Question

An auto mechanic knows the average time it takes to replace a defective car radiator is...

An auto mechanic knows the average time it takes to replace a defective car radiator is 70 minutes with a standard deviation of 12 minutes. This average is based on a random sample of 50. Which Excel statements will construct a 90% confidence interval for the time needed to replace a defective car radiator?

a.

=70+1.96*12/SQRT(50), and =70-1.96*12/SQRT(50)

b.

=70+1.96*3.46/SQRT(50), and =70-1.96*3.46/SQRT(50)

c.

=70+1.645*3.46/SQRT(50), and =70-1.645*3.46/SQRT(50)

d.

=70+1.645*12/SQRT(50), and =70-1.645*12/SQRT(50)

e.

None of the above.

Homework Answers

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 70

Population standard deviation = = 12

Sample size = n = 50

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z 0.05 = 1.645

Confidence interval is = Z/2* ( /n)

= 70 + 1.645 * (12 / 50) and 70 - 1.645 * (12 / 50)

d)

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
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...
The total amount of time it takes a JC Auto Repair mechanic to complete a full-service,...
The total amount of time it takes a JC Auto Repair mechanic to complete a full-service, 16-point oil change is distributed as a uniform random variable, ranging from 18 to 32 minutes. What is the standard deviation of service time? Report your answer to 2 decimal places using conventional rounding rules. ANSWER: What is the probability that the amount of time to complete a full service, 16-point oil change will be less than 28 minutes? Report your answer to 4...
The total amount of time it takes a JC Auto Repair mechanic to complete a full-service,...
The total amount of time it takes a JC Auto Repair mechanic to complete a full-service, 16-point oil change is distributed as a uniform random variable, ranging from 18 to 32 minutes. What is the standard deviation of service time? Report your answer to 2 decimal places using conventional rounding rules. ANSWER: What is the probability that the amount of time to complete a full service, 16-point oil change will be less than 28 minutes? Report your answer to 4...
A car manufacturer has determined it takes an average time of 56 minutes to produce a...
A car manufacturer has determined it takes an average time of 56 minutes to produce a car. The population standard deviation is assumed to be 6 minutes. The company pays a bonus to the workers for every car produced in 48 minutes or less. Assuming that the production time is normally distributed, answer the following questions. Let X = production time of a randomly selected car. Round all probabilities to four decimal places and times to two decimal places. a)...
A car manufacturer has determined it takes an average time of 51 minutes to produce a...
A car manufacturer has determined it takes an average time of 51 minutes to produce a car. The population standard deviation is assumed to be 7 minutes. The company pays a bonus to the workers for every car produced in 43 minutes or less. Assuming that the production time is normally distributed, answer the following questions. Let X = production time of a randomly selected car. (Round probabilities to four decimals and times to two decimals.) a) What is the...
Suppose you are interested in measuring the amount of time, on average, it takes you to...
Suppose you are interested in measuring the amount of time, on average, it takes you to make your commute to school. You've estimated that the average time is 39 minutes with a standard deviation of 5.611 minutes. Assuming that your estimated parameters are correct and the commute time is normally distributed, what is the probability that the average commute time of 12 random days is greater than 40.72 minutes? 1) 0.8598 2) 0.1441 3) 0.3796 4) 0.8559 5) 0.6204 A...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
b) If we know that the length of time it takes a college student to fingd...
b) If we know that the length of time it takes a college student to fingd a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot. c) The mean score of all pro golfers for a particular course has...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT