Asked by Jason L
It is known that the amount of time needed to change the oil in a car is normally distributed with a standard deviation of 5 minutes. A random sample of 100 oil changes yielded a sample mean of 22 minutes. Compute the 99% confidence interval estimate of the mean of the population.
Also determine the necessary sample size if you wish to be 99% confident and can tolerate an error of 1 minute.
Formula:
CI99 = mean + or - 2.575(sd divided by √n)
...where + or - 2.575 represents the 99% confidence interval using a z-table, sd = standard deviation, √ = square root, and n = sample size.
With your data:
CI99 = 22 + or - 2.575(5/√100)
Finish the calculation for your confidence interval estimate.
Formula for the second part:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 2.575 using a z-table to represent the 99% confidence interval, sd = 5, E = 1, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
Also determine the necessary sample size if you wish to be 99% confident and can tolerate an error of 1 minute.
Formula:
CI99 = mean + or - 2.575(sd divided by √n)
...where + or - 2.575 represents the 99% confidence interval using a z-table, sd = standard deviation, √ = square root, and n = sample size.
With your data:
CI99 = 22 + or - 2.575(5/√100)
Finish the calculation for your confidence interval estimate.
Formula for the second part:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 2.575 using a z-table to represent the 99% confidence interval, sd = 5, E = 1, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
Answers
There are no AI answers yet. The ability to request AI answers is coming soon!
There are no human answers yet. A form for humans to post answers is coming very soon!