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Chapter 10: Problem 85

Calculate the estimate for the standard error of the difference between two proportions for each of the following cases:a. \( n_{1}=40, p_{1}^{\prime}=0.8, n_{2}=50,\) and \(p_{2}^{\prime}=0.8\) b. \( n_{1}=33, p_{1}^{\prime}=0.6, n_{2}=38,\) and \(p_{2}^{\prime}=0.65\)

Short Answer

Expert verified
a. The estimated standard error for the difference between the two proportions for case a is approximately 0.101. For case b, it is approximately 0.217, rounded to three decimal places.

Step by step solution

01

Standard deviation for proportion 1

First, we need to calculate the standard deviation for proportion 1. For this, we apply the formula for the standard deviation of a proportion, which is \(\sqrt{p_{1} * (1 - p_{1}) / n_{1}}\). Let's apply this formula for case a, where \( p_{1} = 0.8 \) and \( n_{1} = 40 \). Hence, \( S_{1} = \sqrt{(0.8 * 0.2) / 40} \)
02

Standard deviation for proportion 2

Now, we will calculate the standard deviation for proportion 2 by using the same method as we did for proportion 1. For case a, \( p_{2} = 0.8 \) and \( n_{2} = 50 \). Hence, we calculate \( S_{2} = \sqrt{(0.8 * 0.2) / 50} \)
03

Calculate the Standard Error of Difference

Now that we have the standard deviations of both proportions, we can calculate the standard error of difference. For this, we square both standard deviations, add them, and then take the square root of the result. Following this, we get the formula \( SED = \sqrt{s_{1}^{2} + s_{2}^{2}} \). For case a, \( SED = \sqrt{S_{1}^{2} + S_{2}^{2}} \)
04

Repeat steps for case b

We then repeat steps one to three for case b where, \( n_{1} = 33, p_{1} = 0.6, n_{2} = 38, p_{2} = 0.65 \). Therefore, we calculate the standard deviations as \( S_{1} = \sqrt{(0.6 * 0.4) / 33} \), \( S_{2} = \sqrt{(0.65 * 0.35) / 38} \). Then the standard error of difference is calculated as, \( SED = \sqrt{S_{1}^{2} + S_{2}^{2}} \)

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Key Concepts

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

Proportion Difference
Understanding the difference between two proportions is crucial when analyzing statistical data. Proportions reflect part of a whole, expressed as a fraction or percentage, within a population. For example, if we survey a community and find that 80% favor a particular policy, this 80% is a proportion. But what if we want to compare this proportion with another community's views? This is where calculating the difference between two proportions comes into play.
  • When comparing these proportions, it involves two separate populations. We want to know how significantly different these proportions are.
  • The difference in proportions helps in hypothesis testing or determining how much one proportion exceeds or lags behind the other.
When you calculate the difference, you're preparing to evaluate how varied your outcomes are. This is essential for drawing conclusions in research.
Standard Deviation
Standard deviation is a measure of variance or dispersion in data. It tells us how spread out numbers are in a dataset compared to the mean. In the context of proportions, we modify this concept slightly to accommodate the specific nature of proportion data. When dealing with binary data (like yes/no responses), it's crucial to understand its variability with proper calculations.
Every standard deviation of a proportion takes into account both the success and failure probabilities, symbolized as \( p \) and \( 1 - p \), respectively.
  • The formula used is: \( \sqrt{p * (1 - p) / n} \).
  • This formula helps in quantifying the uncertainty or fluctuation in proportions.
A lower standard deviation indicates low variance, while a higher one indicates high variance in data.
Formula Application
Applying formulas is a cornerstone in mathematics and statistics. For estimating the standard error of the difference between two proportions, we rely on a succession of well-established formulas.
  • First, calculate the standard deviation of each proportion individually using the formula: \( S = \sqrt{p * (1 - p) / n} \).
  • For each proportion, \( p \) is the success probability, and \( n \) the sample size.
  • Once you have the standard deviations \( S_1 \) and \( S_2 \), each squared, add them together.
  • Finally, take the square root of this sum to get the standard error difference (\( SED \)), which is \( \sqrt{S_1^2 + S_2^2} \).
The correct application of these formulas ensures a reliable estimate of the standard error difference, a crucial aspect of statistical comparison.
Step by Step Solution
When solving statistical problems, a systematic, step-by-step approach clarifies the process, making it easier to understand each component involved. In solving for the standard error of the difference between two proportions, begin by comprehensively breaking down the problem.
1. **Understanding Proportions**: Clearly identify your sample sizes \( n_1, n_2 \) and success probabilities \( p_1, p_2 \) for both groups.2. **Standard Deviation Calculation**: Use the formula \( S = \sqrt{p * (1 - p) / n} \) to find the standard deviation for each group.3. **Combining Results**: Once both standard deviations \( S_1 \) and \( S_2 \) are identified, square them, sum them up, and take the square root, resulting in the standard error of the difference (\( SED \)).
  • This calculated value helps assess how much the sample proportions are expected to vary around the true difference in population proportions.
Completing each step carefully will solidify your understanding and ensure precision in your calculated results.

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Most popular questions from this chapter

In determining the "goodness" of a test question, a teacher will often compare the percentage of better students who answer it correctly with the percentage of poorer students who answer it correctly. One expects that the proportion of better students who will answer the question correctly is greater than the proportion of poorer students who will answer it correctly. On the last test, 35 of the students with the top 60 grades and 27 of the students with the bottom 60 grades answered a certain question correctly. Did the students with the top grades do significantly better on this question? Use \(\alpha=0.05.\) a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

A study in Pediatric Emergency Care compared the injury severity between younger and older children. One measure reported was the Injury Severity Score (ISS). The standard deviation of ISS's for 37 children 8 years or younger was \(23.9,\) and the standard deviation for 36 children older than 8 years was \(6.8 .\) Assume that ISS's are normally distributed for both age groups. At the 0.01 level of significance, is there sufficient reason to conclude that the standard deviation of ISS's for younger children is larger than the standard deviation of ISS's for older children?

State the null and alternative hypotheses that would be used to test the following claims: a. The difference between the means of the two populations is more than 20 lb. b. The mean of population \(A\) is less than 50 more than the mean of population B. c. The difference between the two populations is at least \$500. d. The average size yard for neighborhood \(A\) is no more than 30 square yards greater than the average size yard in neighborhood B.

A manufacturer designed an experiment to compare the differences between men and women with respect to the times they require to assemble a product. A total of 15 men and 15 women were tested to determine the time they required, on average, to assemble the product. The time required by the men had a standard deviation of 4.5 minutes, and the time required by the women had a standard deviation of 2.8 minutes. Do these data show that the amount of time needed by men is more variable than the amount of time needed by women? Use \(\alpha=0.05\) and assume the times are approximately normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Determine the critical region and critical value(s) that would be used to test the following hypotheses using the classical approach when \(F \star\) is used as the test statistic. a. \(\quad H_{o}: \sigma_{1}^{2}=\sigma_{2}^{2}\) versus \(H_{a}: \sigma_{1}^{2}>\sigma_{2}^{2},\) with \(n_{1}=10, n_{2}=16\) and \(\alpha=0.05\) b. \(\quad H_{o}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}=1\) versus \(H_{a}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}} \neq 1,\) with \(n_{1}=25, n_{2}=31\) and \(\alpha=0.05\)b. \(\quad H_{o}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}=1\) versus \(H_{a}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}} \neq 1,\) with \(n_{1}=25, n_{2}=31\) and \(\alpha=0.05\) c. \(\quad H_{o}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}=1\) versus \(H_{a}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}>1,\) with \(n_{1}=10, n_{2}=10\) and \(\alpha=0.01\) d. \(\quad H_{o}: \sigma_{1}=\sigma_{2}\) versus \(H_{a}: \sigma_{1}<\sigma_{2},\) with \(n_{1}=25, n_{2}=16\) and \(\alpha=0.01\)
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