Older people often have a hard time finding work. AARP reported on the number of weeks it takes a worker aged 55 plus to find a job. The data on number of weeks spent searching for a job contained in the file JobSearch are consistent with the AARP findings \((A A R P\) Bulletin, April 2008). a. Provide a point estimate of the population mean number of weeks it takes a worker aged 55 plus to find a job. b. \(\quad\) At \(95 \%\) confidence, what is the margin of error? c. What is the \(95 \%\) confidence interval estimate of the mean? d. Discuss the degree of skewness found in the sample data. What suggestion would you make for a repeat of this study?

Short Answer

Expert verified
The point estimate of the population mean, denoted by \(\bar{X}\), can be found by calculating the sample mean. The 95% confidence margin of error is found using the formula 1.96 * (s / sqrt(n)), where s is the sample standard deviation and n is the sample size. The 95% confidence interval estimate of the mean is given by \((\bar{X} - \text{Margin of Error}, \bar{X} + \text{Margin of Error})\). To analyze the skewness of the sample data, calculate the skewness statistic, considering future studies should increase sample size, investigate factors causing skewness, and possibly use transformations or non-parametric methods if skewness is significant.

Step by step solution

01

Calculate the sample mean

Using the data in the "JobSearch" file, add up all the number of weeks taken by each individual to find a job and divide by the total number of individuals in the sample. This will give you the sample mean, denoted by \(\bar{X}\).
02

Calculate the sample standard deviation

Using the same data, calculate the sample standard deviation, denoted by s. This is done by first finding the variance. To calculate the variance, subtract the sample mean \(\bar{X}\) from each data point, square the results, and then find the average of these squared differences. Finally, take the square root of the variance to find the standard deviation.
03

Find the critical value for a 95% confidence interval

We are asked to find the margin of error and confidence interval at a 95% confidence level. To do this, we need to find the critical value, which is the z-score corresponding to the desired level of confidence. For a 95% confidence interval, the critical value is approximately 1.96.
04

Calculate the margin of error

To calculate the margin of error, use the formula: Margin of Error = Critical Value (z) * (Sample Standard Deviation / sqrt(sample size)) In our case, we have: Margin of Error = 1.96 * (s / sqrt(n)) where s is the sample standard deviation and n is the sample size.
05

Calculate the 95% confidence interval estimate of the mean

The confidence interval estimate is the range within which we expect the true population mean to lie with 95% confidence. To calculate this interval, use the formula: Confidence Interval = sample mean ± margin of error In our case, the confidence interval will be: \((\bar{X} - \text{Margin of Error}, \bar{X} + \text{Margin of Error})\)
06

Analyze the skewness and provide suggestions

To analyze the skewness of the sample data, first, calculate the skewness statistic using appropriate software or statistical tools. A skewness value near zero indicates a roughly symmetric distribution. Positive skewness indicates that the distribution is skewed to the right, while negative skewness indicates a left-skewed distribution. For a future study on this subject, consider the following suggestions: 1. Increase the sample size to improve the precision of estimates and reduce the margin of error. 2. Investigate factors that could potentially cause skewness in the data and explore partitioning the data to analyze the subgroups separately. 3. Consider using transformations or non-parametric methods if the sample data display significant skewness, which could violate the normality assumptions for calculating confidence intervals.

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Key Concepts

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

Point Estimate
A point estimate provides a single value as an estimate of a population parameter. In the context of job search analysis for older workers, the point estimate represents an estimate of the population mean - that is, the average number of weeks all older workers might expect to spend looking for a job. To calculate this from a sample, we add all the weeks reported by individuals and divide by the number of respondents. This calculated mean, often denoted as \( \bar{X} \), is our point estimate. It's a valuable figure because it gives us a reference point for the central tendency of the dataset. Using point estimates allows researchers and policy makers to make informed guesses about the population from the sample data, though it is key to remember that a point estimate is susceptible to sampling error and does not reflect any measure of variability on its own.
Confidence Interval
The confidence interval adds context to the point estimate by providing a range within which we can say, with a certain level of confidence, that the true population parameter lies. For older workers' job search data, constructing a 95% confidence interval around the sample mean tells us that if we were to draw many samples and construct intervals in the same way, 95 out of 100 of those intervals would contain the actual average weeks it would take for older workers to find jobs.

The formula for this interval is the sample mean \( \bar{X} \) plus and minus the margin of error, with the margin defined as the critical value from the standard normal distribution (often 1.96 for 95% confidence) multiplied by the sample standard deviation divided by the square root of the sample size. This interval is crucial because rather than just estimating a single value, it acknowledges the possibility of variation and provides a plausible range for the population parameter.
Sample Standard Deviation
Sample standard deviation (denoted as s) is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. When analyzing job search time for older workers, calculating the sample standard deviation is essential for two reasons: It helps us understand the variability of the job search duration among individuals in the sample, and it's vital for the calculation of both the margin of error and the confidence interval.

We calculate s by determining the variance, which involves subtracting the sample mean from each observation, squaring these differences, averaging them out, and finally taking the square root. Understanding the standard deviation gives context to the point estimate, as it identifies how much the individual data points are diverging from the average.
Data Skewness
Data skewness is a measure of the asymmetry of the probability distribution of a real-valued random variable. In simple terms, skewness tells us if the data is evenly distributed around the mean, or if it's leaning towards the left (negative skew) or right (positive skew).

In the context of the job search for older workers, analyzing skewness is pivotal to interpret how evenly the job search duration is spread among the sample. Skewness can influence the interpretation of the mean; for example, a positively skewed distribution may have a mean that is higher than the median, which would indicate a presence of outliers on the high end (people taking a very long time to find a job).

It's important to assess skewness because significant skewness in data can impact the validity of the statistical analyses (like the construction of confidence intervals) that assume normal distribution. Researchers may need to consider data transformation or different analytical techniques to accommodate or understand the skewness in their data, especially for developing interventions or policy recommendations.

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Most popular questions from this chapter

Disney's Hannah Montana: The Movie opened on Easter weekend in April 2009. Over the three-day weekend, the movie became the number-one box office attraction (The Wall Street Journal, April 13,2009 ). The ticket sales revenue in dollars for a sample of 25 theaters is as follows. $$\begin{array}{rrrrr} 20,200 & 10,150 & 13,000 & 11,320 & 9,700 \\ 8,350 & 7,300 & 14,000 & 9,940 & 11,200 \\ 10,750 & 6,240 & 12,700 & 7,430 & 13,500 \\ 13,900 & 4,200 & 6,750 & 6,700 & 9,330 \\ 13,185 & 9,200 & 21,400 & 11,380 & 10,800 \end{array}$$ a. What is the \(95 \%\) confidence interval estimate for the mean ticket sales revenue per theater? Interpret this result. b. Using the movie ticket price of \(\$ 7.16\) per ticket, what is the estimate of the mean number of customers per theater? c. The movie was shown in 3118 theaters. Estimate the total number of customers who saw Hannah Montana: The Movie and the total box office ticket sales for the threeday weekend.

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30 a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of \(.02 ?\) Use \(95 \%\) confidence. b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population? c. What is the \(95 \%\) confidence interval for the proportion of smokers in the population?

Nielsen Media Research conducted a study of household television viewing times during the 8 P.M. to 11 P.M. time period. The data contained in the file named Nielsen are consistent with the findings reported (The World Almanac, 2003). Based upon past studies, the population standard deviation is assumed known with \(\sigma=3.5\) hours. Develop a \(95 \%\) confidence interval estimate of the mean television viewing time per week during the 8 P.M. to 11 P.M. time period.

According to statistics reported on \(\mathrm{CNBC}\), a surprising number of motor vehicles are not covered by insurance (CNBC, February 23,2006 ). Sample results, consistent with the CNBC report, showed 46 of 200 vehicles were not covered by insurance. a. What is the point estimate of the proportion of vehicles not covered by insurance? b. Develop a \(95 \%\) confidence interval for the population proportion.

A Phoenix Wealth Management/Harris Interactive survey of 1500 individuals with net worth of \(\$ 1\) million or more provided a variety of statistics on wealthy people (Business Week, September 22,2003 ). The previous three-year period had been bad for the stock market, which motivated some of the questions asked. a. The survey reported that \(53 \%\) of the respondents lost \(25 \%\) or more of their portfolio value over the past three years. Develop a \(95 \%\) confidence interval for the proportion of wealthy people who lost \(25 \%\) or more of their portfolio value over the past three years. b. The survey reported that \(31 \%\) of the respondents feel they have to save more for retirement to make up for what they lost. Develop a \(95 \%\) confidence interval for the population proportion. c. Five percent of the respondents gave \(\$ 25,000\) or more to charity over the previous year. Develop a \(95 \%\) confidence interval for the proportion who gave \(\$ 25,000\) or more to charity. d. Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to \(\bar{p} ?\) When the same sample is being used to estimate a variety of proportions, which of the proportions should be used to choose the planning value \(p^{*} ?\) Why do you think \(p^{*}=.50\) is often used in these cases?

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