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Chapter 7: Problem 60

In a simple random sample of 1200 Americans age 20 and over, the proportion with diabetes was found to be \(0.115\) (or \(11.5 \%)\). a. What is the standard error for the estimate of the proportion of all Americans age 20 and over with diabetes? b. Find the margin of error, using a \(95 \%\) confidence level, for estimating this proportion. c. Report the \(95 \%\) confidence interval for the proportion of all Americans age 20 and over with diabetes. d. According to the Centers for Disease Control and Prevention, nationally, \(10.7 \%\) of all Americans age 20 or over have diabetes. Does the confidence interval you found in part c support or refute this claim? Explain.

Short Answer

Expert verified
Standard error: approximately 0.0083. Margin of error at 95% confidence level: approximately 0.01628. Confidence interval: approximately 0.09872 to 0.13128. The computed confidence interval supports the claim that 10.7% of all Americans age 20 or over have diabetes.

Step by step solution

01

Compute Standard Error

To compute the standard error of the proportion, use the formula \(\sqrt{p(1-p)/n}\) where \(p\) is the sample proportion and \(n\) is the sample size. Plugging in \(p = 0.115\) and \(n = 1200\), we find the standard error to be approximately \(0.0083\).
02

Compute the Margin of Error

The margin of error for a 95% confidence level can be computed as the product of the standard error and the z-score corresponding to a 95% confidence level (which is approximately \(1.96\)). With a standard error of about \(0.0083\) from the previous step, the margin of error is approximately \(1.96 \times 0.0083 = 0.01628\).
03

Compute the Confidence Interval

The confidence interval can be expressed as \(p \pm\) (margin of error). Plug in the calculated results from the previous steps, we get a confidence interval of approximately \(0.115 \pm 0.01628\), or an interval from \(0.09872\) to \(0.13128\) (rounded to five decimal places).
04

Analyze the Results

With the given figure of 10.7% or \(0.107\), it falls within the computed confidence interval (\(0.09872\) to \(0.13128\)). Therefore, the confidence interval supports the claim from the Centers for Disease Control and Prevention that \(10.7 \%\) of all Americans age 20 or over have diabetes.

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Key Concepts

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

Standard Error
The standard error provides insight into the variability of a sample proportion. It's like taking a snapshot of how much variation you might expect if you were to sample again. The formula to calculate the standard error for a proportion is given by:
  • \( \text{SE} = \sqrt{\frac{p(1-p)}{n}} \)
In this formula, \(p\) represents the sample proportion, and \(n\) is the sample size.
For our example, where the proportion \(p = 0.115\) and the sample size \(n = 1200\), we find
  • \( \text{SE} = \sqrt{\frac{0.115 \times (1 - 0.115)}{1200}} \approx 0.0083 \)
This tells us how much we might expect the sample proportion to vary from the true population proportion. A smaller standard error indicates more confidence in the accuracy of the sample proportion.
Margin of Error
The margin of error extends the standard error to account for the desired level of confidence. For a 95% confidence level, we use a z-score of approximately 1.96. This z-score represents how far we stretch on either side of the sample proportion to create the interval.
To find the margin of error, we multiply the standard error by the z-score:
  • \( \text{Margin of Error} = 1.96 \times \text{SE} \)
Using our standard error of about 0.0083, the margin of error is:
  • \( 1.96 \times 0.0083 \approx 0.01628 \)
This margin of error tells us how much our sample proportion may deviate when we generalize to the population. It's an essential piece of the confidence interval puzzle, highlighting the range's boundaries.
Sample Proportion
The sample proportion is the backbone of our confidence interval calculations. It鈥檚 simply the fraction of the sample with the characteristic of interest. In our example, the sample proportion is 11.5%, or 0.115.
This proportion serves as our best estimate of the population proportion. However, to determine the range within which the true population proportion lies, we must calculate the confidence interval. This interval stretches from the sample proportion minus the margin of error to the sample proportion plus the margin of error:
  • \( \text{Confidence Interval} = p \pm \text{Margin of Error} \)
For our calculation, this becomes:
  • \( 0.115 \pm 0.01628 \)
  • Resulting in an interval from 0.09872 to 0.13128.
This range suggests where the true population proportion likely falls, enhancing our understanding and ensuring more grounded decisions.

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Most popular questions from this chapter

Bob Ross hosted a weekly television show, The Joy of Painting, on PBS in which he taught viewers how to paint. During each episode, he produced a complete painting while teaching viewers how they could produce a similar painting. Ross completed 30,000 paintings in his lifetime. Although it was an art instruction show, PBS estimated that only \(10 \%\) of viewers painted along with Ross during his show based on surveys of viewers. For each of the following, also identify the population and explain your choice. a. Is the number 30,000 a parameter or a statistic? b. Is the number \(10 \%\) a parameter or a statistic?

a. If a rifleman's gunsight is adjusted incorrectly, he might shoot bullets consistently close to 2 feet left of the bull's-eye target. Draw a sketch of the target with the bullet holes. Does this show lack of precision or bias? b. Draw a second sketch of the target if the shots are both unbiased and precise (have little variation). The rifleman's aim is not perfect, so your sketches should show more than one bullet hole.

Suppose it is known that \(20 \%\) of students at a certain college participate in a textbook recycling program each semester. a. If a random sample of 50 students is selected, do we expect that exactly \(20 \%\) of the sample participates in the textbook recycling program? Why or why not? b. Suppose we take a sample of 500 students and find the sample proportion participating in the recycling program. Which sample proportion do you think is more likely to be closer to \(20 \%\) : the proportion from a sample size of 50 or the proportion from a sample size of \(500 ?\) Explain your reasoning.

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To determine if patrons are satisfied with performance quality, a theater surveys patrons at an evening performance by placing a paper survey inside their programs. All patrons receive a program as they enter the theater. Completed surveys are placed in boxes at the theater exits. On the evening of the survey, 500 patrons saw the performance. One hundred surveys were completed, and \(70 \%\) of these surveys indicated dissatisfaction with the performance. Should the theater conclude that patrons were dissatisfied with performance quality? Explain.
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