Question

5. a.) A random sample of 45 door-to-door salespersons were asked how long on average they...

5.

a.)

A random sample of 45 door-to-door salespersons were asked how long on average they were able to talk to the potential customer. Their answers revealed a mean of 8.5 minutes with a variance of 9 minutes. Which Excel statements will construct a 95% confidence interval for the time it takes a salesperson to talk to a potential customer.

a.

=8.5+1.96*9/SQRT(45), and =8.5-1.96*9/SQRT(45)

b.

=8.5+1.96*3/SQRT(45), and =8.5-1.96*3/SQRT(45)

c.

=45+1.96*3/SQRT(9), and =45-1.96*3/SQRT(9)

d.

=45+0.95*3/SQRT(9), and =45-0.95*3/SQRT(9)

e.

None of the above.

A random sample of 45 door-to-door salespersons were asked how long on average they were able to talk to the potential customer. Their answers revealed a mean of 8.5 minutes with a variance of 9 minutes. What is the point estimate for the average conversation length?

a.

45 salespersons

b.

9 minutes

c.

8.5 minutes

d.

5 minutes

e.

None of the above.

Homework Answers

Answer #1

5)

Solution :

(a)

Given that,

= 8.5

2 = 9

= 3

n = 45

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

+ / -  Z/2* ( /n)

.b

= 8.5+1.96*3/SQRT(45), and =8.5-1.96*3/SQRT(45)

Option b) is correct.

(b)

= 8.5

The point estimate for the average conversation length = 8.5 minutes .

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A random sample of 45 door-to-door salespersons were asked how long on average they were able...
A random sample of 45 door-to-door salespersons were asked how long on average they were able to talk to the potential customer. Their answers revealed a mean of 8.5 minutes with a variance of 9 minutes. Which Excel statement will calculate the margin of error? a. =9-8.5 b. =9-5 c. =8.5-5 d. =1.96*3/SQRT(45) e. =1.96*9/SQRT(45) f. None of the above.
A random sample of 13 students were asked how long it took them to complete a...
A random sample of 13 students were asked how long it took them to complete a certain exam. The mean length of time was 111.5 minutes, with a standard deviation of 83.0 minutes. Find the lower bound of the 90% confidence interval for the true mean length of time it would take for all students to complete the exam.
a random sample of 13 students were asked how long it took them to complete a...
a random sample of 13 students were asked how long it took them to complete a certain exam. The mean length of time was 122.1 minutes what the standard deviation of 66.6 minutes. Final lower bound of 90% confidence interval for the true mean length of time it would take for all students to complete the exam . Round to one decimal place
A random sample of 13 students were asked how long it took them to complete a...
A random sample of 13 students were asked how long it took them to complete a certain exam. The mean length of time was 117.6 minutes, with a standard deviation of 69.2 minutes. Find the lower bound of the 90% confidence interval for the true mean length of time it would take for all students to complete the exam. Round to one decimal place (for example: 108.1). Write only a number as your answer. Do not write any units.
A random sample of 13 students were asked how long it took them to complete a...
A random sample of 13 students were asked how long it took them to complete a certain exam. The mean length of time was 107.3 minutes, with a standard deviation of 82.5 minutes. Find the lower bound of the 90% confidence interval for the true mean length of time it would take for all students to complete the exam. Round to one decimal place (for example: 108.1). Write only a number as your answer. Do not write any units.
A simple random sample of ten college freshmen were asked how many hours of sleep they...
A simple random sample of ten college freshmen were asked how many hours of sleep they typically got per night. The results were 8.5 9.5 8 6.5 7.5 9 6 24 8.5 9.5 Notice that one joker said that he sleeps 24 hours a day. 99% confidence interval for the mean amount of sleep from the remaining values 6.72 Leave the outlier in and construct the 99% confidence interval. Round the answers to two decimal places. If the outlier is...
Sleeping outlier: A simple random sample of nine college freshmen were asked how many hours of...
Sleeping outlier: A simple random sample of nine college freshmen were asked how many hours of sleep they typically got per night. The results were 8.5 24 9 8.5 8 6.5 6 7.5 9.5 Notice that one joker said that he sleeps 24 a day. (a) The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 99.9% confidence interval for the mean amount of sleep from the remaining values. Round the answers to at...
Sleeping outlier: A simple random sample of eight college freshmen were asked how many hours of...
Sleeping outlier: A simple random sample of eight college freshmen were asked how many hours of sleep they typically got per night. The results were: 8.5 7.5 6.5 9 24 6 8.5 8 Notice that one joker said that he sleeps 24 hours a day. The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a confidence interval for the mean amount of sleep from the remaining values. Round the answers to two decimal places....
A random sample of 10 adults and women were chosen to determine if there is a...
A random sample of 10 adults and women were chosen to determine if there is a significant difference in the number of minutes men and women talk on the phone. The following data was collected, these numbers are the average numbers of minutes spent on the phone per day. Alpha = 0.05 1 2 3 4 5 6 7 8 9 10 Men 15 20 22 33 34 38 43 55 72 98 Women 23 43 53 61 74 85...
1. a. How large should the random sample of ink cartridges be so that at 92%...
1. a. How large should the random sample of ink cartridges be so that at 92% confidence interval, the maximum allowable error between the estimate obtained from the sample and the true population mean is within 0.1mm. Previous studies have indicated that the standard deviation of the lengths of the cartridges produced to be 2mm.​ 6 marks A sample of 400 Health statistics students taking a nationwide licensure examination revealed that an average score of 560 and a sample standard...