/*! 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 44 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
After performing these calculations, you should get the \(99 \%\) confidence interval for \(p_{1}-p_{2}\). You need to substitute the calculated standard error in the confidence interval formula to get final interval.

Step by step solution

01

Calculate the Standard Error

The standard error (SE) is calculated using the formula: \[ SE = \sqrt{ \frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2} } \] Substituting the known values we get:\[ SE = \sqrt{ \frac{0.55(1-0.55)}{300} + \frac{0.62(1-0.62)}{200} } \]
02

Compute the difference between sample proportions

Compute the difference between the sample proportions:\[ \hat{p}_1 - \hat{p}_2 = 0.55 - 0.62 = -0.07 \]
03

Calculate the confidence interval

The Confidence Interval is calculated using the formula: \[CI = (\hat{p}_1 - \hat{p}_2) \pm Z * SE\] where \(Z\) is the Z score for the desired level of confidence (2.576 for \(99\%\)). Substituting the known values we get: \[((-0.07) - 2.576*SE, (-0.07) + 2.576*SE)\]

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.

Standard Error
The standard error (SE) is an important concept when constructing a confidence interval for the difference in proportions. It helps us understand the variability of the sample proportion difference, which means how much this difference might vary if we took different samples from the population. The formula used for SE in the context of two proportions is
  • \(SE = \sqrt{ \frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2} }\)
Here, \(\hat{p_1}\) and \(\hat{p_2}\) are the sample proportions, and \(n_1\), \(n_2\) are the sample sizes. The square root incorporates both samples into one measure of spread.

By substituting the given values into the formula, you calculate the SE, which is essential for further steps when determining precise confidence intervals. This standard error acts much like the standard deviation but focuses on the proportion differences.
Sample Proportions
Sample proportions (\(\hat{p_1}\) and \(\hat{p_2}\)) are estimates of population proportions based on sample data. They represent the fraction of the sample that shares a certain characteristic, such as approval of a product or support for a policy.

The difference in sample proportions is calculated as
  • \(\hat{p_1} - \hat{p_2} = 0.55 - 0.62 = -0.07\)
This negative result indicates that the second sample proportion is larger than the first. Understanding the difference in these proportions helps assess whether there is a meaningful difference in the population.

It’s important to remember that sample proportions won't always reflect true population proportions precisely due to possible sampling error.
Z Score
The Z score in confidence interval calculations is a critical factor. It relates to how confident we want to be that the interval contains the true difference between population proportions. Higher confidence levels are associated with larger Z scores.

For a 99% confidence level, the Z score is 2.576. This score reflects the number of standard errors we need to stretch out on each side of our point estimate (the difference in sample proportions) to attain the desired confidence.
  • Using this Z score, the confidence interval is derived with the formula:\[CI = (\hat{p}_1 - \hat{p}_2) \pm Z * SE\]
Here, "±" indicates that the interval is calculated by adding and subtracting the Z score of the SE from the estimated difference. This creates a range where the true difference in population proportions is likely to fall.
Confidence Level
The confidence level indicates the degree of certainty we have that the calculated confidence interval includes the true difference in population proportions. A 99% confidence level means we can be bravely sure that 99 out of 100 times, the interval we compute will contain the true difference.

Choosing a high confidence level like 99% means the interval will be wider because we want to capture the possible true difference with greater certainty.
  • It signifies a trade-off: higher confidence means a broader range and less precision, but more assurance that the range is correct.
  • It’s vital in contexts where making mistakes could be costly, like medication effects, where being wrong could lead to serious consequences.
Ultimately, a 99% confidence level provides a substantial assurance that our results are meaningful and applicable to the real-world scenario we're examining.

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

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. a. Construct a 98\% confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using a \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store \(\mathrm{B}\) ?

Sixty-five percent of all male voters and \(40 \%\) of all female voters favor a particular candidate. A sample of 100 male voters and another sample of 100 female voters will be polled. What is the probability that at least 10 more male voters than female voters will favor this candidate?

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. \(\begin{array}{lllllllll}\text { Sample 1: } & 47.7 & 46.9 & 51.9 & 34.1 & 65.8 & 61.5 & 50.2 & 40.8 \\ & 53.1 & 46.1 & 47.9 & 45.7 & 49.0 & & & \\ \text { Sample 2: } & 50.0 & 47.4 & 32.7 & 48.8 & 54.0 & 46.3 & 42.5 & 40.8 \\ & 39.0 & 68.2 & 48.5 & 41.8 & & & & \end{array}\) a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at a \(1 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

Gamma Corporation is considering the installation of governors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of \(2 \mathrm{mpg}\) with a \(90 \%\) confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of \(s_{d}=3 \mathrm{mpg}\) is reasonable. How many cars should be tested? (Note that the critical value of \(t\) will depend on \(n\), so it will be necessary to use trial and error.)

The following information was obtained from two independent samples selected from two populations with unequal and unknown population standard deviations. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$ Test at a \(2.5 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

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.