part 2 -
An entertainment company owns and operates movie theaters in Wyoming. The president of the company is concerned that film rentals are hurting the business. They directed a staff member to estimate the total number of films rented by households in wyoming in a particular month. A phone survey involving a random sample of 300 homes was conducted with the following results - xbar = 2.4 films. Sx = 1.6 films. Then the 90% confidence interval estimate for the total number of films rented by the 211,000 households in that particular month is?
Part 2 - The president of the company is uncomfortable with the precision of the estimates derived for the sample mean. They are not willing to tolerate a very large error. e = .04, how large would the sample size have to be if they specify a 98% confidence level?
I'll get you started and see if you can finish this one on your own.
CI90 = mean + or - z (sd divided by √n)
...where + or - z represents the 90% confidence interval using a z-table, sd = standard deviation, √ = square root, and n = sample size.
Determine what the 90% interval would be using a z-table, then substitute what you know into the formula.
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 represent the 98% confidence interval using a z-table, E = .04, ^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.
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