Question

1-A manager of an e-commerce company would like to determine average delivery time of the products. A sample of 25 customers is taken. The average delivery time in the sample was four days. Suppose the delivery times are normally distributed with a standard deviation of 1.2 days.

a) Provide a 95 % confidence interval for the mean delivery time.

b) The manager claims that the average delivery time of their products does not exceed 3 days. Write the null and alternative hypothesis regarding to the claim of the manager.

c) Test the manager’s claim at 95 % confidence level.

d) Write the conclusion of your result 3)

2- For an effective parental skill study, a researcher asked: How many hours do your kids watch the television during a typical week in Barcelona? The mean of 100 Kids (ages 6-11) spend about 28 hours a week in front of the TV. Suppose the study follows a normal distribution with standard deviation 5.

a) Estimate the mean of all kids (ages 6-11) in Barcelona, using 99% confidence interval. (show all the calculations)

b) Write the conclusion of your result

Answer 1

1.) Given

sample size n = 25

sample mean = 4 days

population std dev = 1.2 days

a) 95% CI =

Z0.025 = 1.96

=( 4 - 1.96*1.2/sqrt(25) , 4 + 1.96*1.2/sqrt(25) )

= (3.527, 4.470)

b) The manager claims that the average delivery time of their products does not exceed 3 days. Write the null and alternative hypothesis regarding to the claim of the manager.

H0:

H1:

c)

test statistic Z = = = 1.863

critical value = -Z0.05 = -1.645

since test statistic Z > -Z0.05 we fail to reject H0 and there is no significant evidence that mean delivery time is less than 3.

The 95% CI = (3.527 , 4.470) does not contain 3 and range does not contain values < 3, hence there is no significant evidence that mean delivery time is less than 3.

d) There is no significant evidence that mean delivery time is less than 3.

2.)

Given

sample size n = 100

sample mean = 28 hours

population std dev = 5 hours

99% CI =

Z0.005 = 2.58

( 28 - 2.58*5/sqrt(100) ,28 + 2.58*5/sqrt(100) ) = (26.71, 29.29)

b) We are 95% confident that the population mean time spent by kids in a week will lie in between (26.71, 29.29).