Question

In: Statistics and Probability

13. A television manufacturer takes a random sample of 30 customers and determines the times (after...

13.
A television manufacturer takes a random sample of 30 customers and determines the times (after purchasing) until the televisions need servicing have a mean of 6 years and a standard deviation of 1.2 years.
a) Determine the 95% confidence interval for the mean and explain whether it is reasonable for the company to claim the mean time until servicing is needed is 6.4 years.

b) Repeat a), but use a 90% interval.
c) Suppose instead a larger sample of 100 customers was taken and it had the same mean and standard deviation as the sample of 30. Determine both the 90% and 95% confidence intervals. Does either interval support the company’s claim that the mean time is 6.4 years until servicing is needed?

Solutions

Expert Solution

( a )

yes , it is reasonable for the company to claim the mean time until servicing is needed is 6.4 years. because

6.4 lies between 5.552 and 6.448

( b )

No , it is not reasonable for the company to claim the mean time until servicing is needed is 6.4 years. because

6.4 does not lies between 5.628 ad 6.372

( c )

No , it is not reasonable for the company to claim the mean time until servicing is needed is 6.4 years. because

6.4 does not lies between 5.762 and 6.238

No , it is not reasonable for the company to claim the mean time until servicing is needed is 6.4 years. because

6.4 does not lies between 5.801 and 6.199


Related Solutions

A quality control specialist for a restaurant chain takes a random sample of size 13 to...
A quality control specialist for a restaurant chain takes a random sample of size 13 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.54. Assume the underlying population is normally distributed. We wish to construct a 95% confidence interval for the true population mean for the amount of soda served. What is the error bound? (Round your answer to two decimal places.)
A certain gym selected a random sample of 10 customers and monitored the number of times...
A certain gym selected a random sample of 10 customers and monitored the number of times each customer used the workout facility in a​ one-month period. The data are shown in the table below. The​ gym's managers are considering a promotion in which they reward frequent users with a small gift. They have decided that they will only give gifts to those customers whose number of visits in a​ one-month period is 1 standard deviation above the mean. Find the...
The Hiatus retail outlet takes a random sample of 25 customers from a segment population of...
The Hiatus retail outlet takes a random sample of 25 customers from a segment population of 1,000 with a mean average transaction size of $80 normally distributed with a known population standard deviation of $20 per transaction. Find The 90% confidence interval for transaction size, and The 95% confidence interval for transaction size, and The 99% confidence interval for transaction size. What do these results indicate for management?
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the sample is from a normal population. Note that σ is unknown. 21 22 22 17 21 17 23 20 20 24 9 22 16 21 22 21 The sample mean is 19.875 and the sample standard deviation is 3.65. Which of the following represents the 80 percent confidence interval for µ? Select one: a. [13.75, 25.25] b. [18.65, 21.10] c. [19.55, 20.425] d. [18.8,...
Suppose a random sample of 30 customers is taken to test a company’s claim that 85%...
Suppose a random sample of 30 customers is taken to test a company’s claim that 85% of customers are satisfied with their dog food. Assume trials are independent. What is the probability at least 22 customers are satisfied?
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 26 16...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 26 16 25 18 17 24 18 23 14 20 10 18 19 13 17 16 Click here for the Excel Data File Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population. (Round your standard deviation answer to 4 decimal places and t-value to 3 decimal places. Round your answers to 3 decimal places.)    The 90% confidence...
Educational Television In a random sample of 186 people, 138 said that they watched educational television....
Educational Television In a random sample of 186 people, 138 said that they watched educational television. Find the 94%confidence interval of the true proportion of people who watched educational television. Use a graphing calculator. Round the answers to at least three decimal places.
(a) Suppose that a random sample of 387 television ads in the United Kingdom reveals that...
(a) Suppose that a random sample of 387 television ads in the United Kingdom reveals that 131 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor. (Round your answers to 3 decimal places.)    pˆp^ = [ ] The 95 percent confidence interval is [ , ]. (b) Suppose a random sample of 493 television ads in the United States reveals that 134...
In a random sample of n = 500 families owning television sets in the city of...
In a random sample of n = 500 families owning television sets in the city of Amman, Jordan, it is found that x = 340 subscribe to Netflix service. 1- Find a 95% confidence interval for the true proportion of families with television sets in this city that subscribe to Netflix service. 2- If ˆp is used as an estimate of p, can we be 95% confident that the error will not exceed 0.04. Explain Part (b)- The weights, in...
A simple random sample of 30 items resulted in a sample mean of 30. The population...
A simple random sample of 30 items resulted in a sample mean of 30. The population standard deviation is 15. a. Compute the 95% confidence interval for the population mean. Round your answers to one decimal place. ( ,  ) b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places. ( ,  )
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT