/*! 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 60 Construct a \(99 \%\) confidence... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Construct a \(99 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=300 \quad \hat{p}_{1}=.55 \quad n_{2}=200 \quad \hat{p}_{2}=.62 $$

Short Answer

Expert verified
The \(99\%\) confidence interval for the difference in population proportions \(p_{1} - p_{2}\) is approximately \(-0.214\) to \(0.074\).

Step by step solution

01

Calculate difference in sample proportions

First, we need to compute the difference in the sample proportions. This is simply \(\hat{p}_{1} - \(\hat{p}_{2}\), or \(.55 - .62 = -.07.\)
02

Calculate the standard error

The standard error for the difference in two proportions is calculated as: \[ \sqrt { \frac {\hat{p}_{1} (1- \hat{p}_{1})} {n_{1}} + \frac {\hat{p}_{2} (1- \hat{p}_{2})} {n_{2}} } \] By inserting the numbers: \[ \sqrt { \frac {.55(1 - .55)} {300} + \frac {.62(1 - .62)} {200} } = 0.056. \]
03

Determine the z-score for the desired confidence level

Given a \(99 \%\) confidence interval, we consult a z-table or use a calculator to find the z-score associated with this confidence level. The z-score for a \(99 \%\) confidence level is approximately \(2.576\).
04

Apply the formula for the confidence interval

The confidence interval is calculated as: \[ (\hat{p}_{1} - \hat{p}_{2}) \pm Z_{\alpha/2} * standard \: error. \] Therefore, the \(99 \%\) CI is: \[ (-0.07) \pm 2.576 (0.056), \] which is approximately \(-0.214\) to \(0.074\).

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
Sample proportions are a way to estimate the proportion of a characteristic in a population using a smaller, more manageable subset called a sample. For instance, if you want to understand how many people prefer tea over coffee in a city, instead of asking everyone, you can take a sample.
Taking a sample involves selecting a group of people and finding the proportion of them who prefer tea. If 165 out of 300 people prefer tea, the sample proportion (\( \hat{p} \)) is 0.55.
In this exercise, we compute sample proportions for two different samples:
  • Sample 1 has \( \hat{p}_1 = 0.55 \) with a sample size \( n_1 = 300 \).
  • Sample 2 has \( \hat{p}_2 = 0.62 \) with a sample size \( n_2 = 200 \).
These sample proportions help us understand the characteristic or preference of the two distinct groups.
Finally, the difference \( \hat{p}_1 - \hat{p}_2 \) indicates how these proportions compare to each other.
Standard Error
The standard error measures variability. Specifically, it tells us how much we expect the sample proportion to differ from the true population proportion.
It provides insight into how well our sample represents the population.
The key idea behind standard error for two sample proportions is combining the variability from two groups.
This is calculated using the formula: \[ \sqrt{ \frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2} } \]In our context, using values
  • \( \hat{p}_1 = 0.55 \), \( n_1 = 300 \), \( \hat{p}_2 = 0.62 \), \( n_2 = 200 \).
This gives a standard error of \( 0.056 \).
The smaller the standard error, the more stable our estimate, helping make better predictions about population differences.
Z-Score
A z-score gauges how far a data point is from the mean, measured in standard deviations.
For confidence intervals, z-scores point to the number of standard deviations a statistic, like a sample proportion, is from the estimated population parameter.
For a given confidence level, a specific z-score corresponds.
For example, a 99% confidence level results in a z-score of approximately \( 2.576 \).
This means we can be 99% confident that the true difference in proportions lies within the range calculated using this z-score.
  • Higher confidence levels yield higher z-scores.
  • Higher z-scores denote a wider interval.
Keeping this in mind helps us understand what range our difference in sample proportions could have when calculating the interval.
Confidence Level
The confidence level indicates how sure we are that the interval contains the true population parameter.
Its common values include 90%, 95%, and 99%.
  • A 99% confidence level means we are sure, almost all of the time, that our interval covers the actual difference in proportions between two groups.
The interval is calculated as\[(\hat{p}_1 - \hat{p}_2) \pm Z_{\alpha/2} \times \text{standard error}\]In our case:
  • The difference is \(-0.07\).
  • Z-score for a 99% level is approximately \(2.576\).
  • The standard error is \(0.056\).
So, the interval becomes approximately \(-0.214\) to \(0.074\).
This implies, with 99% confidence, the true difference of proportions lies within this range, helping us make informed decisions about our hypothesis.

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

Two competing airlines, Alpha and Beta, fly a route between Des Moines, Iowa, and Wichita, Kansas. Each airline claims to have a lower percentage of flights that arrive late. Let \(p_{1}\) be the proportion of Alpha's flights that arrive late and \(p_{2}\) the proportion of Beta's tlights that arrive late. a. You are asked to observe a random sample of arrivals for each airline to estimate \(p_{1}-p_{2}\) with a \(90 \%\) confidence level and a margin of error of estimate of \(.05 .\) How many arrivals for each airline would you have to observe? (Assume that you will observe the same number of arrivals, \(n\), for each airline. To be sure of taking a large enough sample, use \(p_{1}=p_{2}=.50\) in your calculations for \(n .\) ) b. Suppose that \(p_{1}\) is actually \(.30\) and \(p_{2}\) is actually \(.23 .\) What is the probability that a sample of 100 flights for each airline ( 200 in all) would yield \(\hat{p}_{1} \geq \hat{p}_{2}\) ?

A town that recently started a single-stream recycling program provided 60-gallon recycling bins to 25 randomly selected households and 75-gallon recycling bins to 22 randomly selected households. The total volume of recycling over a 10-week period was measured for each of the households. The average total volumes were 382 and 415 gallons for the households with the 60 - and 75 -gallon bins, respectively. The sample standard deviations were \(52.5\) and \(43.8\) gallons, respectively. Assume that the 10 -week total volumes of recycling are approximately normally distributed for both groups and that the population standard deviations are equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recycling for the households with the 60 - and 75 -gallon bins. b. Using the \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average volume of all households that have 75 -gallon containers?

As mentioned in Exercise \(10.29\), a company claims that its medicine, Brand A, provides faster relief from pain than another company's medicine, Brand \(\mathrm{B}\). A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes.$$ \begin{array}{cccc} \hline \text { Brand } & \text { Sample Size } & \begin{array}{c} \text { Mean of } \\ \text { Relief Times } \end{array} & \begin{array}{c} \text { Standard Deviation } \\ \text { of Relief Times } \end{array} \\ \hline \text { A } & 25 & 44 & 11 \\ \text { B } & 22 & 49 & 9 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at the \(1 \%\) significance level whether the mean relief time for Brand \(\mathrm{A}\) is less than that for Brand B. c. Suppose that the sample standard deviations were \(13.3\) and \(7.2\) minutes, respectively. Redo parts a and \(\mathrm{b}\). Discuss any changes in the results.

An economist was interested in studying the impact of the recession on dining out, including drive-thru meals at fast food restaurants. A random sample of forty-eight families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95\). Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have the same population standard deviation. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in diningout spending levels for the two populations. b. Using the \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

Conduct the following tests of hypotheses, assuming that the populations of paired differences are normally distributed a. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=26, \quad \bar{d}=9.6, \quad s_{d}=3.9, \quad \alpha=.05\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=15, \quad \bar{d}=8.8, \quad s_{d}=4.7, \quad \alpha=.01\) c. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=20, \quad \bar{d}=-7.4, \quad s_{d}=2.3, \quad \alpha=.10\)

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.