In A Study Of The Income Of US Factory Workers: A Random Sam

Question

In A Study Of The Income Of US Factory Workers A Random Sample O 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? In a study of the starting salary of college graduates with degrees in Accounting, a random sample of 40 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. A telephone poll of 900 American adults asked "where would you rather go in your spare time?" One response, by 250 adults, was "a movie." Compute and explain a 95% confidence interval estimate of the proportion of all American adults who would respond "a movie."

Dr. Jack HW Helper · Accepted Answer

The construction and interpretation of confidence intervals are fundamental concepts in inferential statistics, allowing researchers and analysts to estimate population parameters based on sample data. This paper discusses the calculations of confidence intervals for population mean and proportions, illustrating how variations in confidence levels influence the width of the intervals and their implications in practical scenarios. Confidence Interval for Population Mean: Factory Workers Income The first problem centers on estimating the true mean income of U.S. factory workers based on a sample of 100 workers. Given that the population standard deviation is known and sample data is normally distributed, the confidence interval can be directly calculated using the z-distribution. The sample mean ($\bar{x}$) is $35,000, and the population standard deviation ($\sigma$) is $4,500. For a 90% confidence level, the z-value is approximately 1.645; for 95%, approximately 1.96; and for 99%, approximately 2.576. The formula for the confidence interval is: CI = $\bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$ Calculating each: 90% CI: $35,000 \pm 1.645 \times \frac{4,500}{\sqrt{100}} = 35,000 \pm 1.645 \times 450 = 35,000 \pm 740.25$, which yields (34,259.75, 35,740.25). 95% CI: $35,000 \pm 1.96 \times 450 = 35,000 \pm 882$, resulting in (34,118, 35,882). 99% CI: $35,000 \pm 2.576 \times 450 = 35,000 \pm 1,159.2$, resulting in (33,840.8, 36,159.2). These intervals give a range where the true population mean income is likely to lie with specified confidence. As the confidence level increases, the z-value increases, resulting in a wider interval that reflects greater uncertainty about the estimate. This behavior is reasonable because a higher confidence level demands a broader range to ensure the true parameter is included, embodying the trade-off between precision and confidence. Confidence Interval for Population Mean: College Graduate Salaries The second pr

In A Study Of The Income Of US Factory Workers: A Random Sam

In A Study Of The Income Of US Factory Workers A Random Sample O

Paper For Above instruction

Confidence Interval for Population Mean: Factory Workers Income

Confidence Interval for Population Mean: College Graduate Salaries

Confidence Interval for Population Proportion: Response to Survey Question

Discussion on Changes in Interval Width with Confidence Levels

Conclusion

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