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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

Suppose it is known that \(60 \%\) of employees at a company use a Flexible Spending Account (FSA) benefit. a. If a random sample of 200 employees is selected, do we expect that exactly \(60 \%\) of the sample uses an FSA? Why or why not? b. Find the standard error for samples of size 200 drawn from this population. What adjustments could be made to the sampling method to produce a sample proportion that is more precise?

From Formula \(7.2\), an estimate for margin of error for a \(95 \%\) confidence interval is \(m=2 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\mathrm{n}\) is the required sample size and \(\hat{p}\) is the sample proportion. Since we do not know a value for \(\hat{p}\), we use a conservative estimate of \(0.50\) for \(\hat{p}\). Replace \(\hat{p}\) with \(0.50\) in the formula and simplify.

A recent Monmouth University poll found that 675 out of 1008 randomly selected people in the United States felt that college and universities with big sports programs placed too much emphasis on athletics over academics. Assuming the conditions for using the CLT were met, use the Minitab output provided to answer these questions. $$\begin{aligned}&\text { Descriptive Statistics }\\\&\begin{array}{rrrr}\mathrm{N} & \text { Event } & \text { Sample p } & 95 \% \mathrm{Cl} \text { for } \mathrm{p} \\\\\hline 1008 & 675 & 0.669643 & (0.639648,0.698643)\end{array}\end{aligned}$$ a. Complete this sentence: I am \(95 \%\) confident that the population proportion believing that colleges and universities with big sports programs place too much emphasis on athletics over academics is between _____ and _______. Report each number as a percentage rounded to one decimal place. b. Suppose a sports blogger wrote an article claiming that the majority of Americans believe that colleges and university with big sports programs place too much emphasis on athletics over academics. Does this confidence interval support the blogger's claim? Explain your reasoning.

According to a 2017 survey conducted by Netflix, \(46 \%\) of couples have admitted to "cheating" on their significant other by streaming a TV show ahead of their partner. Suppose a random sample of 80 Netflix subscribers is selected. a. What percentage of the sample would we expect have "cheated" on their partner? b. Verify that the conditions for the Central Limit Theorem are met. c. What is the standard error for this sample proportion? d. Complete the sentence: We expect ____ \(\%\) of streaming couples to admit to Netflix "cheating," give or take _____ \(\% .\)

Statistics student Hector Porath wanted to find out whether gender and the use of turn signals when driving were independent. He made notes when driving in his truck for several weeks. He noted the gender of each person that he observed and whether he or she used the turn signal when turning or changing lanes. (In his state, the law says that you must use your turn signal when changing lanes, as well as when turning.) The data he collected are shown in the table. $$\begin{array}{|l|l|l|}\hline & \text { Men } & \text { Women } \\\\\hline \text { Turn signal } & 585 & 452 \\ \hline \text { No signal } & 351 & 155 \\\\\hline & 936 & 607 \\ \hline\end{array}$$ a. What percentage of men used turn signals, and what percentage of women used them? b. Assuming the conditions are met (although admittedly this was not a random selection), find a \(95 \%\) confidence interval for the difference in percentages. State whether the interval captures 0, and explain whether this provides evidence that the proportions of men and women who use turn signals differ in the population. c. Another student collected similar data with a smaller sample size: $$\begin{array}{|l|c|c|}\hline & \text { Men } & \text { Women } \\\\\hline \text { Turn Signal } & 59 & 45 \\ \hline \text { No Signal } & 35 & 16 \\\\\hline & 94 & 61 \\ \hline\end{array}$$ First find the percentage of men and the percentage of women who used turn signals, and then, assuming the conditions are met, find a \(95 \%\) confidence interval for the difference in percentages. State whether the interval captures 0 , and explain whether this provides evidence that the percentage of men who use turn signals differs from the percentage of women who do so. d. Are the conclusions in parts \(\mathrm{b}\) and \(\mathrm{c}\) different? Explain.
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