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Chapter 6: Problem 34

In a survey of 1000 US adults, twenty percent say they never exercise. This is the highest level seen in five years. \({ }^{7}\) Find and interpret a \(99 \%\) confidence interval for the proportion of US adults who say they never exercise. What is the margin of error, with \(99 \%\) confidence?

Short Answer

Expert verified
The 99% confidence interval for the proportion of US adults who say they never exercise is approximately (0.16744, 0.23256), and the margin of error at 99% confidence level is approximately 卤0.03256 or 卤3.3%.

Step by step solution

01

Define Variables

In this problem, the sample size \(n\) is 1000 and the observed proportion who never exercise, often represented by \( \hat{p} \), is 0.20 or 20%.
02

Test Condition

Check if the sample proportion follows a normal distribution. The condition is satisfied if \(np > 5\) and \(n(1 - p) > 5\). Substituting the values of \(n\) and \( \hat{p} \), we have \(1000 * 0.20 = 200 > 5\) and \(1000 * (1 - 0.20) = 800 > 5\). So, the condition for normal distribution is satisfied here.
03

Find Standard Error

Calculate the standard error using the formula: \(SE = \sqrt{ \hat{p}(1 - \hat{p}) / n}\). So, \(SE = \sqrt{0.20 * 0.80 / 1000} = 0.01265\). The standard error measures the variability of the sample proportions.
04

Find Z-score

For a 99% confidence interval, the standard (Z) score is 2.576 as per the Z-table.
05

Calculate Confidence Interval and Margin of Error

The confidence interval is given by \( \hat{p} 卤 Z * SE\). So, it is \(0.20 卤 2.576 * 0.01265\), which simplifies to \((0.20 - 0.03256, 0.20 + 0.03256)\), or \((0.16744, 0.23256)\). This means we are 99% confident that the true proportion of US adults who say they never exercise is within this interval. The Margin of Error (MOE) is determined by the calculation: \(Z * SE\), or \(2.576 * 0.01265 = 0.03256\), which means the estimate of the proportion of adults who say they never exercise is accurate within about 卤3.3%.

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Key Concepts

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

Understanding Proportion
In statistics, a proportion represents a part of a total or a fraction that indicates the size of a sub-group relative to the entire group. In the context of the given exercise, the proportion is the percentage of adults who say they never exercise. A key application of proportions is to predict or estimate characteristics of a particular population based on sampled data. For instance, the 20% given in this survey is a sample proportion, denoted by \( \hat{p} = 0.20 \). This is derived from the survey of 1000 US adults, which helps estimate the overall behavior of the entire US adult population.Using samples to estimate proportions involves recognizing that these figures could vary from sample to sample. Thus, the primary goal is to estimate these proportions within a range (confidence interval) that includes the true proportion with a specified probability.
Margin of Error Simplified
The margin of error (MOE) is the maximum expected difference between the sample proportion \( \hat{p} \) and the actual population proportion \( p \). It provides a range within which the true proportion is likely to fall, considering the sample data.For the exercise, with a 99% confidence level, the margin of error is calculated as \( Z \times SE \), where \( Z \) is the Z-score from the Z-table and \( SE \) is the standard error. In this context, the 99% confidence level reflects a higher certainty in the estimate, leading to a larger MOE due to the wider range it accounts for. With the calculation, the margin of error turns out to be \( 0.03256 \), or approximately 卤3.3%. This means our sample estimate of 20% is accurate within a range of 3.3%, giving an estimate anywhere from 16.7% to 23.3%.
Demystifying Standard Error
The standard error (SE) is a measure of how much the sample proportion \( \hat{p} \) would vary from the actual population proportion \( p \) if we were to repeat the survey multiple times. It quantifies the precision of the sample's estimation鈥攂asically how trustworthy it is based on the sample size.In our exercise, the standard error is calculated using the formula: \[ SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \]For the values given, \( \hat{p} = 0.20 \) and \( n = 1000 \) yield an SE of \( 0.01265 \). A smaller standard error suggests that the sample proportion is a more accurate estimate of the population proportion. Since the SE is smaller when the sample size \( n \) is larger or when the proportion \( \hat{p} \) is close to 0.5, it emphasizes the importance of having a reasonably large sample size for reliable estimates.

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Most popular questions from this chapter

A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lcc} \hline \text { Case } & \text { Situation 1 } & \text { Situation 2 } \\ \hline 1 & 77 & 85 \\ 2 & 81 & 84 \\ 3 & 94 & 91 \\ 4 & 62 & 78 \\ 5 & 70 & 77 \\ 6 & 71 & 61 \\ 7 & 85 & 88 \\ 8 & 90 & 91 \\ \hline \end{array} $$

Of the 50 states in the Unites States, Alaska has the largest percentage of males and Rhode Island has the largest percentage of females. (Interestingly, Alaska is the largest state and Rhode Island is the smallest). According to the 2010 US Census, the population of Alaska is \(52.0 \%\) male and the population of Rhode Island is \(48.3 \%\) male. If we randomly sample 300 people from Alaska and 300 people from Rhode Island, what is the approximate distribution of \(\hat{p}_{a}-\hat{p}_{r i}\), where \(\hat{p}_{a}\) is the proportion of males in the Alaskan sample and \(\hat{p}_{r i}\) is the proportion of males in the Rhode Island sample?

Exercise B.5 on page 305 introduces a study examining the effect of diet cola consumption on calcium levels in women. A sample of 16 healthy women aged 18 to 40 were randomly assigned to drink 24 ounces of either diet cola or water. Their urine was collected for three hours after ingestion of the beverage and calcium excretion (in mg) was measured. The summary statistics for diet cola are \(\bar{x}_{C}=56.0\) with \(s_{C}=4.93\) and \(n_{C}=8\) and the summary statistics for water are \(\bar{x}_{W}=49.1\) with \(s_{W}=3.64\) and \(n_{W}=8\). Figure 6.26 shows dotplots of the data values. Test whether there is evidence that diet cola leaches calcium out of the system, which would increase the amount of calcium in the urine for diet cola drinkers. In Exercise B.5, we used a randomization distribution to conduct this test. Use a t-distribution here, after first checking that the conditions are met and explaining your reasoning. The data are stored in ColaCalcium.

In Exercises 6.153 to \(6.158,\) if random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from population \(A\) with proportion 0.70 and samples of size 75 from population \(B\) with proportion 0.60

Professor A and Professor \(\mathrm{B}\) are teaching sections of the same introductory statistics course and decide to give common exams. They both have 25 students and design the exams to produce a grade distribution that follows a bell curve with mean \(\mu=75\) and standard deviation \(\sigma=10\) (a) Suppose students are randomly assigned to the two classes and the instructors are equally effective. Describe the center, spread, and shape of the distribution of the difference in class means, \(\bar{x}_{A}-\bar{x}_{B},\) for the common exams. (b) Based on the distribution in part (a), how often should one of the class means differ from the other class by three or more points? (Hint: Look at both the tails of the distribution.) (c) How do the answers to parts (a) and (b) change if the exams are much harder than expected so the distribution for each class is \(N(60,10)\) rather that \(N(75,10) ?\)
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