of 100 customers.
This means that [tex]n = 100[/tex]
3 tested at the null hypothesis
This means that [tex]\mu = 3[/tex]
The average length of time it took the customers in the sample to check out was 3.1 minutes.
This means that [tex]X = 3.1[/tex]
The population standard deviation is known at 0.5 minutes.
This means that [tex]\sigma = 0.5[/tex]
Value of the test-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}[/tex]
[tex]z = 2[/tex]
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.