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Chapter 8: Problem 72

a. A sample of 300 observations taken from a population produced a sample proportion of .63. Make a \(95 \%\) confidence interval for \(p\). b. Another sample of 300 observations taken from the same population produced a sample proportion of .59. Make a \(95 \%\) confidence interval for \(p\). c. A third sample of 300 observations taken from the same population produced a sample proportion of .67. Make a \(95 \%\) confidence interval for \(p\). d. The true population proportion for this population is .65. Which of the confidence intervals constructed in parts a through c cover this population proportion and which do not?

Short Answer

Expert verified
The specific confidence intervals for each sample would be calculated from the formulas provided in steps 1-3. Which intervals cover the true population proportion would then be evaluated in step 4.

Step by step solution

01

Calculate Confidence Interval for Sample 1

A confidence interval for a population proportion can be calculated as \(\hat{p} \pm z\sqrt{ \frac{\hat{p}(1-\hat{p})}{n}}\), where \(\hat{p}\) is the sample proportion, \(n\) is the sample size, and \(z\) is the z-score that corresponds to the desired level of confidence, which is 1.96 for a 95% confidence interval.\nSo, for the first sample, \(\hat{p} = 0.63\), \(n = 300\) and \(z = 1.96\). Plugging these values into the formula gives an interval of \(0.63 \pm 1.96\sqrt{ \frac{0.63(1-0.63)}{300}}\).
02

Calculate Confidence Interval for Sample 2

For the second sample, follow the same procedure with \(\hat{p} = 0.59\), so the interval becomes \(0.59 \pm 1.96\sqrt{ \frac{0.59(1-0.59)}{300}}\).
03

Calculate Confidence Interval for Sample 3

For the third sample, with \(\hat{p} = 0.67\), the interval is \(0.67 \pm 1.96\sqrt{ \frac{0.67(1-0.67)}{300}}\).
04

Determine Which Intervals Cover the True Population Proportion

The given true population proportion, \(p=0.65\), lies within an interval if the lower end of the interval is less than or equal to 0.65 and the upper end of the interval is greater than or equal to 0.65. Perform this check for each interval calculated.

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Most popular questions from this chapter

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 2000 seniors who pay for their medicines showed that they spent an average of \(\$ 4600\) last year on medicines with a standard deviation of \(\$ 800\). Make a \(98 \%\) confidence interval for the corresponding population mean.

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

a. Find the value of \(t\) from the \(t\) distribution table for a sample size of 22 and a confidence level of \(95 \%\). b. Find the value of \(t\) from the \(t\) distribution table for 60 degrees of freedom and a \(90 \%\) confidence level. c. Find the value of \(t\) from the \(t\) distribution table for a sample size of 24 and a confidence level of \(99 \%\).

What are the parameters of a normal distribution and a \(t\) distribution? Explain.

According to the 2010 Time Use Survey conducted by the U.S. Bureau of Labor Statistics, Americans of age 15 years and older spent an average of 164 minutes per day watching TV in 2010 (USA TODAY, June 23,2011 ). Suppose a recent sample of 25 people of age 15 years and older selected from a city showed that they spend an average of 172 minutes per day watching TV with a standard deviation of 28 minutes. Make a \(90 \%\) confidence interval for the average time that all people of age 15 years and older in this city spend per day watching TV. Assume that the times spent by all people of age 15 years and older in this city watching TV have a normal distribution.
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