Question

The waiting time for customers at MacBurger Restaurants follows a normal distribution with a population standard deviation of 5 minute. At the Warren Road MacBurger, the quality assurance department sampled 84 customers and found that the mean waiting time was 13.75 minutes. At the 0.05 significance level, can we conclude that the mean waiting time is less than 15 minutes? Use α = 0.05.

a. State the null hypothesis and the alternate hypothesis.

H₀: μ ≥           

H₁: μ <           

b. State whether the decision rule is true or false: Reject H₀ if z < -1.645.

(Click to select)  True  False

c. Compute the value of the test statistic. (Negative answer should be indicated by a minus sign. Round the final answer to 2 decimal places.)

Test statistic z is            .

d. What is your decision regarding H₀?

(Click to select)  Do not reject  Reject  H₀.

e. What is the p-value? (Round the final answer to 4 decimal places.)

The p-value is  .

Answer 1

given that

populationSD=SD=5

sample size =n=84

sample mean =m=13.75

a)

since we have to test that population mean is less than 15 or not hence

b)

since test is left tailed and level of significance is 0.05 hence cirtical value is given by

P(Z<critical value)=0.05

from Z table P(Z<-1.645) =0.05

Hence critical value =-1.645

Hence we reject H0 if Z<-1.645

so decision rule is TRUE

c)

Now test statistics is given by

d)

since Z =-2.26 < -1.645

Hence we reject H0

e)

test is left tailed so P value is given by

P-value =P(Z<-2.29) =0.0110