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Chapter 8: Problem 9

Calculate the margin of error in estimating a binomial proportion \(p\) using samples of size \(n=100\) and the following values for \(p\) : a. \(p=.1\) b. \(p=.3\) c. \(p=.5\) d. \(p=.7\) e. \(p=.9\) f. Which of the values of \(p\) produces the largest margin of error?

Short Answer

Expert verified
Answer: The largest margin of error occurs when \(p=0.5\), and it is approximately \(0.0980\).

Step by step solution

01

a. Calculate the margin of error for \(p=0.1\)

To calculate the margin of error for \(p=0.1\), substitute the values into the formula: Margin of error = \(1.96 \times \sqrt{\frac{0.1(1-0.1)}{100}}\) By calculating this, we get: Margin of error = \(1.96 \times \sqrt{\frac{0.1(0.9)}{100}} \approx 0.0573\)
02

b. Calculate the margin of error for \(p=0.3\)

To calculate the margin of error for \(p=0.3\), substitute the values into the formula: Margin of error = \(1.96 \times \sqrt{\frac{0.3(1-0.3)}{100}}\) By calculating this, we get: Margin of error = \(1.96 \times \sqrt{\frac{0.3(0.7)}{100}} \approx 0.0904\)
03

c. Calculate the margin of error for \(p=0.5\)

To calculate the margin of error for \(p=0.5\), substitute the values into the formula: Margin of error = \(1.96 \times \sqrt{\frac{0.5(1-0.5)}{100}}\) By calculating this, we get: Margin of error = \(1.96 \times \sqrt{\frac{0.5(0.5)}{100}} \approx 0.0980\)
04

d. Calculate the margin of error for \(p=0.7\)

To calculate the margin of error for \(p=0.7\), substitute the values into the formula: Margin of error = \(1.96 \times \sqrt{\frac{0.7(1-0.7)}{100}}\) By calculating this, we get: Margin of error = \(1.96 \times \sqrt{\frac{0.7(0.3)}{100}} \approx 0.0904\)
05

e. Calculate the margin of error for \(p=0.9\)

To calculate the margin of error for \(p=0.9\), substitute the values into the formula: Margin of error = \(1.96 \times \sqrt{\frac{0.9(1-0.9)}{100}}\) By calculating this, we get: Margin of error = \(1.96 \times \sqrt{\frac{0.9(0.1)}{100}} \approx 0.0573\)
06

f. Determine which value of \(p\) produces the largest margin of error

Comparing the calculated margins of error for each value of \(p\): \(p=0.1\) -> Margin of error \(\approx 0.0573\) \(p=0.3\) -> Margin of error \(\approx 0.0904\) \(p=0.5\) -> Margin of error \(\approx 0.0980\) \(p=0.7\) -> Margin of error \(\approx 0.0904\) \(p=0.9\) -> Margin of error \(\approx 0.0573\) It is clear that \(p=0.5\) produces the largest margin of error, which is approximately \(0.0980\).

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Most popular questions from this chapter

Acid rain, caused by the reaction of certain air pollutants with rainwater, appears to be a growing problem in the northeastern United States. (Acid rain affects the soil and causes corrosion on exposed metal surfaces.) Pure rain falling through clean air registers a pH value of 5.7 (pH is a measure of acidity: 0 is acid; 14 is alkaline). Suppose water samples from 40 rainfalls are analyzed for \(\mathrm{pH}\) and \(\bar{x}\) and \(s\) are equal to 3.7 and \(.5,\) respectively Find a \(99 \%\) confidence interval for the mean \(\mathrm{pH}\) in rainfall and interpret the interval. What assumption must be made for the confidence interval to be valid?

Independent random samples were selected from populations 1 and 2 . The sample sizes, means, and variances are as follows: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 35 & 49 \\\\\text { Sample Mean } & 12.7 & 7.4 \\\\\text { Sample Variance } & 1.38 & 4.14\end{array}$$ a. Find a \(95 \%\) confidence interval for estimating the difference in the population means \(\left(\mu_{1}-\mu_{2}\right) .\) b. Based on the confidence interval in part a, can you conclude that there is a difference in the means for the two populations? Explain.

One of the major costs involved in planning a summer vacation is the cost of lodging. Even within a particular chain of hotels, costs can vary substantially depending on the type of room and the amenities offered. \(^{4}\) Suppose that we randomly select 50 billing statements from each of the computer databases of the Marriott, Radisson, and Wyndham hotel chains, and record the nightly room rates. $$\begin{array}{lccc} & \text { Marriott } & \text { Radisson } & \text { Wyndham } \\\\\hline \text { Sample average } & \$ 170 & \$ 145 & \$ 150 \\\\\text { Sample standard deviation } & 17.5 & 10 & 16.5\end{array}$$ a. Describe the sampled population(s). b. Find a point estimate for the average room rate for the Marriott hotel chain. Calculate the margin of error. c. Find a point estimate for the average room rate for the Radisson hotel chain. Calculate the margin of error. d. Find a point estimate for the average room rate for the Wyndham hotel chain. Calculate the margin of error. e. Display the results of parts \(\mathrm{b}, \mathrm{c},\) and d graphically, using the form shown in Figure \(8.5 .\) Use this display to compare the average room rates for the three hotel chains.

Refer to the Interpreting Confidence Intervals applet. a. Suppose that you have a random sample of size \(n=50\) from a population with unknown mean \(\mu\) and known standard deviation \(\sigma=35 .\) Calculate the half width of a \(95 \%\) confidence interval for \(\mu\). What would the width of this interval be? b. Use the button to create a single confidence interval for \(\mu\). What is the width of this interval? Compare your results to the calculation you did in part a.

Find a \(99 \%\) lower confidence bound for the binomial proportion \(p\) when a random sample of \(n=400\) trials produced \(x=196\) successes.
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