/*! 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 37 Time/CNN voter polls monitored p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 37

Time/CNN voter polls monitored public opinion for the presidential candidates during the 2000 presidential election campaign. One Time/CNN poll conducted by Yankelovich Partners, Inc., used a sample of 589 likely voters (Time, June 26,2000 ). Assume the population proportion for a presidential candidate is \(p=.50 .\) Let \(\bar{p}\) be the sample proportion of likely voters favoring the presidential candidate. a. Show the sampling distribution of \(\bar{p}\) b. What is the probability the Time/CNN poll will provide a sample proportion within 卤.04 of the population proportion? c. What is the probability the Time/CNN poll will provide a sample proportion within 卤.03 of the population proportion? d. What is the probability the Time/CNN poll will provide a sample proportion within 卤.02 of the population proportion?

Short Answer

Expert verified
a. Approx. normal distribution with mean 0.50 and standard error 0.0206. Probability within 卤0.04: 0.9460; within 卤0.03: 0.8558; within 卤0.02: 0.6680.

Step by step solution

01

Define the Sampling Distribution of Sample Proportion

The sampling distribution of the sample proportion \( \bar{p} \) is approximately normal as long as the sample size \( n \) is large enough. The mean of this distribution is equal to the population proportion \( p \). The standard deviation (standard error) of the sample proportion is calculated as follows:\[\sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}}.\]In this case, \( p = 0.50 \) and \( n = 589 \).
02

Calculate the Standard Error

Substitute the given values into the standard error formula:\[\sigma_{\bar{p}} = \sqrt{\frac{0.50(1-0.50)}{589}} = \sqrt{\frac{0.25}{589}} \approx 0.0206.\]
03

Probability of Sample Proportion within 卤0.04

To find the probability the sample proportion \( \bar{p} \) is within 卤0.04 of the population proportion \( p \), we calculate:\[p(0.50 - 0.04 < \bar{p} < 0.50 + 0.04) = P(0.46 < \bar{p} < 0.54)\]Convert this range to a standard normal distribution (Z-score):\[P\left( \frac{0.46 - 0.50}{0.0206} < Z < \frac{0.54 - 0.50}{0.0206} \right)\]\[P(-1.942 < Z < 1.942).\]
04

Use Z-tables for Probability Calculation (卤0.04)

Using the standard normal distribution table, find the probability corresponding to \( Z = 1.942 \):\[ P(Z < 1.942) \approx 0.9730 \]\[ P(Z < -1.942) \approx 0.0270 \]Thus, the probability is\[ 0.9730 - 0.0270 = 0.9460. \]
05

Repeat Process for 卤0.03

Calculate the range for 卤0.03:\[P(0.50 - 0.03 < \bar{p} < 0.50 + 0.03) = P(0.47 < \bar{p} < 0.53)\]Convert to Z-scores:\[P\left( \frac{0.47 - 0.50}{0.0206} < Z < \frac{0.53 - 0.50}{0.0206} \right)\]\[P(-1.456 < Z < 1.456).\]
06

Use Z-tables for Probability Calculation (卤0.03)

Using Z-tables:\[ P(Z < 1.456) \approx 0.9279 \]\[ P(Z < -1.456) \approx 0.0721 \]Thus, the probability is\[ 0.9279 - 0.0721 = 0.8558. \]
07

Repeat Process for 卤0.02

Calculate the range for 卤0.02:\[P(0.50 - 0.02 < \bar{p} < 0.50 + 0.02) = P(0.48 < \bar{p} < 0.52)\]Convert to Z-scores:\[P\left( \frac{0.48 - 0.50}{0.0206} < Z < \frac{0.52 - 0.50}{0.0206} \right)\]\[P(-0.971 < Z < 0.971).\]
08

Use Z-tables for Probability Calculation (卤0.02)

Using Z-tables:\[ P(Z < 0.971) \approx 0.8340 \]\[ P(Z < -0.971) \approx 0.1660 \]Thus, the probability is\[ 0.8340 - 0.1660 = 0.6680. \]

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 91影视!

Key Concepts

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

Sample Proportion
The sample proportion \( \bar{p} \) reflects the success proportion in a given sample. It helps us to make inferences about the population. To determine \( \bar{p} \), you count how many times an attribute appears in the sample and divide by the total sample size. In voting polls, this might mean counting how many people support a candidate and dividing by the total number of surveyed voters.
For instance, if out of a sample of 589 voters, 294 support a candidate, the sample proportion would be \( \bar{p} = \frac{294}{589} \approx 0.50 \).
This is crucial because it allows statisticians to predict how similar or different the sample is compared to the whole population's behavior or characteristic proportion.
Standard Error
The standard error gives an idea about how much our sample proportion \( \bar{p} \) might fluctuate if we took several samples. It is a measure of the dispersion or variability around the estimated sample proportion.
The standard error is calculated using the formula: \[\sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}}\] where \( p \) is the known population proportion, and \( n \) is the sample size. In our scenario, with \( p = 0.50 \) and \( n = 589 \), the standard error becomes approximately 0.0206.
This tells us that if we repeatedly took random samples, the proportion of support would vary around the true population proportion by about 卤0.0206.
Z-score
The Z-score helps us understand where our sample proportion falls within the normal distribution. It standardizes the sample proportion by showing how many standard errors away it is from the population proportion.
To calculate the Z-score for a sample proportion, use the following formula: \[ Z = \frac{\bar{p} - p}{\sigma_{\bar{p}}} \] This formula helps in determining the probability of finding a sample proportion within a specified range. For example, if we want to know the likelihood of \( \bar{p} \) being within 卤0.04 of \( p = 0.50 \), we check how far 0.46 and 0.54 are from 0.50 using the calculated standard error of 0.0206.
Understanding the Z-score helps in comparing diverse sample results effectively since it translates results into a common scale that can be referenced against standard normal distribution probabilities.

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

The average price of a gallon of unleaded regular gasoline was reported to be \(\$ 2.34\) in northern Kentucky (The Cincinnati Enquirer, January 21,2006 ). Use this price as the population mean, and assume the population standard deviation is \(\$ .20\) a. What is the probability that the mean price for a sample of 30 service stations is within \(\$ .03\) of the population mean? b. What is the probability that the mean price for a sample of 50 service stations is within \(\$ .03\) of the population mean? c. What is the probability that the mean price for a sample of 100 service stations is within \(\$ .03\) of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within \(\$ .03\) of the population mean?

A production process is checked periodically by a quality control inspector. The inspector selects simple random samples of 30 finished products and computes the sample mean product weights \(\bar{x}\). If test results over a long period of time show that \(5 \%\) of the \(\bar{x}\) values are over 2.1 pounds and \(5 \%\) are under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process?

BusinessWeek conducted a survey of graduates from 30 top MBA programs (Business- Week, September 22,2003 . On the basis of the survey, assume that the mean annual salary for male and female graduates 10 years after graduation is \(\$ 168,000\) and \(\$ 117,000,\) respectively. Assume the standard deviation for the male graduates is \(\$ 40,000,\) and for the female graduates it is \(\$ 25,000\) a. What is the probability that a simple random sample of 40 male graduates will provide a sample mean within \(\$ 10,000\) of the population mean, \(\$ 168,000 ?\) b. What is the probability that a simple random sample of 40 female graduates will provide a sample mean within \(\$ 10,000\) of the population mean, \(\$ 117,000 ?\) c. In which of the preceding two cases, part (a) or part (b), do we have a higher probability of obtaining a sample estimate within \(\$ 10,000\) of the population mean? Why? d. What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than \(\$ 4000\) below the population mean?

A simple random sample of 5 months of sales data provided the following information: \\[ \begin{array}{lrrrrr} \text {Month:} & 1 & 2 & 3 & 4 & 5 \\ \text {Units Sold:} & 94 & 100 & 85 & 94 & 92 \end{array} \\] a. Develop a point estimate of the population mean number of units sold per month. b. Develop a point estimate of the population standard deviation.

A population has a mean of 200 and a standard deviation of \(50 .\) Suppose a simple random sample of size 100 is selected and \(\bar{x}\) is used to estimate \(\mu\) a. What is the probability that the sample mean will be within 卤5 of the population mean? b. What is the probability that the sample mean will be within 卤10 of the population mean?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation
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.