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Chapter 8: Problem 44

A national opinion poll found that \(44 \%\) of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. The result was based on a small sample. (a) How large an SRS is required to obtain a margin of error of 0.03 (that is, \(\pm 3 \%\) ) in a \(99 \%\) confidence interval? Answer this question using the previous poll's result as the guessed value for \(\hat{p}\). (b) Answer the question in part (a) again, but this time use the conservative guess \(\hat{p}=0.5 .\) By how much do the two sample sizes differ?

Short Answer

Expert verified
The required sample sizes are 1818 for \(\hat{p} = 0.44\) and 1843 for \(\hat{p} = 0.5\), differing by 25.

Step by step solution

01

Understand the Problem

We need to determine the sample size (n) for a given margin of error (ME) in a confidence interval. Part (a) uses a guessed value of \(\hat{p} = 0.44\) and part (b) uses a conservative guess of \(\hat{p} = 0.5\). The confidence level is 99you'll find the sample size formulas and solve for each part.
02

Recall the Formula for Sample Size

The formula for the required sample size in terms of the proportion \(\hat{p}\) and margin of error \(E\) for a confidence level is: \[ n = \left( \frac{Z^* \sqrt{\hat{p}(1-\hat{p})}}{E} \right)^2\]where \(Z^*\) is the critical value from the standard normal distribution for the given confidence level.
03

Determine Z-Value for 99% Confidence Level

For a 99% confidence level, the critical value \(Z^*\) is approximately 2.576.
04

Solve Part (a) Using \(\hat{p} = 0.44\)

Substitute \(\hat{p} = 0.44\), \(E = 0.03\), and \(Z^* = 2.576\) into the formula:\[n = \left( \frac{2.576 \times \sqrt{0.44 \times 0.56}}{0.03} \right)^2\]Calculate the expression under the square root:\[\sqrt{0.44 \times 0.56} = 0.4964\]Now compute:\[n = \left( \frac{2.576 \times 0.4964}{0.03} \right)^2 = \left( 42.6475 \right)^2 \approx 1818\]Thus, an approximate sample size of 1818 is required.
05

Solve Part (b) Using \(\hat{p} = 0.5\)

Substitute \(\hat{p} = 0.5\), \(E = 0.03\), and \(Z^* = 2.576\) into the formula:\[n = \left( \frac{2.576 \times \sqrt{0.5 \times 0.5}}{0.03} \right)^2\]Calculate the expression under the square root:\[\sqrt{0.5 \times 0.5} = 0.5\]Now compute:\[n = \left( \frac{2.576 \times 0.5}{0.03} \right)^2 = \left( 42.9333 \right)^2 \approx 1843\]Thus, an approximate sample size of 1843 is required when \(\hat{p} = 0.5\).
06

Determine the Difference in Sample Sizes

Subtract the sample sizes obtained in parts (a) and (b):\[1843 - 1818 = 25\]The two sample sizes differ by 25.

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Key Concepts

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

Margin of Error
The margin of error is a measure of how much you expect your survey's results to reflect the true population value. It provides a range within which you can expect the true proportion to fall. Suppose you see a survey result with a margin of error of \(\pm 3\%\). This means the true value could be 3% more or 3% less than the reported percentage.
In the example problem, the margin of error is set at 0.03 or 3%. This means if the survey found a 45% approval rate, the true approval could be somewhere between 42% and 48%.
It's essential to choose an appropriate margin of error because it affects how large a sample you need for your study. Less margin means more accuracy, but it usually requires a larger sample.
Confidence Interval
A confidence interval is a range of values used to estimate a population parameter. It combines the sample estimate and its margin of error. If your confidence interval is 99%, it means that if you repeated your survey numerous times, 99% of the calculated confidence intervals would contain the true population parameter.
In our exercise, the confidence interval is aimed at 99%. This is very high and means you want to be very sure of your results. Higher confidence levels lead to wider intervals. Consequently, you need a larger sample size to maintain a reasonable precision in your estimates.
Getting the balance right between confidence level, precision, and sample size is a critical part of planning a statistical study.
Critical Value
The critical value, often denoted as \(Z^*\), is a factor that reflects how many standard deviations you must go out from the mean to cover the desired confidence level. The critical value is essential in calculating the margin of error and determining the sample size.
For a 99% confidence level, the critical value is approximately 2.576. This means you have a 0.5% probability in each tail of the distribution beyond this number.
Knowing the critical value is important because it influences how large your confidence interval will be. A higher critical value (due to a higher confidence level) will demand a larger sample size to achieve the same margin of error, underscoring the trade-off between certainty and feasibility.
Simple Random Sample
A simple random sample (SRS) is a method of selecting a sample from a larger population in such a way that every possible sample of the desired size has an equal chance of being chosen. It avoids biases in the selection process and is fundamental for achieving valid results that are representative of the population.
In the context of our problem, obtaining a simple random sample helps ensure that the findings are as close as possible to reflecting the opinions of the entire population.
SRS is vital in calculating sample size correctly because deviations or biases could lead to inaccurate estimates, impacting both the confidence interval and margin of error.

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Most popular questions from this chapter

A Gallup Poll found that only \(28 \%\) of American adults expect to inherit money or valuable possessions from a relative. The poll's margin of error was 卤3 percentage points at a \(95 \%\) confidence level. This means that (a) the poll used a method that gets an answer within \(3 \%\) of the truth about the population \(95 \%\) of the time. (b) the percent of all adults who expect an inheritance is between \(25 \%\) and \(31 \%\) (c) if Gallup takes another poll on this issue, the results of the second poll will lie between \(25 \%\) and \(31 \%\) (d) there's a \(95 \%\) chance that the percent of all adults who expect an inheritance is between \(25 \%\) and \(31 \%\). (e) Gallup can be \(95 \%\) confident that between \(25 \%\) and \(31 \%\) of the sample expect an inheritance.

The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a reading test scored from 0 to \(500 . \mathrm{A}\) score of 243 is a "basic" reading level and a score of 281 is "proficient." Scores for a random sample of 1470 eighth- graders in Atlanta had \(\bar{x}=240\) with standard deviation \(42.17 .^{24}\) (a) Calculate and interpret a \(99 \%\) confidence interval for the mean score of all Atlanta eighth-graders. (b) Based on your interval from part (a), is there good evidence that the mean for all Atlanta eighth-graders is less than the basic level? Explain.

You have measured the systolic blood pressure of an SRS of 25 company employees. A \(95 \%\) confidence interval for the mean systolic blood pressure for the employees of this company is \((122,138) .\) Which of the following statements is true? (a) \(95 \%\) of the sample of employees have a systolic blood pressure between 122 and 138 . (b) \(95 \%\) of the population of employees have a systolic blood pressure between 122 and 138 . (c) If the procedure were repeated many times, \(95 \%\) of the resulting confidence intervals would contain the population mean systolic blood pressure. (d) If the procedure were repeated many times, \(95 \%\) of the time the population mean systolic blood pressure would be between 122 and 138 . (e) If the procedure were repeated many times, \(95 \%\) of the time the sample mean systolic blood pressure would be between 122 and 138 .

PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste \(\mathrm{PTC}\) is inherited. About \(75 \%\) of Italians can taste \(\mathrm{PTC}\), for example. You want to estimate the proportion of Americans who have at least one Italian grandparent and who can taste PTC. (a) How large a sample must you test to estimate the proportion of PTC tasters within 0.04 with \(90 \%\) confidence? Answer this question using the \(75 \%\) estimate as the guessed value for \(\hat{p}\). (b) Answer the question in part (a) again, but this time use the conservative guess \(\hat{p}=0.5 .\) By how much do the two sample sizes differ?

A New York Times/CBS News Poll asked a random sample of U.S. adults the question, "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the \(95 \%\) confidence interval for the population proportion who favor such an amendment is (0.63,0.69) (a) Interpret the confidence interval. (b) What is the point estimate that was used to create the interval? What is the margin of error? (c) Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate this claim.
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