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Chapter 8: Problem 13

The following sample data are from a normal population: 10,8,12,15,13,11,6,5 a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation? c. With \(95 \%\) confidence, what is the margin of error for the estimation of the population mean? d. What is the \(95 \%\) confidence interval for the population mean?

Short Answer

Expert verified
The point estimate of the population mean is 10, and the point estimate of the population standard deviation is 3.49. The margin of error for the estimation of the population mean with a 95% confidence level is 2.872. The 95% confidence interval for the population mean is \((7.128, 12.872)\).

Step by step solution

01

1. Find the Sample Mean

To find the point estimate for the population mean, we need to calculate the sample mean. The sample mean is the average of all the values in the sample. We will add all the values together and then divide by the number of values in the sample: \(\bar{x} = \frac{\sum{x}}{n}\) where \(x\) represents each data point and \(n\) is the number of data points.
02

2. Find the Sample Standard Deviation

To find the point estimate for the population standard deviation, we need to calculate the sample standard deviation. The formula for the sample standard deviation is: \(s = \sqrt{\frac{\sum{(x-\bar{x})^2}}{n-1}}\) where \(x\) represents each data point, \(\bar{x}\) is the sample mean, and \(n\) is the number of data points.
03

3. Calculate the Margin of Error

The margin of error is given by the formula: \(E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\) where \(E\) is the margin of error, \(t_{\alpha/2}\) is the t-score corresponding to a 95% confidence level, \(s\) is the sample standard deviation, and \(n\) is the number of data points. We will use a t-distribution table to find the t-score.
04

4. Find the 95% Confidence Interval for the Population Mean

The 95% confidence interval for the population mean is given by the formula: \(\bar{x} \pm E\) where \(\bar{x}\) is the sample mean and \(E\) is the margin of error. Now let's calculate these values using the given data.
05

1. Calculate the Sample Mean

Using the given data: \[10, 8, 12, 15, 13, 11, 6, 5\] We can find the sample mean: \(\bar{x} = \frac{10+8+12+15+13+11+6+5}{8} = \frac{80}{8} = 10\) The point estimate of the population mean is 10.
06

2. Calculate the Sample Standard Deviation

Using the same data, we calculate the sample standard deviation: \(s = \sqrt{\frac{(10-10)^2+(8-10)^2+(12-10)^2+(15-10)^2+(13-10)^2+(11-10)^2+(6-10)^2+(5-10)^2}{8-1}}\) \(s = \sqrt{\frac{0+4+4+25+9+1+16+25}{7}} = \sqrt{\frac{84}{7}} = 3.49\) The point estimate of the population standard deviation is 3.49.
07

3. Calculate the Margin of Error

The given data has 8 elements, so the degrees of freedom is \(n-1 = 8-1=7\). We will use a t-distribution table to find the t-score. With 7 degrees of freedom and 95% confidence level, the t-score is 2.365. Now we will calculate the margin of error: \(E = 2.365 \cdot \frac{3.49}{\sqrt{8}} \approx 2.872\) The margin of error for the estimation of the population mean with 95% confidence is 2.872.
08

4. Calculate the 95% Confidence Interval for the Population Mean

Now we will calculate the 95% confidence interval for the population mean: \(\bar{x} \pm E = 10 \pm 2.872\) This gives us the confidence interval of \((7.128, 12.872)\). The 95% confidence interval for the population mean is \((7.128, 12.872)\).

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Most popular questions from this chapter

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent Mercer survey showed that \(52 \%\) of U.S. employers were likely to require higher employee contributions for health care coverage in 2009 (Business Week, February 16,2009 ). Suppose the survey was based on a sample of \(800 \mathrm{com}\) panies. Compute the margin of error and a \(95 \%\) confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009

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An online survey by ShareBuilder, a retirement plan provider, and Harris Interactive reported that \(60 \%\) of female business owners are not confident they are saving enough for retirement (SmallBiz, Winter 2006). Suppose we would like to do a follow-up study to determine how much female business owners are saving each year toward retirement and want to use \(\$ 100\) as the desired margin of error for an interval estimate of the population mean. Use \(\$ 1100\) as a planning value for the standard deviation and recommend a sample size for each of the following situations. a A \(90 \%\) confidence interval is desired for the mean amount saved. b. \(A 95 \%\) confidence interval is desired for the mean amount saved. c. \(\quad\) A \(99 \%\) confidence interval is desired for the mean amount saved. d. When the desired margin of error is set, what happens to the sample size as the confidence level is increased? Would you recommend using a \(99 \%\) confidence interval in this case? Discuss.

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