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91Ó°ÊÓ

The Bureau of Labor Statistics (www.bls.gov/opub \(/ \mathrm{ted} / 2014 / \mathrm{ted}_{-} 20141112 . \mathrm{htm},\) retrieved December \(\left.13,2016\right)\) reported that \(3.8 \%\) of college graduates were unemployed in October 2013 and \(3.1 \%\) of college graduates were unemployed in October 2014 . Suppose that the reported percentages were based on independently selected representative samples of 500 college graduates in each of these two years. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportions of college graduates who were unemployed in these 2 years.

Short Answer

Expert verified
We are 95% confident that the true difference in proportions of college graduates who were unemployed in 2013 and 2014 lies between -0.0038 and 0.0178. Since the interval contains 0, we cannot conclude that there is a statistically significant difference in the unemployment rates of college graduates for these two years.

Step by step solution

01

Identify the given data

We are given the following data: - The proportional unemployment rate in 2013 (p1) = 3.8%, or 0.038 - The proportional unemployment rate in 2014 (p2) = 3.1%, or 0.031 - The sample size for 2013 (n1) = 500 - The sample size for 2014 (n2) = 500
02

Find the difference in sample proportions

First, we need to find the difference in the sample proportions. This can be calculated as follows: \(p' = p_1 - p_2\) \(p' = 0.038 - 0.031\) \(p' = 0.007\)
03

Calculate the standard error of the difference in proportions

Now, we calculate the standard error of the difference in proportions using the following formula: \(\mathrm{SE}(p') = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\) \(\mathrm{SE}(p') = \sqrt{\frac{(0.038)(0.962)}{500} + \frac{(0.031)(0.969)}{500}}\) \(\mathrm{SE}(p') \approx 0.0055\)
04

Find the critical z-score for a 95% confidence interval

For a 95% confidence interval, the critical z-score (Z) is 1.96.
05

Calculate the margin of error

Now, we calculate the margin of error using the following formula: \(ME = Z \times \mathrm{SE}(p')\) \(ME = 1.96 \times 0.0055\) \(ME \approx 0.0108\)
06

Construct the 95% confidence interval for the difference in proportions

Finally, construct the confidence interval using the difference in sample proportions (p') and the margin of error (ME): Confidence interval = \((p' - ME, p' + ME)\) Confidence interval = \((0.007 - 0.0108, 0.007 + 0.0108)\) Confidence interval \(\approx (-0.0038, 0.0178)\)
07

Interpret the confidence interval

We can interpret this 95% confidence interval as follows: We are 95% confident that the true difference in proportions of college graduates who were unemployed in 2013 and 2014 lies between -0.0038 and 0.0178. Since the interval contains 0, we cannot conclude that there is a statistically significant difference in the unemployment rates of college graduates for these two years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference in Proportions
The difference in proportions refers to the disparity between two group percentages, in this case, the unemployment rates of college graduates in different years. In statistics, understanding the difference in proportions allows us to compare two groups and determine if a meaningful difference exists between them. We start by calculating the difference in sample proportions, which is the subtraction of one sample proportion from another. In our exercise:
  • 2013 unemployment rate: 3.8% or 0.038
  • 2014 unemployment rate: 3.1% or 0.031
  • Difference in proportion, \( p' = 0.038 - 0.031 = 0.007 \)
This simple subtraction gives us the initial measure of how much the unemployment rates differed between 2013 and 2014 among the sampled college graduates. This difference, although small, forms the basis for further statistical analysis.
Unemployment Rate
The unemployment rate is an essential economic indicator that measures the percent of the labor force that is jobless and actively seeking employment. For this exercise, we focus on a specific demographic: college graduates. Understanding employment statistics for graduates is crucial, as it reflects the health of the job market for this educated group. In 2013, the unemployment rate for college graduates was reported at 3.8%, and it decreased to 3.1% in 2014.
These percentages were derived from samples of 500 graduates each year, providing a snapshot of the employment landscape for that cohort. Changes in the unemployment rate can indicate economic trends, shifts in the job market, or the effectiveness of economic policies.
Sample Size
The sample size is the number of observations used to compute estimates like the unemployment rate. In statistical studies, the choice of sample size is crucial as it impacts the precision and accuracy of the estimates. In the given problem, each year's unemployment rate was based on a sample of 500 college graduates.
Using a larger sample size generally increases the reliability of the estimate as it more accurately reflects the entire population. However, practical limitations such as time and cost often restrict sample sizes to manageable numbers. In our situation, a sample size of 500 is a significant number that provides a good balance, allowing for meaningful data analysis without being too costly or time-consuming to collect.
Critical Z-score
A critical z-score is a value from the standard normal distribution used in hypothesis testing and confidence intervals. For a 95% confidence interval, the critical z-score is 1.96. This value determines the number of standard deviations a data point must be from the mean to be considered statistically significant.
In our exercise, the critical z-score helps calculate the confidence interval for the difference in unemployment proportions between 2013 and 2014. By multiplying the standard error by the critical z-score, we derive the margin of error, which helps construct the confidence interval. The choice of a 95% confidence level means we're 95% certain that the interval contains the true difference in unemployment proportions. This level of confidence is common in statistical analysis, balancing reliability with meaningful conclusions.

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Most popular questions from this chapter

Women diagnosed with breast cancer whose tumors have not spread may be faced with a decision between two surgical treatments-mastectomy (removal of the breast) or lumpectomy (only the tumor is removed). In a long-term study of the effectiveness of these two treatments, 701 women with breast cancer were randomly assigned to one of two treatment groups. One group received mastectomies, and the other group received lumpectomies and radiation. Both groups were followed for 20 years after surgery. It was reported that there was no statistically significant difference in the proportion surviving for 20 years for the two treatments (Associated Press, October 17,2002 ). Suppose that this conclusion was based on a \(90 \%\) confidence interval for the difference in treatment proportions. Which of the following three statements is correct? Explain why you chose this statement. Statement 1: Both endpoints of the confidence interval were negative. Statement 2: The confidence interval included \(0 .\) Statement 3: Both endpoints of the confidence interval were positive.

A headline that appeared in Woman's World stated "Black Currant Oil Curbs Hair Loss!" (Woman's World, April 4, 2016). This claim was based on an experiment described in the paper "Effect of a Nutritional Supplement on Hair Loss in Women" (Journal of Cosmetic Dermatology [2015]: 76-82). In this experiment, women with stage 1 hair loss were assigned at random to one of two groups. One group was a control group who did not receive a nutritional supplement. Of the 39 women in this group, 20 showed increased hair density at the end of the study period. Those in the second group received a nutritional supplement that included fish oil, black currant oil, vitamin E, vitamin C, and lycopene. Of the 80 women in the supplement group, 70 showed increased hair density at the end of the study period. a. Is there convincing evidence that the proportion with increased hair density is greater for the supplement treatment than for the control treatment? Test the appropriate hypotheses using a 0.01 significance level. b. Write a few sentences commenting on the headline that appeared in Woman's World. c. Based on the description of the actual experiment and the result from your hypothesis test in Part (a), suggest a more appropriate headline.

According to the U.S. Census Bureau (www.census.gov), the percentage of U.S. residents living in poverty in 2015 was \(12.2 \%\) for men and \(14.8 \%\) for women. These percentages were estimates based on data from large representative samples of men and women. Suppose that the sample sizes were 1200 for men and 1000 for women. You would like to use the survey data to estimate the difference in the proportion living in poverty in 2015 for men and women. (Hint: See Example \(11.2 .\).) a. Answer the four key questions (QSTN) for this problem. What method would you consider based on the answers to these questions? b. Use the five-step process for estimation problems \(\left(\mathrm{EMC}^{3}\right)\) to calculate and interpret a \(90 \%\) large- sample confidence interval for the difference in the proportion living in poverty in 2015 for men and women.

A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel's web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is greater for those who reserve a room online? Test the appropriate hypotheses using a significance level of \(0.05 .\) (Hint: See Example \(11.4 .)\)

In the experiment described in the article "Study Points to Benefits of Knee Replacement Surgery Over Therapy Alone" (The New York Times, October 21,2015 ), adults who were considered candidates for knee replacement were followed for one year. Suppose that 200 patients were randomly assigned to one of two groups. One hundred were assigned to a group that had knee replacement surgery followed by therapy and the other half were assigned to a group that did not have surgery but did receive therapy. After one year, \(86 \%\) of the patients in the surgery group and \(68 \%\) of the patients in the therapy only group reported pain relief. Is there convincing evidence that the proportion experiencing pain relief is greater for the surgery treatment than for the therapy treatment? Use a significance level of 0.05

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