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Chapter 8: Problem 7

Calculate the margin of error in estimating a binomial proportion for each of the following values of \(n\). Use \(p=.5\) to calculate the standard error of the estimator. a. \(n=30\) b. \(n=100\) c. \(n=400\) d. \(n=1000\)

Short Answer

Expert verified
Answer: The margins of error for the given sample sizes are approximately: - 0.1790 for a sample size of 30 - 0.0980 for a sample size of 100 - 0.0490 for a sample size of 400 - 0.0310 for a sample size of 1000.

Step by step solution

01

Formula for Standard Error

To calculate the standard error of a binomial proportion, we use the formula: $$SE = \sqrt{\frac{p(1-p)}{n}}$$ where \(SE\) stands for standard error, \(p\) is the probability of success, and \(n\) is the sample size.
02

Calculate the Standard Error for Each Sample Size

Given that \(p=0.5\) for each of the sample sizes, we plug the values of \(p\) and \(n\) into the formula from Step 1 for each of the cases: a. \(n=30:\) $$SE_{30} = \sqrt{\frac{0.5(1-0.5)}{30}} = \sqrt{\frac{0.25}{30}} \approx 0.0913$$ b. \(n=100:\) $$SE_{100} = \sqrt{\frac{0.5(1-0.5)}{100}} = \sqrt{\frac{0.25}{100}} \approx 0.05$$ c. \(n=400:\) $$SE_{400} = \sqrt{\frac{0.5(1-0.5)}{400}} = \sqrt{\frac{0.25}{400}} \approx 0.025$$ d. \(n=1000:\) $$SE_{1000} = \sqrt{\frac{0.5(1-0.5)}{1000}} = \sqrt{\frac{0.25}{1000}} \approx 0.0158$$
03

Determine the Z-score for the Desired Confidence Level

A common confidence level used is \(95\%\). To determine the corresponding Z-score, we look it up in a Z-table or use a calculator. For a \(95\%\) confidence level, the Z-score is approximately \(1.96\).
04

Calculate the Margin of Error for Estimating the Binomial Proportion for Each Sample Size

Using the Z-score and standard errors from Steps 2 and 3, we calculate the margin of error by multiplying them: a. \(n=30:\) $$ME_{30} = 1.96 \times SE_{30} = 1.96 \times 0.0913 \approx 0.1790$$ b. \(n=100:\) $$ME_{100} = 1.96 \times SE_{100} = 1.96 \times 0.05 \approx 0.0980$$ c. \(n=400:\) $$ME_{400} = 1.96 \times SE_{400} = 1.96 \times 0.025 \approx 0.0490$$ d. \(n=1000:\) $$ME_{1000} = 1.96 \times SE_{1000} = 1.96 \times 0.0158 \approx 0.0310$$ Thus, we have calculated the margin of error for estimating a binomial proportion using \(p = 0.5\) for the sample sizes of \(30\), \(100\), \(400\), and \(1000\).

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Most popular questions from this chapter

Don't Americans know that eating pizza and french fries leads to being overweight? In the same American Demographics article referenced in Exercise \(8.98,\) a survey of women who are the main meal preparers in their households reported these results: \(\cdot$$90 \%\) know that obesity causes health problems. \(\cdot$$80 \%\) know that high fat intake may lead to health problems. \(\cdot$$86 \%\) know that cholesterol is a health problem. \(\cdot$$88 \%\) know that sodium may have negative effects on health. a. Suppose that this survey was based on a random sample of 750 women. How accurate do you expect the percentages given above to be in estimating the actual population percentages? (HINT: If these are the only four percentages for which you need a margin of error, a conservative estimate for \(p\) is \(p \approx .80 .)\) b. If you want to decrease your sampling error to \(\pm 1 \%,\) how large a sample should you take?

An experiment was conducted to compare two diets \(\mathrm{A}\) and \(\mathrm{B}\) designed for weight reduction. Two groups of 30 overweight dieters each were randomly selected. One group was placed on diet \(\mathrm{A}\) and the other on diet \(\mathrm{B},\) and their weight losses were recorded over a 30 -day period. The means and standard deviations of the weight-loss measurements for the two groups are shown in the table. Find a \(95 \%\) confidence interval for the difference in mean weight loss for the two diets. Interpret your confidence interval. $$\begin{array}{ll}\text { Diet } A & \text { Diet B } \\\\\hline \bar{x}_{A}=21.3 & \bar{x}_{B}=13.4 \\\s_{A}=2.6 & s_{B}=1.9\end{array}$$

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Born between 1980 and \(1990,\) Generation Next have lived in a post-Cold War world and a time of relative economic prosperity in America, but they have also experienced September 11 th and the fear of another attack, two Gulf Wars, the tragedy at Columbine High School, Hurricane Katrina, and the increasing polarization of public discourse. More than any who came before, Generation Next is engaged with technology, and the vast majority is dependent upon it. \({ }^{15}\) Suppose that a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Construct a \(98 \%\) confidence interval for the difference in the proportions of female and male students who decided to attend college in order to make more money. b. What does it mean to say that you are "98\% confident"? c. Based on the confidence interval in part a, can you conclude that there is a difference in the proportions of female and male students who decided to attend college in order to make more money?

A random sampling of a company's monthly operating expenses for \(n=36\) months produced a sample mean of \(\$ 5474\) and a standard deviation of \(\$ 764\). Find a \(90 \%\) upper confidence bound for the company's mean monthly expenses.
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