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?
Jason --
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To estimate the total number of films rented by households in Wyoming in a particular month, we can use the sample mean and standard deviation from the phone survey to construct a confidence interval.
Given:
Sample mean (x̄) = 2.4 films
Sample standard deviation (Sx) = 1.6 films
Number of households in Wyoming = 211,000
To calculate the 90% confidence interval estimate for the total number of films rented, we can use the formula:
CI = x̄ ± Z * (Sx / √n)
Where:
CI = Confidence interval
x̄ = Sample mean
Z = Z-score (corresponding to the desired confidence level)
Sx = Sample standard deviation
n = Sample size
First, we need to determine the Z-score for the desired confidence level of 90%. Looking up the Z-score from the Z-table, we find that for a 90% confidence level, the corresponding Z-score is approximately 1.645.
Plugging in the given values into the formula, we have:
CI = 2.4 ± 1.645 * (1.6 / √300)
Now, we can calculate the confidence interval:
CI = 2.4 ± 1.645 * (1.6 / √300)
= 2.4 ± 0.1785
The lower bound of the confidence interval is given by:
2.4 - 0.1785 = 2.2215
The upper bound of the confidence interval is given by:
2.4 + 0.1785 = 2.5785
Therefore, the 90% confidence interval estimate for the total number of films rented by the 211,000 households in that particular month is approximately 2.22 to 2.58 films.
Now, moving on to Part 2:
The president of the company wants to specify a 98% confidence level and a smaller margin of error (e = 0.04). To determine the required sample size, we can use the formula:
n = (Z^2 * S^2) / e^2
Where:
n = Required sample size
Z = Z-score (corresponding to the desired confidence level)
S = Sample standard deviation
e = Margin of error
First, we need to determine the Z-score for the desired confidence level of 98%. Looking up the Z-score from the Z-table, we find that for a 98% confidence level, the corresponding Z-score is approximately 2.33.
Plugging in the given values into the formula, we have:
n = (2.33^2 * 1.6^2) / 0.04^2
Now, we can calculate the required sample size:
n = (5.4289 * 2.56) / 0.0016
= 8.7175 / 0.0016
= 5457.19
Therefore, the sample size would have to be approximately 5458 households if the company specifies a 98% confidence level and wants a margin of error of 0.04.