Question

Question Workspace Exhibit 9-2 The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. Refer to Exhibit 9-2. The p-value is

Answer 1

Answer:

$z=\frac{3.1-3}{\frac{0.5}{\sqrt{100}}}=2$  

Since is a right-sided test the p value would be:  

$p_v =P(z>2)=0.02275$  

Step-by-step explanation:

Data given and notation  

$\bar X=3.1$ represent the sample mean    

$\sigma=0.5$ represent the population standard deviation

$n=100$ sample size  

$\mu_o =3$ represent the value that we want to 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 true mean is higher than 3 minutes , the system of hypothesis would be:  

Null hypothesis:$\mu \leq 3$  

Alternative hypothesis:$\mu > 3$  

Since we know the population deviation, 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)  

Calculate the statistic  

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

$z=\frac{3.1-3}{\frac{0.5}{\sqrt{100}}}=2$  

P-value  

Since is a right-sided test the p value would be:  

$p_v =P(z>2)=0.02275$