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Chapter 8: Problem 56

Independent random samples of \(n_{1}=1265\) and \(n_{2}=1688\) observations were selected from binomial populations 1 and \(2,\) and \(x_{1}=849\) and \(x_{2}=910\) successes were observed. a. Find a \(99 \%\) confidence interval for the difference \(\left(p_{1}-p_{2}\right)\) in the two population proportions. What does "99\% confidence" mean? b. Based on the confidence interval in part a, can you conclude that there is a difference in the two binomial proportions? Explain.

Short Answer

Expert verified
1. Compute the sample proportions for each population: \(p_1 = \frac{x_1}{n_1} = \frac{849}{1265} = 0.6715\) \(p_2 = \frac{x_2}{n_2} = \frac{910}{1688} = 0.5392\) 2. Calculate the standard error of the difference: \(SE(p_1 - p_2) = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} = \sqrt{\frac{0.6715(1-0.6715)}{1265} + \frac{0.5392(1-0.5392)}{1688}} = 0.0292\) 3. Determine the z-score for a 99% confidence interval: 2.576 4. Calculate the 99% confidence interval for the difference in proportions: \((p_1 - p_2) \pm (Z_{\alpha/2} \times SE(p_1 - p_2)) = (0.6715 - 0.5392) \pm (2.576 \times 0.0292) = 0.1323 \pm 0.0752\) The 99% confidence interval is (0.0571, 0.2075). 5. Analyze the confidence interval: The confidence interval does not contain 0 (0.0571 < 0 < 0.2075). Therefore, we can conclude with 99% confidence that there is a difference in the binomial proportions. 6. Answer part b: Based on the confidence interval calculated in part a, the interval does not contain 0. Thus, we can conclude with 99% confidence that there is a difference in the binomial proportions of the two populations.

Step by step solution

01

Compute sample proportions

From the given sample data, we need to compute sample proportions for each population: \(p_1 = \frac{x_1}{n_1} = \frac{849}{1265}\) \(p_2 = \frac{x_2}{n_2} = \frac{910}{1688}\)
02

Calculate standard error of the difference

The standard error of the difference between the two proportions is given by the formula: \(SE(p_1 - p_2) = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\)
03

Determine z-score for 99% confidence interval

To compute the 99% confidence interval, we need to find the z-score corresponding to .005 in each tail (since we want to have 99% of the probability distribution in the middle). The z-score for a .005-tail is approximately 2.576.
04

Calculate the 99% confidence interval

Now that we have the z-score and the standard error, we can compute the confidence interval for the difference (\(p_1 - p_2\)) as: \((p_1 - p_2) \pm (Z_{\alpha/2} \times SE(p_1 - p_2))\) Plug in the values to obtain the 99% confidence interval.
05

Analyzing the confidence interval

The 99% confidence interval tells us that we are 99% confident that the true difference between the two proportions lies within the calculated range. If the confidence interval contains 0, it means that the two proportions might be equal, and we cannot conclude that there is a difference in the binomial proportions. On the other hand, if the confidence interval does not contain 0, then we can conclude with a 99% confidence level that there is a difference in the binomial proportions.
06

Answer part b

Based on the confidence interval calculated in part a, assess if the interval contains 0 or not. If the interval contains 0, no conclusion can be drawn about the difference in the binomial proportions with 99% confidence. If the interval does not contain 0, we can conclude with 99% confidence that there is a difference in the binomial proportions.

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Key Concepts

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

Binomial Proportions
Understanding binomial proportions is key to solving problems involving two outcomes. Binomial proportions refer to the probability of one specific outcome occurring out of a set of possibilities in repeated trials. For example, if you are tossing a coin, the binomial proportions would relate to the probability of getting a head versus a tail.

In statistical problems, each trial is assumed to be independent, meaning the outcome of one trial doesn't affect the other. This is why in the given exercise, we gather data on successes and failures like whether a condition is met or not, and calculate the respective proportions for these outcomes.

When working with binomial proportions, it's crucial to differentiate between theoretical proportions and sample proportions. The sample proportions provide us with an estimate of the theoretical proportions within a population, which we'll discuss in the next section.
Sample Proportion
The sample proportion is a key concept that helps us understand the likelihood of a particular event happening in our sample group. In the given exercise, we calculate the proportion of successes ( ) for each of the two populations:

  • For the first population: \( p_1 = \frac{x_1}{n_1} = \frac{849}{1265} \)
  • For the second population: \( p_2 = \frac{x_2}{n_2} = \frac{910}{1688} \)


Each sample proportion is a simple calculation that provides a ratio of successes to the total number of observations. It serves as an estimate of the true proportion in the population from which the sample was drawn. Understanding this concept is important because it forms the basis for determining confidence intervals and making statistical inferences about populations.
Standard Error
The standard error of a statistic reflects the amount of variation or "spread" we can expect in the calculations of sample statistics, like our sample proportion. It's incredibly useful for understanding how much our sample statistic might differ from the true population value.

In our exercise, we determine the standard error of the difference between the two sample proportions using this formula:

\[ SE(p_1 - p_2) = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \]

This calculation gives us insight into the reliability of our estimate of the proportion difference. A smaller standard error indicates that the sample proportion is likely a better estimate of the true proportion in the population. When computing confidence intervals, the standard error plays a critical role in understanding the range within which the true proportion may lie.
Z-Score
The z-score is a statistical measure that tells us how many standard deviations a data point is from the mean. It becomes particularly relevant when calculating confidence intervals, as it helps us understand how likely it is that a sample proportion lies within a certain range of the mean proportion.

For a 99% confidence interval, the z-score that corresponds to this level of confidence is approximately 2.576. This number is derived from the standard normal distribution table. Essentially, it indicates that 99% of the data points lie within 2.576 standard deviations from the mean in a standard normal distribution.

When used with the standard error in this exercise, the z-score helps us define the range of our 99% confidence interval for the difference \( (p_1 - p_2) \) between two proportions. This interval provides a range where we can be 99% confident that the true difference in population proportions lies, based on our sample data.

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Most popular questions from this chapter

Multimedia Kids Do our children spend as much time enjoying the outdoors and playing with family and friends as previous generations did? Or are our children spending more and more time glued to the television, computer, and other multimedia equipment? A random sample of 250 children between the ages of 8 and 18 showed that 170 children had a TV in their bedroom and that 120 of them had a video game player in their bedroom. a. Estimate the proportion of all 8 - to 18 -year-olds who have a TV in their bedroom, and calculate the margin of error for your estimate. b. Estimate the proportion of all 8 - to 18 -year-olds who have a video game player in their bedroom, and calculate the margin of error for your estimate.

Suppose you wish to estimate a population mean based on a random sample of \(n\) observations, and prior experience suggests that \(\sigma=12.7\). If you wish to estimate \(\mu\) correct to within 1.6 , with probability equal to .95, how many observations should be included in your sample?

What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{9}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) and a standard deviation of \(0.73^{\circ} .\) a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value \(98.6^{\circ},\) the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

Do well-rounded people get fewer colds? A study on the Chronicle of Higher \(E d u\) cation was conducted by scientists at Carnegie Mellon University, the University of Pittsburgh, and the University of Virginia. They found that people who have only a few social outlets get more colds than those 16 who are involved in a variety of social activities. Suppose that of the 276 healthy men and women tested, \(n_{1}=96\) had only a few social outlets and \(n_{2}=105\) were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: a. Construct a \(99 \%\) confidence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected finding?

Explain what is meant by "margin of error" in point estimation.
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