Thirty-three small communities in Connecticut (population near 10,000 each) gave an average of x = 138.5 reported cases of larceny per year. Assume that σ is known to be 42.9 cases per year.
(a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.)
lower limit
126.2
Correct: Your answer is correct.
upper limit
150.8
Correct: Your answer is correct.
margin of error
12.3
Correct: Your answer is correct.
(b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.)
lower limit
91.2
Incorrect: Your answer is incorrect.
upper limit
185.8
Incorrect: Your answer is incorrect.
margin of error
47.3
Incorrect: Your answer is incorrect.
(c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.)
lower limit
upper limit
margin of error
62.2
Incorrect: Your answer is incorrect.
(d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase?
As the confidence level increases, the margin of error increases.
As the confidence level increases, the margin of error remains the same.
As the confidence level increases, the margin of error decreases.
(e) Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?
As the confidence level increases, the confidence interval increases in length.
As the confidence level increases, the confidence interval remains the same length.
As the confidence level increases, the confidence interval decreases in length.
To find the confidence intervals and margin of error for this problem, you can use the formula:
Confidence interval = x ± (Z * σ / √n)
Where:
- x is the sample mean
- Z is the Z-score corresponding to the desired confidence level
- σ is the known population standard deviation
- n is the sample size
Let's solve each part of the problem:
(a) For a 90% confidence level, the Z-score corresponding to this confidence level is 1.645 (found in the Z-table). Given x = 138.5, σ = 42.9, and n = 10,000, we can calculate the confidence interval as follows:
Confidence interval = 138.5 ± (1.645 * 42.9 / √10,000) = 138.5 ± 6.2
The lower limit is 138.5 - 6.2 = 132.3, and the upper limit is 138.5 + 6.2 = 144.7. Therefore, the 90% confidence interval is [132.3, 144.7], and the margin of error is 6.2.
(b) For a 95% confidence level, the Z-score corresponding to this confidence level is 1.96. Using the same values as in part (a), we can calculate the confidence interval as follows:
Confidence interval = 138.5 ± (1.96 * 42.9 / √10,000) = 138.5 ± 8.3
The lower limit is 138.5 - 8.3 = 130.2, and the upper limit is 138.5 + 8.3 = 146.8. Therefore, the 95% confidence interval is [130.2, 146.8], and the margin of error is 8.3.
(c) For a 99% confidence level, the Z-score corresponding to this confidence level is 2.576. Using the same values as in part (a), we can calculate the confidence interval as follows:
Confidence interval = 138.5 ± (2.576 * 42.9 / √10,000) = 138.5 ± 10.9
The lower limit is 138.5 - 10.9 = 127.6, and the upper limit is 138.5 + 10.9 = 149.4. Therefore, the 99% confidence interval is [127.6, 149.4], and the margin of error is 10.9.
(d) As the confidence levels increase, the margins of error increase. This is because the Z-score increases as the desired confidence level increases, resulting in a larger range around the sample mean. The larger range means a larger margin of error.
(e) As the confidence levels increase, the confidence intervals increase in length. This is because higher confidence levels require a larger range around the sample mean to capture a higher percentage of the population. As a result, the confidence intervals become wider with higher confidence levels.