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Chapter 8: Problem 28

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.

Short Answer

Expert verified
In summary, for the desired margin of error of $100 and using the planning value for the standard deviation of $1100, the required sample sizes for the different confidence intervals are as follows: 30 for a 90% confidence interval, 48 for a 95% confidence interval, and 89 for a 99% confidence interval. As the confidence level increases, so does the required sample size. The decision on whether to use a 99% confidence interval depends on the study's context and the available resources.

Step by step solution

01

a. 90% Confidence Interval

1. Find the Z-score value for a 90% confidence interval We will use a standard normal distribution table to find the Z-score for a 90% confidence interval. The Z-score corresponding to a two-tailed 90% confidence interval is 1.645. 2. Use the formula to calculate the sample size Let's plug the values into the formula: \(n = (\frac{1.645 * 1100}{100})^2\) \(n = 29.54\) Since we can't have a fraction of a person, we round up to the next whole number: n = 30 So, a sample size of 30 is required for a 90% confidence interval.
02

b. 95% Confidence Interval

1. Find the Z-score value for a 95% confidence interval We will use a standard normal distribution table to find the Z-score for a 95% confidence interval. The Z-score corresponding to a two-tailed 95% confidence interval is 1.960. 2. Use the formula to calculate the sample size Let's plug the values into the formula: \(n = (\frac{1.960 * 1100}{100})^2\) \(n = 47.04\) Since we can't have a fraction of a person, we round up to the next whole number: n = 48 So, a sample size of 48 is required for a 95% confidence interval.
03

c. 99% Confidence Interval

1. Find the Z-score value for a 99% confidence interval We will use a standard normal distribution table to find the Z-score for a 99% confidence interval. The Z-score corresponding to a two-tailed 99% confidence interval is 2.576. 2. Use the formula to calculate the sample size Let's plug the values into the formula: \(n = (\frac{2.576 * 1100}{100})^2\) \(n = 88.21\) Since we can't have a fraction of a person, we round up to the next whole number: n = 89 So, a sample size of 89 is required for a 99% confidence interval.
04

d. Relationship between sample size and confidence level

As the confidence level increases, the required sample size also increases. This is because a higher confidence level requires a wider confidence interval to encompass the true population parameter. Hence, a larger sample size is needed to provide enough data to maintain the desired margin of error at a higher confidence level. In this case, the required sample size increased from 30 for a 90% confidence interval to 89 for a 99% confidence interval. The marginal benefit of increased confidence comes at the cost of a significant increase in the required sample size. As for the recommendation, it depends on the context of the study and the resources available. If it is important to have a very high level of confidence in the estimations, then a 99% confidence interval might be warranted. However, if resources (time, money, etc.) are limited, it might be better to opt for a lower confidence level with a more manageable sample size, such as a 95% confidence interval.

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Most popular questions from this chapter

Is your favorite TV program often interrupted by advertising? CNBC presented statistics on the average number of programming minutes in a half-hour sitcom (CNBC, February 23,2006)\(.\) The following data (in minutes) are representative of its findings. $$\begin{array}{lll} 21.06 & 22.24 & 20.62 \\ 21.66 & 21.23 & 23.86 \\ 23.82 & 20.30 & 21.52 \\ 21.52 & 21.91 & 23.14 \\ 20.02 & 22.20 & 21.20 \\ 22.37 & 22.19 & 22.34 \\ 23.36 & 23.44 & \end{array}$$ Assume the population is approximately normal. Provide a point estimate and a \(95 \%\) confidence interval for the mean number of programming minutes during a half-hour television sitcom.

The range for a set of data is estimated to be 36 a. What is the planning value for the population standard deviation? b. At \(95 \%\) confidence, how large a sample would provide a margin of error of \(3 ?\) c. At \(95 \%\) confidence, how large a sample would provide a margin of error of \(2 ?\)

Find the \(t\) value(s) for each of the following cases. a. Upper tail area of .025 with 12 degrees of freedom b. Lower tail area of .05 with 50 degrees of freedom c. Upper tail area of .01 with 30 degrees of freedom d. Where \(90 \%\) of the area falls between these two \(t\) values with 25 degrees of freedom e. Where \(95 \%\) of the area falls between these two \(t\) values with 45 degrees of freedom

A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desired margin of error is .03 at \(98 \%\) confidence. a. How large a sample should be selected if it is anticipated that roughly \(70 \%\) of the firm's card holders carry a nonzero balance at the end of the month? b. How large a sample should be selected if no planning value for the proportion could be specified?

The average cost of a gallon of unleaded gasoline in Greater Cincinnati was reported to be \(\$ 2.41(\text {The Cincinnati Enquirer, } \text { February } 3,2006\) ). During periods of rapidly changing prices, the newspaper samples service stations and prepares reports on gasoline prices frequently. Assume the standard deviation is \(\$ .15\) for the price of a gallon of unleaded regular gasoline, and recommend the appropriate sample size for the newspaper to use if it wishes to report a margin of error at \(95 \%\) confidence. a. Suppose the desired margin of error is \(\$ .07\) b. Suppose the desired margin of error is \(\$ .05\) c. Suppose the desired margin of error is \(\$ .03\)
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