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Hard times In June \(2010,\) a random poll of 800 working men found that \(9 \%\) had taken on a second job to help pay the bills. (www.careerbuilder.com) a) Estimate the true percentage of men that are taking on second jobs by constructing a \(95 \%\) confidence interval. b) A pundit on a TV news show claimed that only \(6 \%\) of working men had a second job. Use your confidence interval to test whether his claim is plausible given the poll data.

Short Answer

Expert verified
(a) (0.07, 0.11); (b) The claim is not plausible since 6% is not within the confidence interval.

Step by step solution

01

Identify Given Information

First, identify the numbers given in the exercise. We have 800 working men surveyed, and 9% of them reported having a second job. The confidence level provided is 95%.
02

Calculate Sample Proportion

Calculate the sample proportion of working men with second jobs. Divide the percentage by 100. \[ \hat{p} = \frac{9}{100} = 0.09 \]
03

Calculate Standard Error

The standard error (SE) for the sample proportion is calculated using the formula:\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Where \(n\) is the sample size of 800.Thus, \[ SE = \sqrt{\frac{0.09 \times (1-0.09)}{800}} \approx 0.0102 \]
04

Find the Z-Score for 95% Confidence

For a 95% confidence interval, the Z-score is approximately 1.96.
05

Construct the Confidence Interval

Calculate the margin of error using the formula:\[ ME = Z \times SE \]Where Z is the Z-score for 95% confidence.\[ ME = 1.96 \times 0.0102 \approx 0.02 \]Thus, the confidence interval can be calculated as:\[ \hat{p} \pm ME = 0.09 \pm 0.02 \]This gives a confidence interval:\[ (0.07, 0.11) \]
06

Evaluate the Pundit's Claim

The pundit claims that 6% of working men have second jobs. Check if 0.06 is within the calculated confidence interval of (0.07, 0.11). Since 0.06 is not within this range, his claim is not plausible.

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Key Concepts

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

Sample Proportion
One of the basic concepts in estimating a confidence interval for a population proportion is determining the sample proportion. The sample proportion, often represented by \(\hat{p}\), is a straightforward calculation derived from your raw data. In this case, the exercise involves a poll of 800 working men, with 9% of them indicating they have a second job. To convert this percentage into a proportion, you simply divide by 100. Hence, the sample proportion \(\hat{p}\) can be calculated as \(0.09\). Having this value gives you the first piece of the puzzle when constructing a confidence interval. The sample proportion acts as an estimate of the true population proportion, reflecting a snapshot of what's happening in the wider population. Understanding your sample proportion is key since it forms the baseline average around which your confidence interval is constructed. Key reminders:
  • Sample proportion is derived directly from your data.
  • Always translate percentages into proportions by dividing by 100.
Standard Error
Standard error represents the spread or variability in the sampling distribution of the sample proportion. Calculating the standard error is fundamental to understanding how your sample's proportion might vary if you were to take several similar samples from the population.To find the standard error for a sample proportion, use the formula:\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \(n\) is the sample size. For our exercise, with a sample size \(n = 800\) and a sample proportion \(\hat{p} = 0.09\), the calculation becomes: \[ SE = \sqrt{\frac{0.09 \times (1-0.09)}{800}} \approx 0.0102 \]The standard error gives you an idea of how much you can expect the sample proportion to fluctuate from the true population proportion. A smaller standard error generally indicates that the sample proportion is a more accurate reflection of the population proportion. Key insights:
  • Standard error quantifies how much your sample proportion might vary.
  • A lower standard error means higher precision of your estimate.
Z-score
In statistics, the Z-score is a measure that represents the number of standard deviations a data-point is from the mean. In the context of confidence intervals, the Z-score helps us to cover the desired confidence level around a statistic, such as the sample proportion. For a confidence interval, the Z-score reflects how far you need to go on either side of the sample proportion to capture a specific percentage of the data. For example, a 95% confidence interval implies that you are aiming to capture the true population proportion 95 times out of 100. The Z-score value for a 95% confidence level is typically 1.96. This means you're assuming that your sample proportion, plus or minus this number of standard deviations, will cover the true population proportion 95% of the time. Understanding the role of the Z-score:
  • The Z-score affects the width of your confidence interval.
  • Different confidence levels will utilize different Z-scores (e.g., 1.96 for 95%, 1.64 for 90%).
Margin of Error
The margin of error reflects the range within which the true population proportion likely falls, considering the variability and your chosen level of confidence. It's calculated using both the standard error and the Z-score.The formula for determining the margin of error (ME) is: \[ ME = Z \times SE \] Where Z is your Z-score and SE is the standard error. In the given example, with a Z-score of 1.96 and a standard error of approximately 0.0102, the margin of error becomes: \[ ME = 1.96 \times 0.0102 \approx 0.02 \]This value is then used to construct the confidence interval, around your sample proportion. Using the sample proportion 0.09, the confidence interval is: \(0.09 \pm 0.02\), which translates to \((0.07, 0.11)\).Helpful points:
  • Margin of error determines the extent of your interval's range.
  • It is contingent on both Z-score and standard error.

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Most popular questions from this chapter

Stop signs Highway safety engineers test new road signs, hoping that increased reflectivity will make them more visible to drivers. Volunteers drive through a test course with several of the new- and old-style signs and rate which kind shows up the best. a) Is this a one-tailed or a two-tailed test? Why? b) In this context, what would a Type I error be? c) In this context, what would a Type II error be? d) In this context, what is meant by the power of the test? e) If the hypothesis is tested at the \(1 \%\) level of significance instead of \(5 \%,\) how will this affect the power of the test? f) The engineers hoped to base their decision on the reactions of 50 drivers, but time and budget constraints may force them to cut back to 20. How would this affect the power of the test? Explain.

Significant again? A new reading program may reduce the number of elementary school students who read below grade level. The company that developed this program supplied materials and teacher training for a large-scale test involving nearly 8500 children in several different school districts. Statistical analysis of the results showed that the percentage of students who did not meet the grade-level goal was reduced from \(15.9 \%\) to \(15.1 \%\) The hypothesis that the new reading program produced no improvement was rejected with a P-value of 0.023 a) Explain what the P-value means in this context. b) Even though this reading method has been shown to be significantly better, why might you not recommend that your local school adopt it?

Loans Before lending someone money, banks must decide whether they believe the applicant will repay the loan. One strategy used is a point system. Loan officers assess information about the applicant, totaling points they award for the person's income level, credit history, current debt burden, and so on. The higher the point total, the more convinced the bank is that it's safe to make the loan. Any applicant with a lower point total than a certain cutoff score is denied a loan. We can think of this decision as a hypothesis test. Since the bank makes its profit from the interest collected on repaid loans, their null hypothesis is that the applicant will repay the loan and therefore should get the money. Only if the person's score falls below the minimum cutoff will the bank reject the null and deny the loan. This system is reasonably reliable, but, of course, sometimes there are mistakes. a) When a person defaults on a loan, which type of error did the bank make? b) Which kind of error is it when the bank misses an opportunity to make a loan to someone who would have repaid it? c) Suppose the bank decides to lower the cutoff score from 250 points to \(200 .\) Is that analogous to choosing a higher or lower value of \(\alpha\) for a hypothesis test? Explain. d) What impact does this change in the cutoff value have on the chance of each type of error?

Ads A company is willing to renew its advertising contract with a local radio station only if the station can prove that more than \(20 \%\) of the residents of the city have heard the ad and recognize the company's product. The radio station conducts a random phone survey of 400 people. a) What are the hypotheses? b) The station plans to conduct this test using a \(10 \%\) leve of significance, but the company wants the significance level lowered to \(5 \%\). Why? c) What is meant by the power of this test? d) For which level of significance will the power of this test be higher? Why? e) They finally agree to use \(\alpha=0.05,\) but the company proposes that the station call 600 people instead of the 400 initially proposed. Will that make the risk of Type II error higher or lower? Explain.

Errors For each of the following situations, state whether a Type I, a Type II, or neither error has been made. Explain briefly. a) A bank wants to see if the enrollment on their website is above \(30 \%\) based on a small sample of customers. It tests \(\mathrm{H}_{0}: p=0.3\) vs. \(\mathrm{H}_{\mathrm{A}}: p>0.3\) and rejects the null hypothesis. Later the bank finds out that actually \(28 \%\) of all customers enrolled. b) A student tests 100 students to determine whether other students on her campus prefer Coke or Pepsi and finds no evidence that preference for Coke is not \(0.5 .\) Later, a marketing company tests all students on campus and finds no difference. c) A pharmaceutical company tests whether a drug lifts the headache relief rate from the \(25 \%\) achieved by the placebo. The test fails to reject the null hypothesis because the P-value is 0.465. Further testing shows that the drug actually relieves headaches in \(38 \%\) of people.

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