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Chapter 7: Problem 2

Find \(\hat{p}\) and \(\hat{q}\) for each situation. $$ \begin{array}{l}{\text { a. } n=60 \text { and } X=35} \\ {\text { b. } n=95 \text { and } X=43} \\ {\text { c. } 68 \%} \\ {\text { d. } 55 \%} \\\ {\text { e. } 12 \%}\end{array} $$

Short Answer

Expert verified
a. \(\hat{p} = \frac{7}{12}, \hat{q} = \frac{5}{12}\); b. \(\hat{p} = \frac{43}{95}, \hat{q} = \frac{52}{95}\); c. \(\hat{p} = 0.68, \hat{q} = 0.32\); d. \(\hat{p} = 0.55, \hat{q} = 0.45\); e. \(\hat{p} = 0.12, \hat{q} = 0.88\).

Step by step solution

01

Understanding the Problem

The problem asks us to find the estimates \( \hat{p} \) and \( \hat{q} \) for different scenarios. Here, \( \hat{p} \) represents the sample proportion, and \( \hat{q} = 1 - \hat{p} \) represents the complement proportion.
02

Solving Part A

Given \( n = 60 \) and \( X = 35 \). The sample proportion is given by \( \hat{p} = \frac{X}{n} = \frac{35}{60} \). First, simplify this fraction: \( \frac{35}{60} = \frac{7}{12} \). Thus, \( \hat{p} = \frac{7}{12} \). Calculating \( \hat{q} = 1 - \hat{p} = 1 - \frac{7}{12} = \frac{5}{12} \).
03

Solving Part B

Given \( n = 95 \) and \( X = 43 \). Compute \( \hat{p} = \frac{X}{n} = \frac{43}{95} \). Simplification is not possible for \( \frac{43}{95} \), so we keep it as \( \hat{p} = \frac{43}{95} \). Then, \( \hat{q} = 1 - \hat{p} = 1 - \frac{43}{95} = \frac{52}{95} \).
04

Solving Part C

Given a percentage of 68\%. The proportion \( \hat{p} = \frac{68}{100} = 0.68 \). \( \hat{q} = 1 - \hat{p} = 1 - 0.68 = 0.32 \).
05

Solving Part D

Given a percentage of 55\%. Calculate \( \hat{p} = \frac{55}{100} = 0.55 \). \( \hat{q} = 1 - \hat{p} = 1 - 0.55 = 0.45 \).
06

Solving Part E

Given a percentage of 12\%. Determine \( \hat{p} = \frac{12}{100} = 0.12 \). Consequently, \( \hat{q} = 1 - \hat{p} = 1 - 0.12 = 0.88 \).

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Key Concepts

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

Complement Proportion
When working with proportions, it's crucial to understand the concept of complement proportion. In essence, if you know the proportion of a given event, its complement is simply the proportion of the event NOT occurring. Mathematically, if the proportion or probability of an event happening is denoted by \( \hat{p} \), then the complement proportion \( \hat{q} \) is represented by the expression:
  • \( \hat{q} = 1 - \hat{p} \)
For example, if you know that there is a proportion \( \hat{p} = 0.6 \) of choosing "heads" in a coin toss, the complement proportion of choosing "tails" would be \( \hat{q} = 1 - 0.6 = 0.4 \). This concept helps ensure that both probabilities add up to 1, confirming they account for all possibilities in this simple probability space.
Percentages
Understanding percentages is fundamental in grasping proportions and probabilities. Percentages portray a number as a part of 100. If something is 25% of a whole, it means it is 25 out of 100 parts of that whole.To convert a percentage to a decimal which can be used as a sample proportion, you divide the percentage by 100. For instance:
  • 68% becomes \( \frac{68}{100} = 0.68 \)
  • 55% becomes \( \frac{55}{100} = 0.55 \)
  • 12% becomes \( \frac{12}{100} = 0.12 \)
Using these conversions, percentages can easily translate into proportions, which can be further analyzed or used for calculation in various statistical or real-world scenarios.
Fraction Simplification
A key skill in math, particularly when calculating proportions, is fraction simplification. Simplification reduces fractions to their simplest form, making them easier to interpret and compare. To simplify a fraction, find the greatest common divisor (GCD) of the numerator and the denominator, then divide both by this number.For instance, consider simplifying the fraction \( \frac{35}{60} \). Both 35 and 60 can be divided by their GCD of 5, yielding:
  • \( \frac{35}{60} = \frac{35 \div 5}{60 \div 5} = \frac{7}{12} \)
Simplified fractions are straightforward to understand and more manageable for further arithmetic operations, ensuring accuracy in mathematical conclusions.

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Most popular questions from this chapter

Find each. $$ \begin{array}{l}{\text { a. } z_{a / 2} \text { for the } 99 \% \text { confidence interval }} \\ {\text { b. } z_{a / 2} \text { for the } 98 \% \text { confidence interval }} \\ {\text { c. } z_{a / 2} \text { for the } 95 \% \text { confidence interval }} \\ {\text { d. } z_{a / 2} \text { for the } 90 \% \text { confidence interval }} \\ {\text { e. } z_{a / 2} \text { for the } 94 \% \text { confidence interval }}\end{array} $$

Cyber Monday Shopping A survey of \(1000 \mathrm{U.S}\). adults found that \(33 \%\) of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the \(95 \%\) confidence inter- val of the true proportion.

Home Ownership Rates The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with \(99 \%\) confidence. Assume the variable is normally distributed. $$ 66.075 .8 \quad 70.9 \quad 73.9 \quad 63.4 \quad 68.5 \quad 73.3 \quad 65.9 $$

Unhealthy Days in Cities The number of unhealthy days based on the AQI (Air Quality Index) for a random sample of metropolitan areas is shown. Con- struct a \(98 \%\) confidence interval based on the data. $$ \begin{array}{lllllllllll}{61} & {12} & {6} & {40} & {27} & {38} & {93} & {5} & {13} & {40}\end{array} $$

Internet Viewing A researcher wishes to estimate the average number of minutes per day a person spends on the Internet. How large a sample must she select if she wishes to be \(90 \%\) confident that the population mean is within 10 minutes of the sample mean? Assume the population standard deviation is 42 minutes.
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