1. In a study of the income of U.S. factory workers, a random sample of 100 workers shows a sample mean of $35,000. Assume that the population standard deviation is $4,500, and that the population is normally distributed.
A) Compute the 90%, 95% and 99% confidence intervals for the unknown population mean.
B) Briefly discuss what happens to the width of the interval estimate as the confidence level increases. Why does this seem reasonable?
2. In a study of the starting salary of college graduates with degrees in Accounting, a random sample of 80 graduates shows a sample mean of $36,000 and a sample standard deviation of $2,500. Assume that the population is normally distributed.
A) Compute and explain a 95% confidence interval estimate of the population mean starting salary for Accounting graduates.
3. A telephone poll of 950 American adults asked "where would you rather go in your spare time?" One response, by 300 adults, was "a movie". Compute and explain a 95% confidence interval estimate of the proportion of all American adults who would respond "a movie".
1A) Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05, .025, & .005, respectively) and their Z scores.
% = mean ± Z SEm
SEm = SD/√n
2A) Same process.
3) Data lacking.
1. A) To compute the confidence intervals, we can use the formula:
Sample Mean ± Margin of Error
where the margin of error is determined by the desired confidence level and the standard deviation of the population.
For a 90% confidence level, we need to find the z-score corresponding to the desired confidence level. The z-score can be found using a standard normal distribution table or calculator. The z-score for a 90% confidence level is approximately 1.645.
Using the formula:
Margin of Error = z * (Population Standard Deviation / sqrt(n))
where z is the z-score, Population Standard Deviation is $4,500, and n is the sample size (100), we can calculate the margin of error for the 90% confidence interval:
Margin of Error = 1.645 * ($4,500 / sqrt(100))
≈ 1.645 * $450
≈ $739.05
Therefore, the 90% confidence interval for the population mean is:
$35,000 ± $739.05
($34,260.95, $35,739.05)
Similarly, you can calculate the 95% and 99% confidence intervals by finding the corresponding z-scores and plugging them into the formula.
For a 95% confidence level, the z-score is approximately 1.96.
For a 99% confidence level, the z-score is approximately 2.576.
B) As the confidence level increases, the width of the interval estimate also increases. This is because a higher confidence level requires a greater margin of error, which results in a wider range of values in the confidence interval. It seems reasonable because higher confidence levels provide stronger assurance that the true population parameter lies within the interval estimate, but at the cost of having a wider range of possible values.
2. A) To compute the confidence interval for the population mean starting salary for Accounting graduates, we can use the formula:
Sample Mean ± Margin of Error
The margin of error is determined by the desired confidence level and the standard deviation of the sample.
Since the sample standard deviation is known ($2,500), we can use the z-distribution to find the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score is approximately 1.96.
Using the formula:
Margin of Error = z * (Sample Standard Deviation / sqrt(n))
where z is the z-score, Sample Standard Deviation is $2,500, and n is the sample size (80), we can calculate the margin of error:
Margin of Error = 1.96 * ($2,500 / sqrt(80))
≈ 1.96 * $278.39
≈ $546.20
Therefore, the 95% confidence interval for the population mean starting salary for Accounting graduates is:
$36,000 ± $546.20
($35,453.80, $36,546.20)
3. To compute the confidence interval estimate of the proportion of all American adults who would respond "a movie", we can use the formula:
Sample Proportion ± Margin of Error
where the margin of error is determined by the desired confidence level and the sample size.
The sample proportion is calculated as the number of "a movie" responses divided by the sample size. In this case: Sample Proportion = 300/950 ≈ 0.3168
Using the formula:
Margin of Error = z * sqrt((Sample Proportion * (1 - Sample Proportion)) / n)
where z is the z-score corresponding to the desired confidence level and n is the sample size, we can calculate the margin of error.
Since the confidence level is 95%, the corresponding z-score is approximately 1.96.
Margin of Error = 1.96 * sqrt((0.3168 * (1 - 0.3168))/950)
≈ 1.96 * 0.0168
≈ 0.0329
Therefore, the 95% confidence interval estimate of the proportion of all American adults who would respond "a movie" is:
0.3168 ± 0.0329
(0.2839, 0.3497)