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.
To determine the 90% confidence interval estimate for the total number of films rented by the 211,000 households in that particular month, follow these steps:
1. Find the z-value corresponding to the desired confidence level. For a 90% confidence level, the z-value is approximately 1.645. You can find this value by consulting a standard normal distribution table or using a calculator.
2. Substitute the given values into the confidence interval formula:
CI90 = x̄ ± z * (Sx / √n)
x̄ = 2.4 (sample mean)
Sx = 1.6 (sample standard deviation)
n = 300 (sample size)
z = 1.645
CI90 = 2.4 ± 1.645 * (1.6 / √300)
3. Calculate the standard error (SE): SE = (Sx / √n).
SE = 1.6 / √300
4. Calculate the margin of error by multiplying the z-value by the standard error: z * SE
Margin of error = 1.645 * (1.6 / √300)
5. Plug in the values to find the confidence interval range:
CI90 = 2.4 ± (1.645 * (1.6 / √300))
Simplify the expression, round the result to an appropriate decimal place, and write the final 90% confidence interval estimate.
To determine the sample size needed to achieve a 98% confidence level with an error margin of E = 0.04, follow these steps:
1. Find the z-value corresponding to the desired confidence level. For a 98% confidence level, the z-value is approximately 2.326.
2. Substitute the known values into the sample size formula:
n = [(z * sd) / E]^2
z = 2.326
sd = 1.6 (sample standard deviation)
E = 0.04
n = [(2.326 * 1.6) / 0.04]^2
3. Simplify the expression and calculate the sample size. Round the result to the next highest whole number, as the sample size must be a whole number.
Remember to double-check the calculations and ensure that you have used the correct z-values and formulas.