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Chapter 3: Problem 7

Fireworks on July \(4^{\text {th }}\). In late June 2012, Survey USA published results of a survey stating that \(56 \%\) of the 600 randomly sampled Kansas residents planned to set off fireworks on July \(4^{t h}\). Determine the margin of error for the \(56 \%\) point estimate using a \(95 \%\) confidence level.

Short Answer

Expert verified
The margin of error is approximately 3.97%.

Step by step solution

01

Understanding the Problem

We need to determine the margin of error for a sample proportion when the sample size and the confidence level are given. In this exercise, the proportion of Kansas residents planning to set off fireworks is given as 56%, the sample size is 600, and the confidence level is 95%.
02

Identify Formula and Values

To find the margin of error for a proportion, we can use the formula: \[ ME = Z \cdot \sqrt{\frac{p(1-p)}{n}} \] where \( p = 0.56 \) (the sample proportion), \( n = 600 \) (the sample size), and \( Z \) is the Z-score corresponding to a 95% confidence level, which is approximately 1.96.
03

Calculate Standard Error

First, we calculate the standard error (SE) of the proportion using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \] Substituting the values, \( SE = \sqrt{\frac{0.56 \times 0.44}{600}} \). Calculate this to find the standard error.
04

Compute the Standard Error

Substituting the numbers into the standard error formula, we have: \( SE = \sqrt{\frac{0.56 \times 0.44}{600}} = \sqrt{\frac{0.2464}{600}} = \sqrt{0.000410667} \approx 0.02026 \).
05

Determine the Margin of Error

The margin of error (ME) is calculated by multiplying the standard error by the Z-score for 95% confidence level. Using \( ME = 1.96 \times 0.02026 \), we get \( ME \approx 0.0397 \).
06

Convert Margin of Error to Percentage

Since we initially started with a percentage, convert the margin of error back to percentage: \( ME \approx 0.0397 \times 100 \approx 3.97\% \). This is the margin of error for the estimate.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values that is likely to contain the true population parameter. In the context of survey results, it helps us understand the reliability of our estimate, such as the percentage of Kansas residents planning to set off fireworks on July 4th.
The confidence level, commonly expressed as a percentage (e.g., 95%), indicates how sure we are that the interval contains the true population parameter. A 95% confidence level means that if we were to take 100 different samples and construct a confidence interval for each one, we expect about 95 of those intervals to contain the true proportion.
In our specific scenario, the 56% represents the sample proportion, and the margin of error gives us a buffer around this estimate. Adding and subtracting the margin of error to the sample proportion provides the confidence interval, offering a range of percentages with confidence that it contains the true population proportion.
Sample Proportion
In survey sampling, the sample proportion is the ratio of individuals in the sample who have a particular characteristic to the total number of individuals in the sample.
For example, in our survey of Kansas residents, 56% is the sample proportion. This means that 56 out of every 100 people in our sample were planning to set off fireworks on the specified date. The sample proportion is often denoted by the letter \( p \) and used as an estimate of the true population proportion.
Understanding sample proportions is crucial because they form the basis for constructing confidence intervals and determining the margin of error. They give us a point estimate that can be further refined by statistical techniques like calculating the standard error or applying the Z-score for more accurate and reliable predictions.
Standard Error
The standard error (SE) is a key statistical metric that measures the variability of a sample proportion from the true population proportion.
In simple terms, the standard error tells us how much we can expect our sample proportion to fluctuate due to random sampling error. It's calculated using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the sample proportion and \( n \) is the sample size.
For our fireworks survey, substituting the values into the formula gives us the standard error. SE is smaller when the sample size is larger, indicating that larger samples provide more precise estimates. By understanding SE, statisticians can assess the reliability of a sample proportion, leading to more informed decisions.
Z-Score
The Z-score is a statistical measurement that tells us how many standard deviations a data point is from the mean within a standard normal distribution.
When calculating confidence intervals, the Z-score helps us determine how far we can stretch from the sample mean to capture a certain percentage of all possible sample means. For a 95% confidence interval, the Z-score is approximately 1.96. This Z-score provides the necessary multiplier to apply to the standard error when calculating the margin of error for our survey data.
Using Z-scores, statistical calculations translate empirical data into a broader picture, such as determining plausible population proportions based on our sample. Understanding how Z-scores work helps in applying confidence measures accurately, enabling us to make statistical inferences with greater confidence.

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Most popular questions from this chapter

Prop 19 in California. In a 2010 Survey USA poll, \(70 \%\) of the 119 respondents between the ages of 18 and 34 said they would vote in the 2010 general election for Prop \(19,\) which would change California law to legalize marijuana and allow it to be regulated and taxed. At a \(95 \%\) confidence level, this sample has an \(8 \%\) margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning." (a) We are \(95 \%\) confident that between \(62 \%\) and \(78 \%\) of the California voters in this sample support Prop \(19 .\) (b) We are \(95 \%\) confident that between \(62 \%\) and \(78 \%\) of all California voters between the ages of 18 and 34 support Prop \(19 .\) (c) If we considered many random samples of 119 California voters between the ages of 18 and 34 , and we calculated \(95 \%\) confidence intervals for each, \(95 \%\) of them will include the true population proportion of Californians who support Prop \(19 .\) (d) In order to decrease the margin of error to \(4 \%\), we would need to quadruple (multiply by 4) the sample size. (e) Based on this confidence interval, there is sufficient evidence to conclude that a majority of California voters between the ages of 18 and 34 support Prop \(19 .\)

Shipping holiday gifts. A December 2010 survey asked 500 randomly sampled Los Angeles residents which shipping carrier they prefer to use for shipping holiday gifts. The table below shows the distribution of responses by age group as well as the expected counts for each cell (shown in parentheses). Method \begin{tabular}{l|cc|cc|cc|c} & \multicolumn{6}{|c|} { Age } & \\ \cline { 2 - 5 } & \multicolumn{2}{|c|} {\(18-34\)} & \multicolumn{2}{|c|} {\(35-54\)} & \multicolumn{2}{|c|} {\(55+\)} & Total \\ \hline USPS & 72 & \((\mathrm{~B} 1)\) & 97 & (102) & 76 & (62) & 245 \\ UPS & 52 & (83) & 76 & \((6 \mathrm{~b})\) & 34 & (41) & 162 \\ FedEx & 31 & (21) & 24 & (27) & 9 & (16) & 64 \\ Something else & 7 & (5) & 6 & (7) & 3 & (4) & 16 \\ Not sure & 3 & (5) & 6 & (5) & 4 & (3) & 13 \\ \hline Total & \multicolumn{2}{|c|} {165} & \multicolumn{2}{|c|} {209} & \multicolumn{2}{|c|} {126} & 500 \end{tabular} (a) State the null and alternative hypotheses for testing for independence of age and preferred shipping method for holiday gifts among Los Angeles residents. (b) Are the conditions for inference using a chi-square test satisfied?

Heart transplant success. The Stanford University Heart Transplant Study was conducted to determine whether an experimental heart transplant program increased lifespan. Each patient entering the program was officially designated a heart transplant candidate, meaning that he was gravely ill and might benefit from a new heart. Patients were randomly assigned into treatment and control groups. Patients in the treatment group received a transplant, and those in the control group did not. The table below displays how many patients survived and died in each 41 group. \begin{tabular}{ccc} \hline & control & treatment \\ \hline alive & 4 & 24 \\ dead & 30 & 45 \\ \hline \end{tabular} A hypothesis test would reject the conclusion that the survival rate is the same in each group, and so we might like to calculate a confidence interval. Explain why we cannot construct such an interval using the normal approximation. What might go wrong if we constructed the confidence interval despite this problem?

True or false, Part. II. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found \(X^{2}=10\) with \(d f=5\) you would fail to reject \(H_{0}\) at the \(5 \%\) significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.

The Civil War. A national survey conducted in 2011 among a simple random sample of 1.507 adults shows that \(56 \%\) of Americans think the Civil War is still relevant to American politics and political life. \({ }^{37}\) (a) Conduct a hypothesis test to determine if these data provide strong evidence that the majority of the Americans think the Civil War is still relevant. (b) Interpret the p-value in this context. (c) Calculate a \(90 \%\) confidence interval for the proportion of Americans who think the Civil War is still relevant. Interpret the interval in this context, and comment on whether or not the confidence interval agrees with the conclusion of the hypothesis test.
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