/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Construct a confidence interval ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Construct a confidence interval for \(p_{1}-p_{2}\) at the given level of confidence. \(x_{1}=109, n_{1}=475, x_{2}=78, n_{2}=325,99 \%\) confidence

Short Answer

Expert verified
The 99% confidence interval for the difference in proportions, \(p_{1} - p_{2}\), can be found using calculated proportions, the standard error, and the margin of error steps.

Step by step solution

01

Calculate sample proportions

First, compute the sample proportions for both groups. The sample proportion is given by \(\hat{p} = \frac{x}{n}\). For group 1: \(\hat{p}_{1} = \frac{109}{475}\). For group 2: \(\hat{p}_{2} = \frac{78}{325}\).
02

Compute the combined standard error

Calculate the standard error for the difference between two proportions using the formula: \[SE = \sqrt{\hat{p}_{1}\left(1-\hat{p}_{1}\right)/n_{1} + \hat{p}_{2}\left(1-\hat{p}_{2}\right)/n_{2}}\].
03

Find the critical value

For a 99% confidence interval, determine the critical value (z*) corresponding to the given confidence level. Using standard Z-tables, the critical value for 99% confidence is approximately 2.576.
04

Calculate the margin of error

Multiply the standard error by the critical value to find the margin of error (ME): \[ME = z* \times SE\].
05

Determine the confidence interval

Finally, calculate the confidence interval for the difference in proportions using the formula: \[(\hat{p}_{1} - \hat{p}_{2}) \pm ME\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Proportions
When comparing two groups, we often want to understand the proportion of occurrences of a certain event in each group. This proportion is called the sample proportion and is calculated using the formula: \(\hat{p} = \frac{x}{n}\). Here, \(x\) is the number of occurrences and \(n\) is the total sample size of the group. By calculating the sample proportions, we can have a better understanding of the event's frequency in each group, which is crucial for further statistical analysis. In our exercise, we have \(\hat{p}_{1} = \frac{109}{475}\) for group 1, and \(\hat{p}_{2} = \frac{78}{325}\) for group 2. Calculating these will give us numerical values that we need for subsequent steps.
Standard Error
The standard error (SE) is a measure of how much the sample proportions are expected to fluctuate due to sampling variability. It helps to quantify the uncertainty in the estimate of the difference between two proportions. The combined standard error for the difference between two proportions can be calculated using the formula: \[\text{SE} = \sqrt{\hat{p}_{1}(1-\hat{p}_{1})/n_{1} + \hat{p}_{2}(1-\hat{p}_{2})/n_{2}}\]. A smaller standard error indicates more precise estimates of the difference in proportions. This concept is important because it directly impacts the width of our confidence interval. The more accurate our estimate, the narrower the confidence interval.
Critical Value
The critical value (z*) corresponds to the desired level of confidence we need for our interval. It is derived from the standard normal distribution and reflects the probability that the true difference in proportions lies within a specified range. For a 99% confidence level, the critical value is approximately \(2.576\). This high value ($2.576) further accounts for a wider range within which we can be 99% confident that the true difference falls. It's important to use this appropriate critical value to multiply it with the standard error while calculating the margin of error.
Margin of Error
The margin of error (ME) provides an upper and lower boundary for our confidence interval. It's calculated by multiplying the critical value by the standard error: \(\text{ME} = z* \times \text{SE}\). This boundary tells us how much the sample proportion difference could vary from the true population proportion difference. A larger margin of error means a wider confidence interval, which indicates greater uncertainty, while a smaller margin of error signifies higher precision.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

Perform the appropriate hypothesis test. If \(n_{1}=31, s_{1}=12, n_{2}=51,\) and \(s_{2}=10,\) test whether \(\sigma_{1}>\sigma_{2}\) at the \(\alpha=0.05\) level of significance.

Assume that the populations are normally distributed. (a) Test whether \(\mu_{1}<\mu_{2}\) at the \(\alpha=0.05\) level of significance for the given sample data. (b) Construct a \(95 \%\) confidence interval about \(\mu_{1}-\mu_{2}\). $$ \begin{array}{lcc} & \text { Sample 1 } & \text { Sample 2 } \\ \hline n & 40 & 32 \\ \hline \bar{x} & 94.2 & 115.2 \\ \hline s & 15.9 & 23.0 \\ \hline \end{array} $$

To illustrate the effects of driving under the influence (DUI) of alcohol, a police officer brought a DUI simulator to a local high school. Student reaction time in an emergency was measured with unimpaired vision and also while wearing a pair of special goggles to simulate the effects of alcohol on vision. For a random sample of nine teenagers, the time (in seconds) required to bring the vehicle to a stop from a speed of 60 miles per hour was recorded. $$ \begin{array}{lccccccccc} \text { Subject } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} \\ \hline \text { Normal, } X_{i} & 4.47 & 4.24 & 4.58 & 4.65 & 4.31 & 4.80 & 4.55 & 5.00 & 4.79 \\ \hline \text { Impaired, } Y_{i} & 5.77 & 5.67 & 5.51 & 5.32 & 5.83 & 5.49 & 5.23 & 5.61 & 5.63 \\ \hline \end{array} $$ (a) Whether the student had unimpaired vision or wore goggles first was randomly selected. Why is this a good idea in designing the experiment? (b) Use a \(95 \%\) confidence interval to test if there is a difference in braking time with impaired vision and normal vision where the differences are computed as "impaired minus normal." Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Assume that the populations are normally distributed. Test the given hypothesis. \(\sigma_{1}>\sigma_{2}\) at the \(\alpha=0.01\) level of significance $$ \begin{array}{ccc} & \text { Sample 1 } & \text { Sample 2 } \\ \hline n & 26 & 19 \\ \hline s & 9.9 & 6.4 \\ \hline \end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.