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Chapter 6: Problem 48

Find the sample size needed to give, with \(95 \%\) confidence, a margin of error within \(\pm 3 \%\) when estimating a proportion. First, find the sample size needed if we have no prior knowledge about the population proportion \(p\). Then find the sample size needed if we have reason to believe that \(p \approx 0.7\). Finally, find the sample size needed if we assume \(p \approx 0.9 .\) Comment on the relationship between the sample size and estimates of \(p\).

Short Answer

Expert verified
The required sample sizes for \(95\%\) confidence and \(\pm 3\%\) margin of error are approximately 1068 for \(p \approx 0.5\), 902 for \(p \approx 0.7\), and 1557 for \(p \approx 0.9\). As \(p\) gets close to 0 or 1, the sample size needed increases to ensure the same level of confidence and margin of error.

Step by step solution

01

Apply the Formula for Confidence Interval

The formula for a confidence interval for proportions is given by \[p \pm Z_{\frac{\alpha}{2}} \sqrt{\frac{p(1-p)}{n}}\], where:- \(p\) is the population proportion,- \(Z_{\frac{\alpha}{2}}\) is the z-value (in this case, \(Z_{\frac{\alpha}{2}}\) for a 95% confidence interval is 1.96 as taken from the Z-table according to the given confidence),- and \(n\) is the sample size. For \(p \approx 0.5\) (maximum variance when we have no prior knowledge), margin of error \(E = 0.03\), and \(Z_{\frac{\alpha}{2}} = 1.96\), we can set up the following equation:\[\pm E = \pm Z_{\frac{\alpha}{2}} \sqrt{\frac{p(1-p)}{n}}\] This becomes: \[0.03 = 1.96\sqrt{\frac{0.5(1-0.5)}{n}}\], which we will solve (simultaneously squaring both sides) for the unknown \(n\) in the next step.
02

Solve for \(n\) When \(p \approx 0.5\)

Squaring both sides of the equation, we get \(0.03^2 = (1.96)^2 \cdot \frac{0.5(1-0.5)}{n}\). Solving for \(n\), we get: \[n = \frac{(1.96)^2 \cdot 0.5(1-0.5)}{0.03^2}\], which simplifies to approximately \(n = 1067.1\). Because we cannot have a .1 sample, we always round up to the nearest whole number in order to maintain the proper level of confidence. Therefore, the sample size needed without prior knowledge about the population proportion is 1068.
03

Apply the Formula When \(p \approx 0.7\)

Proceed in the same process as in steps 1 and 2, but this time substituting \(p=0.7\) into the equation. Thus, setting\[0.03 = 1.96\sqrt{\frac{0.7(1-0.7)}{n}}\] and solving for \(n\), we have \[n = \frac{(1.96)^2 \cdot 0.7(1-0.7)}{0.03^2}\], which simplifies to approximately \(n = 901.5\). Again, we round up to the nearest whole number and thus, for \(p \approx 0.7\), the sample size needed is 902.
04

Apply the Formula When \(p \approx 0.9\)

Proceed in the same process as in steps 1, 2, and 3, but this time substituting \(p=0.9\) into the equation. Thus, setting\[0.03 = 1.96\sqrt{\frac{0.9(1-0.9)}{n}}\]and solving for \(n\), we have \[n = \frac{(1.96)^2 \cdot 0.9(1-0.9)}{0.03^2}\], which simplifies to approximately \(n = 1556.9\). In this case, we round up to the nearest whole number and thus, for \(p \approx 0.9\), the sample size needed is 1557.
05

Comment on the Relationship Between Sample Size and Estimates of \(p\)

Based on the calculations, it is clear that the sample size needed increases as the estimate of \(p\) deviates from 0.5 towards 0 or 1. At \(p=0.5\), the variance is maximized, meaning that there's the highest uncertainty about the expected outcome, hence lesser samples are needed compared to when \(p\) approximates 0.9, which requires more samples to ensure the same level of confidence and margin of error.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values used to estimate the true proportion of a population, derived from sampling data. When we say that a confidence interval has a 95% confidence level, it means if we were to take 100 different samples and compute a confidence interval for each sample, we expect about 95 of those intervals to contain the true proportion.

It's a way of quantifying uncertainty and is calculated based on the sample data and a specified confidence level. The confidence interval formula for proportions involves the sample proportion, the Z-score corresponding to the confidence level, and the sample size. This formula helps us understand how reliable our estimate of the population's true proportion is.
  • The center of the confidence interval is the sample estimate which gives the most likely value.
  • The margin of error shows the range above and below the sample estimate that's expected to encompass the true proportion.
In practice, if the interval of estimation is narrower, we have a more precise estimate of the population parameter.
Margin of Error
The margin of error is a statistical term that expresses the amount of random sampling error in a survey's results. It represents the range within which the true population parameter is expected to lie. In the context of a confidence interval, it is the amount we add and subtract from the sample proportion to find the upper and lower bounds.

For instance, with a margin of error of 卤3% in a proportion estimate, we can be assured that the actual proportion lies within 3 percentage points above or below the surveyed value, with a high level of confidence if the sample is appropriate.
  • Factors affecting the margin of error include sample size and variance within the data.
  • A larger sample size usually results in a smaller margin of error, assuming the level of confidence remains constant.
This means to achieve a smaller margin of error, one may need to increase the sample size if all other factors stay the same.
Proportion Estimation
Proportion estimation involves guessing the percentage of a population having a particular attribute. It's a common task in statistics where a proportion, like a percentage of people who approve of a policy, is estimated.

The accuracy of this estimation heavily relies on the sample size and variability in the results. When no prior information is known about the proportion, the most conservative approach is to assume a proportion of 0.5, maximizing uncertainty and leading to the largest sample size needed for a given margin of error.

As demonstrated in the exercise solution, knowing more about the probable proportion helps us adjust the required sample size. For instance:
  • If we approximate the proportion to be around 0.7, the required sample size reduces.
  • When estimating it to be close to 0.9, the sample size increases again.
This relationship indicates that estimation accuracy can improve when the variance is expected to be lower, requiring fewer samples.

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Most popular questions from this chapter

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 300 from a population with proportion 0.08

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

On page 1 1 in Section \(1.1,\) we describe studies to investigate whether there is evidence of pheromones (subconscious chemical signals) in female tears that affect sexual arousal in men. In one of the studies, \(^{71} 50\) men had a pad attached to the upper lip that contained either female tears or a salt solution dripped down the same female's face. Each subject participated twice, on consecutive days, once with tears and once with saline, randomized for order, and doubleblind. Testosterone levels were measured before sniffing and after sniffing on both days. While normal testosterone levels vary significantly between different men, average levels for the group were the same before sniffing on both days and after sniffing the salt solution (about \(155 \mathrm{pg} / \mathrm{mL}\) ) but were reduced after sniffing the tears (about \(133 \mathrm{pg} / \mathrm{mL}\) ). The mean difference in testosterone levels after sniffing the tears was 21.7 with standard deviation \(46.5 .\) (a) Why did the investigators choose a matchedpairs design for this experiment? (b) Test to see if testosterone levels are significantly reduced after sniffing tears? (c) Can we conclude that sniffing female tears reduces testosterone levels (which is a significant indicator of sexual arousal in men)?

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting invasive breast cancer? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=10\) from a population with \(p=0.2\)
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