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Chapter 8: Problem 28

In a study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were firstborn or only children; in a sample of 100 non-graduates of comparable age and socioeconomic background, the number of firstborn or only children was \(54 .\) Estimate the difference in the proportions of firstborn or only children for the two populations from which these samples were drawn. Give a bound for the error of estimation.

Short Answer

Expert verified
Estimated difference is 0.16 with error bound of 0.125; CI: [0.035, 0.285].

Step by step solution

01

Define the Sample Proportions

Let's denote the sample of college graduates as sample 1, and the sample of non-graduates as sample 2. Compute the sample proportions for firstborn or only children in each group. For sample 1 (graduates):\[ \hat{p}_1 = \frac{126}{180} = 0.7 \]For sample 2 (non-graduates):\[ \hat{p}_2 = \frac{54}{100} = 0.54 \]
02

Calculate the Difference in Sample Proportions

Compute the difference between the two sample proportions:\[ \hat{p}_1 - \hat{p}_2 = 0.7 - 0.54 = 0.16 \]This difference suggests that a higher proportion of firstborn or only children were observed among college graduates compared to non-graduates.
03

Find the Standard Error of the Difference

The standard error of the difference between two proportions is calculated as:\[ SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} \]Where \( n_1 = 180 \) and \( n_2 = 100 \). Substitute the known values:\[ SE = \sqrt{\frac{0.7 \times 0.3}{180} + \frac{0.54 \times 0.46}{100}} \]Calculate each part and square them:\[ \text{For graduates: } \frac{0.7 \times 0.3}{180} \approx 0.001167 \]\[ \text{For non-graduates: } \frac{0.54 \times 0.46}{100} \approx 0.002484 \]Add these:\[ SE = \sqrt{0.001167 + 0.002484} \approx 0.064 \]
04

Determine a Confidence Interval for the Difference

To construct a confidence interval, we need a critical value, typically a Z-value for common confidence levels. For a 95% confidence interval, \( Z \approx 1.96 \).The margin of error (ME) is:\[ ME = Z \times SE = 1.96 \times 0.064 \approx 0.125 \]Thus, the confidence interval is:\[ 0.16 \pm 0.125 = [0.035, 0.285] \]
05

State the Final Estimate and Error Bound

The estimated difference in proportions is \(0.16\) with a confidence interval from \(0.035\) to \(0.285\). Hence, the bound for the error of estimation is \(0.125\).

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Key Concepts

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

Understanding Sample Proportion
When exploring statistics, especially in studies like comparing college graduates and non-graduates, the term 'sample proportion' becomes highly important. In simple terms, the sample proportion is a way to express part of a whole in a fraction or percentage. Here's how it works:
  • A sample is a smaller, manageable version of a population you want to study.
  • Proportion refers to the part of the sample that meets a certain criterion.
  • It's calculated by dividing the number of items meeting the criterion by the total sample size.
In the given exercise, for instance, the proportion of firstborn or only children among graduates is calculated by dividing 126 (the number of firstborn or only children) by 180 (the total number of graduates), which gives us a sample proportion of 0.7 or 70%.
Calculating the Standard Error
The standard error is integral in statistics because it measures the accuracy with which a sample represents a population. Essentially, it's the expected variation of the sample proportion if you were to take many samples from the same population. Standard error is calculated using the formula:\[ SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} \]Where:
  • \(\hat{p}_1 \) and \(\hat{p}_2 \) are the sample proportions for the two groups,
  • \( n_1 \) and \( n_2 \) are the sizes of the two samples.
By calculating the standard error, we're able to assess how much our sample proportion might deviate from the true population proportion.
Applying the Confidence Interval
A confidence interval provides a range of values that is likely to contain the population parameter (in this case, the difference in proportions). It is expressed with a certain level of confidence, typically 95%, meaning if you were to take 100 different samples, about 95 of them would produce a confidence interval containing the true difference in proportions. The formula for the confidence interval in this context is:\[ \text{CI} = (\hat{p}_1 - \hat{p}_2) \pm Z \times SE \]Where:
  • \( \hat{p}_1 - \hat{p}_2 \) is the calculated difference of sample proportions,
  • \( Z \) is the Z-value for the desired confidence level (for 95%, \( Z \approx 1.96 \)),
  • \( SE \) is the standard error.
The confidence interval gives us a practical range to understand how varying the data is and help estimate the true population difference.
Exploring the Difference of Proportions
Understanding the difference in proportions helps in comparing two distinct groups. In our case, this is the comparison between the sample proportions of firstborn or only children in college graduates and non-graduates. This difference highlights whether one group tends to have a particular feature more than the other. Here are key points to note:
  • A positive difference, like 0.16 in the exercise, suggests that a higher proportion of firstborn or only children were found in the group of college graduates compared to non-graduates.
  • It offers insights into population trends, potentially indicating societal patterns or influences.
  • Analyzing these differences helps in drawing conclusions and making predictions about other similar populations.
The difference of proportions serves as a valuable tool for understanding relationships between categories within a dataset.

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Most popular questions from this chapter

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