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Construct a confidence interval for \(p_{1}-p_{2}\) at the given level of confidence. \(x_{1}=28, n_{1}=254, x_{2}=36, n_{2}=301,95 \%\) confidence

Short Answer

Expert verified
The 95% confidence interval for the difference in proportions \( p_{1} - p_{2} \) is \( (-0.0355, 0.06449) \).

Step by step solution

01

- Calculate Sample Proportions

Calculate the sample proportions for both samples. For the first sample, the sample proportion is \(\hat{p}_{1} = \frac{x_{1}}{n_{1}} = \frac{28}{254} \). For the second sample, the sample proportion is \(\hat{p}_{2} = \frac{36}{301} \).
02

- Calculate the Point Estimate of the Difference

Calculate the point estimate of the difference between the two population proportions: \(\hat{p}_{1} - \hat{p}_{2} \).
03

- Calculate the Standard Error

Calculate the standard error for the difference between two proportions using the formula \[ SE = \sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}} \].
04

- Find the Critical Value

Find the critical value for a 95% confidence level. For a 95% confidence interval, the critical value (z-score) is 1.96.
05

- Calculate the Margin of Error

Calculate the margin of error by multiplying the critical value by the standard error: \[ ME = z \times SE \].
06

- Construct the Confidence Interval

Finally, construct the confidence interval using: \[ (\hat{p}_{1} - \hat{p}_{2}) \pm ME \]. This gives the range within which the true difference in population proportions lies with 95% confidence.

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Key Concepts

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

Sample Proportion
The sample proportion is a key concept when working with confidence intervals for the difference in proportions. A sample proportion is simply the ratio of the number of successes in a sample to the total number of observations in that sample.
For example, in the given exercise, the sample proportion for the first sample is calculated as: \[ \hat{p_{1}} = \frac{28}{254} \] and for the second sample, it's: \[ \hat{p_{2}} = \frac{36}{301} \].
These proportions provide a snapshot of what proportion of the sample exhibits the characteristic of interest.
Understanding sample proportions is vital because they serve as the building blocks for estimating the difference between two population proportions.
Point Estimate
A point estimate gives us a single value as our best guess for the difference between two population proportions. In the context of the exercise, the point estimate is the difference between the two sample proportions: \[ \hat{p_{1}} - \hat{p_{2}} \].
By calculating the point estimate, we can get an initial understanding of how much the proportions differ between the two populations sampled.
This value will later be used as the center of our confidence interval, around which we will calculate our margin of error to understand the range in which the true difference likely lies.
Standard Error
The standard error measures the dispersion or variability in the sampling distribution of a statistic. In this context, it assesses the variability of the difference in sample proportions.
To calculate the standard error between two sample proportions, we use the formula: \[ SE = \sqrt{\frac{\hat{p_{1}}(1-\hat{p_{1}})}{n_{1}} + \frac{\hat{p_{2}}(1-\hat{p_{2}})}{n_{2}}} \].
The smaller the standard error, the more precise our estimate will be.
The standard error helps us to create a range (confidence interval) where the true difference between population proportions lies with a specific level of confidence (e.g., 95%).
Critical Value
The critical value is a threshold that determines the range of values that we consider plausible for the difference between the population proportions.
For a 95% confidence level, the critical value corresponds to the z-score that captures the central 95% of a standard normal distribution. Typically, this value is 1.96.
This critical value is used to calculate the margin of error, which adjusts our point estimate to create a confidence interval. By multiplying this critical value by the standard error, we get the margin of error: \[ ME = z \times SE \].
The margin of error is then added to and subtracted from the point estimate to construct the confidence interval: \[ (\hat{p_{1}} - \hat{p_{2}}) \pm ME \].
This interval provides a range where we can be 95% confident that the true difference in proportions lies within.

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Most popular questions from this chapter

\( \begin{array}{cccccccc} \text { Observation } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} \\ \hline X_{i} & 7.6 & 7.6 & 7.4 & 5.7 & 8.3 & 6.6 & 5.6 \\ \hline Y_{i} & 8.1 & 6.6 & 10.7 & 9.4 & 7.8 & 9.0 & 8.5 \end{array} \) (a) Determine \(d_{i}=X_{i}-Y_{i}\) for each pair of data. (b) Compute \(\bar{d}\) and \(s_{d}\) (c) Test if \(\mu_{d}<0\) at the \(\alpha=0.05\) level of significance. (d) Compute a \(95 \%\) confidence interval about the population mean difference \(\mu_{d}\)

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For each study, explain which statistical procedure (estimating a single proportion; estimating a single mean; hypothesis test for a single proportion; hypothesis test for a single mean; hypothesis test or estimation of two proportions, hypothesis test or estimation of two means, dependent or independent) would most likely be used for the research objective given. Assume all model requirements for conducting the appropriate procedure have been satisfied. Is the mean IQ of the students in Professor Dang's statistics class higher than that of the general population, \(100 ?\)

Clifford Adelman, a researcher with the Department of Education, followed a cohort of students who graduated from high school in \(1992,\) monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to compare students who first attended a community college to those who immediately attended and remained at a four-year institution. The sample standard deviation time to complete a bachelor's degree of the 268 students who transferred to a four-year school after attending community college was \(1.162 .\) The sample standard deviation time to complete a bachelor's degree of the 1145 students who immediately attended and remained at a four-year institution was \(1.015 .\) Assuming the time to earn a bachelor's degree is normally distributed, does the evidence suggest the standard deviation time to earn a bachelor's degree is different between the two groups? Use the \(\alpha=0.05\) level of significance.

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