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Chapter 9: Problem 34

The mean age of the 42 presidents of the United States on the day of inauguration is 54.8 years with a standard deviation of 6.2 years. A researcher constructed a \(95 \%\) confidence interval for the mean age of presidents on inauguration day. He wrote that he was \(95 \%\) confident the mean age of the president on inauguration day is between 53.0 and 56.7 years of age. What is wrong with the researcher's analysis?

Short Answer

Expert verified
The correct range should be [52.93, 56.67]. The researcher's lower bound of 53.0 years is slightly off.

Step by step solution

01

Understand the Confidence Interval

The confidence interval provides a range within which the true population mean is expected to lie with a specified level of confidence, in this case, 95%. It is calculated using the sample mean, standard deviation, and the sample size.
02

Identify the Given Values

Given values are: Sample size ( = 42), Sample mean (\(\overline{x}\) = 54.8), Standard deviation (\(s\) = 6.2), and the confidence interval (CI) range: 53.0 to 56.7.
03

Calculate the Standard Error (SE)

The standard error (SE) is calculated by dividing the standard deviation by the square root of the sample size. \(SE = \frac{s}{\sqrt{n}} = \frac{6.2}{\sqrt{42}}\). Calculating this, \(SE \approx 0.9562\).
04

Determine the Critical Value for 95% CI

The critical value (z-score) for a 95% confidence interval is approximately 1.96 for a normal distribution.
05

Construct the Confidence Interval

Using the CI formula: \(CI = \overline{x} \pm (z \times SE)\), we get \(CI = 54.8 \pm (1.96 \times 0.9562)\) Simplifying, \(CI = 54.8 \pm 1.8734\) Therefore, the correct CI range is \([52.9266, 56.6734]\), rounded to 2 decimal places: \([52.93, 56.67]\).
06

Compare with the Given CI

The researcher's CI was 53.0 to 56.7 years. Compared to the calculated CI, the lower bound should be around 52.93 rather than 53.0. This means the researcher slightly miscalculated the minimum value of the CI.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mean age
The mean age is a measure of the central tendency of a set of values, which in this case, represents the average age of the 42 U.S. presidents at their inauguration. The mean age is calculated by adding up all the ages and then dividing by the number of presidents (which is 42). For instance, if we know the mean age is 54.8 years, it serves as the midpoint of our data set when plotting these ages on a graph. It's a critical starting point for understanding the data.
standard deviation
Standard deviation measures the dispersion or spread of a set of values around the mean. For the U.S. presidents' example, a standard deviation of 6.2 years indicates that the ages typically deviate by 6.2 years from the mean. A larger standard deviation would mean the ages are more spread out from the mean, while a smaller standard deviation would indicate that the ages are closer to the mean. Understanding this helps in assessing the variability in presidential inauguration ages.
sample size
The sample size (denoted as 'n') refers to the number of observations or data points in the sample. In our case, the sample size is 42, as it includes all presidents' ages at their inauguration. A larger sample size usually provides more accurate and reliable estimates of the population parameters, as it reduces the standard error. For instance, calculating the mean age from a sample of 42 is more reliable than from a sample of 5.
standard error
Standard error (SE) is a measure of the accuracy with which a sample represents a population. It is calculated by dividing the standard deviation by the square root of the sample size. In our scenario, the SE is approximately 0.9562. It shows how far the sample mean (54.8 years) is likely to be from the true population mean. Lower SE suggests a more accurate estimate, while a higher SE indicates more variability. The formula to find SE is: \(SE = \frac{s}{\sqrt{n}}\).
critical value
The critical value is a factor used to compute the margin of error in a confidence interval. For a 95% confidence interval in a normal distribution, it typically takes the value of approximately 1.96. It indicates the number of standard errors needed to develop the confidence interval. For the presidents' ages, the critical value ensures that our calculated interval covers the true mean 95% of the time. This value can change depending on the confidence level.
normal distribution
Normal distribution, often referred to as the bell curve, is a continuous probability distribution characterized by its symmetric bell-shaped curve. Most values fall close to the mean, with fewer values tailing off symmetrically on either side. The ages at inauguration, assuming they follow a normal distribution, allows us to apply z-scores and use standard tables to find critical values and confidence intervals. The properties of normal distribution facilitate confidence interval construction.

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Most popular questions from this chapter

Packer Fans In a Harris Poll conducted October \(20-25\) 2004,381 of 2114 randomly selected adults who follow professional football said the Green Bay Packers were their favorite team. (a) Verify that the requirements for constructing a confidence interval about \(\hat{p}\) are satisfied. (b) Construct a \(90 \%\) confidence interval for the proportion of adults who follow professional football who say the Green Bay Packers is their favorite team. Interpret this interval. (c) Construct a \(99 \%\) confidence interval for the proportion of adults who follow professional football who say the Green Bay Packers is their favorite team. Interpret this interval. (d) What is the effect of increasing the level of confidence on the width of the interval?

Suppose a certain population, \(A,\) has standard deviation \(\sigma_{A}=5,\) and a second population, \(B,\) has standard deviation \(\sigma_{B}=10 .\) How many times larger than population \(A^{\prime} \mathrm{s}\) sample size does population \(B\) 's need to be to estimate \(\mu\) with the same margin of error? [Hint: Compute \(\left.n_{B} / n_{A} \cdot\right]\)

A Gallup poll conducted May \(20-22,2005,\) asked 1006 Americans, "During the past year, about how many books, either hardcover or paperback, did you read either all or part of the way through?" Results of the survey indicated that \(\bar{x}=13.4\) books and \(s=16.6\) books. Construct a \(99 \%\) confidence interval for the mean number of books Americans read either all or part of during the preceding year. Interpret the interval.

Load the confidence interval for a mean (the impact of a confidence level) applet. (a) Set the shape to normal with mean \(=50\) and Std. Dev. \(=10 .\) Construct at least 1000 confidence intervals with \(n=10 .\) What proportion of the \(95 \%\) confidence intervals contain the population mean? What proportion did you expect to contain the population mean? (b) Repeat part (a). Did the same proportion of intervals contain the population mean? (c) Set the shape to normal with mean \(=50\) and Std. Dev. \(=10 .\) Construct at least 1000 confidence intervals with \(n=50 .\) What proportion of the \(95 \%\) confidence intervals contain the population mean? What proportion did you expect to contain the population mean? Does sample size have any impact on the proportion of intervals that capture the population mean? (d) Compare the width of the intervals for the samples of size \(n=50\) obtained in part (c) to the width of the intervals for the samples of size \(n=10\) obtained in part (a). Which are wider? Why? (b) Repeat part (a). Did the same proportion of intervals contain the population mean?

Pepcid A study of 74 patients with ulcers was conducted in which they were prescribed \(40 \mathrm{mg}\) of Pepcid \(^{\mathrm{TM}}\). After 8 weeks, 58 reported ulcer healing. (a) Obtain a point estimate for the proportion of patients with ulcers receiving Pepcid who will report ulcer healing. (b) Verify that the requirements for constructing a confidence interval about \(\hat{p}\) are satisfied. (c) Construct a \(99 \%\) confidence interval for the proportion of patients with ulcers receiving Pepcid who will report ulcer healing. (d) Interpret the confidence interval.
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