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: ![H_{0} = 3](https://tex.z-dn.net/?f=H_%7B0%7D%20%3D%203)
Alternate: ![H_{a} = 3](https://tex.z-dn.net/?f=H_%7Ba%7D%20%3D%203)
The test-statistic is given by:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
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 ![n = 100](https://tex.z-dn.net/?f=n%20%3D%20100)
3 tested at the null hypothesis
This means that ![\mu = 3](https://tex.z-dn.net/?f=%5Cmu%20%3D%203)
The average length of time it took the customers in the sample to check out was 3.1 minutes.
This means that ![X = 3.1](https://tex.z-dn.net/?f=X%20%3D%203.1)
The population standard deviation is known at 0.5 minutes.
This means that ![\sigma = 0.5](https://tex.z-dn.net/?f=%5Csigma%20%3D%200.5)
Value of the test-statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B3.1%20-%203%7D%7B%5Cfrac%7B0.5%7D%7B%5Csqrt%7B100%7D%7D%7D)
![z = 2](https://tex.z-dn.net/?f=z%20%3D%202)
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.