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Chapter 9: Problem 45

There are 20,000 eligible voters in York County, South Carolina. A random sample of 500 York County voters revealed 350 plan to vote to return Louella Miller to the state senate. Construct a \(99 \%\) confidence interval for the proportion of voters in the county who plan to vote for Ms. Miller. From this sample information, is it reasonable to conclude that Ms. Miller will receive a majority of the votes?

Short Answer

Expert verified
Yes, the confidence interval suggests Ms. Miller will likely receive a majority of the votes.

Step by step solution

01

Identify the Sample Proportion

The sample proportion \( \hat{p} \) is the ratio of voters in the sample who plan to vote for Ms. Miller. Calculated as \( \hat{p} = \frac{350}{500} = 0.7 \).
02

Identify the Z-Score for a 99% Confidence Interval

For a 99% confidence interval, the Z-score is approximately 2.576. This value corresponds to the critical value from the standard normal distribution used for 99% confidence levels.
03

Calculate the Standard Error

The standard error \( SE \) of the sample proportion is calculated using the formula: \( SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } = \sqrt{ \frac{0.7(1 - 0.7)}{500} } = 0.0205 \).
04

Determine the Margin of Error

The margin of error \( ME \) is found by multiplying the Z-score by the standard error: \( ME = Z \times SE = 2.576 \times 0.0205 = 0.0528 \).
05

Construct the Confidence Interval

The 99% confidence interval is calculated as: \( \hat{p} \pm ME \). Thus the confidence interval is \( 0.7 \pm 0.0528 \), or \( (0.6472, 0.7528) \).
06

Analyze the Confidence Interval

Since the lower bound of the confidence interval \(0.6472\) is greater than 0.5, it is reasonable to conclude with 99% confidence that Ms. Miller will receive a majority of the votes.

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Key Concepts

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

Sample Proportion
A sample proportion is a key statistic in estimating the characteristics of a population based on a sample from that population. In the given exercise, the sample proportion represents the fraction of voters in the sample who intend to vote for Ms. Miller.

To find it, we use the equation:\[ \hat{p} = \frac{x}{n} \]where \(x\) is the number of favorable outcomes (voters planning to vote for Ms. Miller) and \(n\) is the total number of observations in the sample. Here, \(x = 350\) and \(n = 500\), so \(\hat{p} = \frac{350}{500} = 0.7\).

In this context:
  • The sample proportion (\(\hat{p}\)) is an estimate of the true proportion of the population.
  • It gives us a point estimate but not the reliability of the estimate.
Standard Error
The standard error measures the variability or dispersion of the sample proportion. It acts as a standard deviation for the sample proportion, showing us how much the proportion estimate could vary from sample to sample.

The formula for standard error (SE) is given by:\[ SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \]This equation takes into account:
  • The sample proportion \(\hat{p}\), which is 0.7.
  • The complement of the sample proportion, \(1 - \hat{p}\), which is 0.3.
  • The sample size \(n\), which is 500.
Upon calculation, \( SE = \sqrt{ \frac{0.7 \times 0.3}{500} } = 0.0205\).

The smaller the standard error, the more precise the estimate is. A larger sample would typically decrease the standard error, increasing confidence in the reliability of the estimate.
Margin of Error
The margin of error gives us an amount by which our sample proportion may differ from the true population proportion. It combines the standard error and a chosen Z-score to offer a range in which we can say, with a certain confidence level, that the true population proportion lies.

The formula for calculating the margin of error (ME) is:\[ ME = Z \times SE \]where:
  • \(Z\) is the Z-score that corresponds to our confidence level (2.576 for 99%).
  • \(SE\) is the standard error, here calculated as 0.0205.
Using these values, the margin of error is \( ME = 2.576 \times 0.0205 = 0.0528 \).

The margin of error tells us that the true population proportion is likely within 0.0528 of our sample proportion. This accounts for the inherent variability in sampling.
Z-Score
The Z-score is a key player in constructing confidence intervals. It represents the number of standard deviations a data point is from the mean in a standard normal distribution.

For confidence intervals, the Z-score determines the confidence level. A higher Z-score corresponds to higher confidence. In this problem, we use a Z-score of 2.576, which is typical for a 99% confidence interval.

The use of Z-score:
  • Helps in calculating the margin of error, thus providing a range for the confidence interval.
  • A Z-score of 2.576 suggests we're capturing most possible sample variations (99% of them).
Z-scores are derived from statistical tables or software, which relate Z-scores to the cumulative probability in the normal distribution. This decision ensures our confidence interval effectively contains the true population proportion with 99% certainty.

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Most popular questions from this chapter

The human relations department of Electronics Inc. would like to include a dental plan as part of the benefits package. The question is: How much does a typical employee and his or her family spend per year on dental expenses? A sample of 45 employees reveals the mean amount spent last year was 1,820, with a standard deviation of 660. a. Construct a 95%$confidence interval for the population mean. b. The information from part (a) was given to the president of Electronics Inc. He indicated he could afford 1,700 of dental expenses per employee. Is it possible that the population mean could be 1,700 ? Justify your answer.

Marty Rowatti recently assumed the position of director of the YMCA of South Jersey. He would like some data on how long current members of the YMCA have been members. To investigate, suppose he selects a random sample of 40 current members. The mean length of membership for the sample is 8.32 years and the standard deviation is 3.07 years. a. What is the mean of the population? b. Develop a \(90 \%\) confidence interval for the population mean. c. The previous director, in the summary report she prepared as she retired, indicated the mean length of membership was now "almost 10 years." Does the sample information substantiate this claim? Cite evidence.

Near the time of an election, a cable news service performs an opinion poll of 1,000 probable voters. It shows that the Republican contender has an advantage of \(52 \%\) to \(48 \% .\) a. Develop a \(95 \%\) confidence interval for the proportion favoring the Republican candidate. b. Estimate the probability that the Democratic candidate is actually leading. c. Repeat the above analysis based on a sample of 3,000 probable voters.

Dylan Jones kept careful records of the fuel efficiency of his new car. After the first nine times he filled up the tank, he found the mean was 23.4 miles per gallon (mpg) with a sample standard deviation of 0.9 mpg. a. Compute the \(95 \%\) confidence interval for his mpg. b. How many times should he fill his gas tank to obtain a margin of error below 0.1 mpg?

A sample of 250 observations is selected from a normal population with a population standard deviation of \(25 .\) The sample mean is 20 . a. Determine the standard error of the mean. b. Explain why we can use formula \((9-1)\) to determine the \(95 \%\) confidence interval. c. Determine the \(95 \%\) confidence interval for the population mean.
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