Wait time for customers at the MacHamburger Shack has been in increasing issue f

Question

Wait time for customers at the MacHamburger Shack has been in increasing issue for you and your employees. Ideally, you, as the manager, would like to have the wait time to be no more than 2 minutes per customer, once the order has been processed. In an effort to understand how long a customer must currently wait after the order is processed, you implement a process of timing each transaction. You collect the following data (in minutes):

Conduct an appropriate test, where you do not know the industry standard deviation for wait time. You would like your wait time to be 2 minutes (with 95% confidence), but do not know if your actual time is more or less than 2 minutes. After you conduct the test answer the following questions:

1) What is the standard error of the mean? ( Round to 3 decimals)

2) What is the 95% Confidence Interval for the mean? ( Round to 3 decimals)

3) What is the p-value for the test?

4) What is the true (population) between means?

5) Based on the p-value and 95% confidence interval, we would do: (Choose options 1-5)

5)We would reject the null hypothesis and conclude the true mean is less than 2 minutes.

Question 6) Based on the p-value and 95% confidence interval, what would we do? (Choose from options 1-4)

4)There is not enough information to answer this question.

WaitTime 1.6 5.8 6.6 3.4 5.6 9.3 5.1 7.4 3.9 1.8 4.6 6.2 6.5 5.4 6.4 4.9 8.6 5.4 4.2 0.8

Explanation / Answer

let X denotes the waiting time of customers

assumption X~N(u,sigma2) where both are unknown.

confidence level is 95%

hence level of significance=alpha=0.05

we would like the wait time to be 2 minutes (with 95% confidence), but do not know if your actual time is more or less than 2 minutes.

hence null hypothesis H0: u=2    vs alternative hypothesis H1: u>2

we have the sample data with size=n=20

sample mean=xbar=5.175

sample standard deviation=s=2.178

sample size=n=20

1) hence the standard error of mean is s/sqrt(n)=2.178/sqrt(20)=0.487 [answer]

2) since the population variance sigma2 is unknown it is estimated by s2

hence the test statistic T=(xbar-u)*sqrt(n)/s under H0 follows a t distribution with df=n-1=19

now P[|T|>talpha/2;n-1]=1-alpha

or, P[|(xbar-u)*sqrt(n)/s|>talpha/2;n-1]=1-alpha

or, P[xbar-s/sqrt(n)*talpha/2;n-1<u<xbar+s/sqrt(n)*talpha/2;n-1]=1-alpha

hence (xbar-s/sqrt(n)*talpha/2;n-1,xbar+s/sqrt(n)*talpha/2;n-1) is the 100(1-alpha)% confidence interval for the mean

hence for 95% confidence interval alpha=0.05

now xbar=5.175 s=2.178   n=20 alpha=0.05   t0.025;19=2.09302 [using MINITAB]

hence the 95% confidence interval is

(5.175-2.178/sqrt(20)*2.09302,5.175+2.178/sqrt(20)*2.09302)=(4.156,6.194) [answer]

3) the test statistic T=(xbar-2)*sqrt(n)/s under H0 follows a t distribution with df n-1=20-1=19

now the value of test statistic is t=(5.175-2)*sqrt(20)/2.178=6.519

since the test is right tailed the p value is p=P[T>6.519]=0 [using MINITAB]

hence p value=0<0.05=level of significance

5) hence based on p value we have p value=0<0.05=level of significance the 95% confidence interval of mean does not contain the point 2.

hence H0 is rejected and u>2 is the result.

hence the correct option is option 4) We would reject the null hypothesis and conclude the true mean is greater than 2 minutes. [answer]

6) hence the wait time is greater than 2 minutes

hence the correct option is

1) 1)The wait time is more than the required standard and the sales process needs improvement. [answer]

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