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Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. Using the same \(p^{\prime}\) and \(n=80,\) how would the error bound change if the confidence level were increased to 98\(\% ?\) Why?

Short Answer

Expert verified
The error bound increases with a 98% confidence level due to the larger Z-score, indicating a wider interval for greater certainty.

Step by step solution

01

Understand the Variables

Given that there are 64 girls and 16 boys in the selected class, the total number of children, \( n \), is 80. The proportion of girls, \( p^{\prime} \), is calculated by dividing the number of girls by the total number of children: \( p^{\prime} = \frac{64}{80} = 0.8 \).
02

Determine the Standard Error

The standard error (SE) of the proportion is calculated using the formula: \( SE = \sqrt{\frac{p^{\prime}(1-p^{\prime})}{n}} \). Substituting \( p^{\prime} = 0.8 \) and \( n = 80 \), we get \( SE = \sqrt{\frac{0.8(0.2)}{80}} = \sqrt{\frac{0.16}{80}} \approx 0.0447 \).
03

Identify the Z-Score for 98% Confidence Level

For a 98% confidence level, the Z-score is approximately 2.33, which can be found using a standard normal (Z) table or suitable calculator.
04

Calculate the Error Bound for 98% Confidence Level

The error bound (EBM) is calculated using the formula: \( EBM = Z \cdot SE \). With \( Z = 2.33 \) and \( SE \approx 0.0447 \), the error bound is \( EBM = 2.33 \times 0.0447 \approx 0.1042 \).
05

Consider Effect of Increased Confidence Level

By increasing the confidence level from a lower value to 98%, the Z-score increases. A higher Z-score results in a larger error bound, meaning the interval becomes wider, reflecting increased certainty in containing the true proportion.

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Key Concepts

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

Proportion
In statistics, a **proportion** refers to a part or fraction of a whole. It is often expressed as a percentage or a decimal. When calculating the proportion in a given sample, you divide the count of the specific interest by the total number in the sample.
For instance, in the ice-skating class example, the proportion of girls is calculated as the number of girls divided by the total number of students in the class.
Steps to find the proportion:
  • Identify the count of the subgroup of interest (e.g., girls in the class — 64).
  • Determine the total sample size (e.g., total children in the class — 80).
  • Use the formula: \( p^{\prime} = \frac{\text{subgroup count}}{\text{total count}} \).
For the given class, \( p^{\prime} = \frac{64}{80} = 0.8 \), meaning 80% of the class are girls.
Standard Error
The **standard error** (SE) measures how much the sample proportion is expected to vary from the true population proportion. It reflects the precision of the sample proportion as an estimate of the population proportion.
For a proportion, the standard error is calculated with the formula:
  • \( SE = \sqrt{\frac{p^{\prime}(1-p^{\prime})}{n}} \)
Where \(p^{\prime}\) is the sample proportion and \(n\) is the sample size.
In our ice-skating class example, \( p^{\prime} = 0.8 \) and \( n = 80 \), so:
  • \( SE = \sqrt{\frac{0.8(0.2)}{80}} = \sqrt{\frac{0.16}{80}} \approx 0.0447 \)
A smaller standard error indicates a more precise estimate.
Z-Score
A **Z-score** tells us how many standard deviations an element is from the mean in a normal distribution. It is crucial in determining confidence intervals. For instance, it helps in adjusting the width of the confidence interval proportionate to the desired confidence level.
To find an appropriate Z-score for a given confidence level, refer to a Z-table or use a statistical calculator.
For a 98% confidence interval, the corresponding Z-score is approximately 2.33. This means we expect the true proportion to lie within 2.33 standard deviations from the sample proportion and the population mean. A higher Z-score reflects a higher confidence level, thus increasing the interval width.
Error Bound
The **error bound** is the additional amount we add and subtract from the sample proportion to create a confidence interval, expressing the uncertainty or margin of error in our estimate. It depends significantly on the Z-score and standard error.
Calculate the error bound margin using:
  • \( EBM = Z \cdot SE \)
In our example, applying a Z-score of 2.33 for 98% confidence level and an SE of approximately 0.0447:
  • \( EBM = 2.33 \times 0.0447 \approx 0.1042 \)
By increasing the confidence level from a lower percentage to 98%, the Z-score, and consequently the error bound, also increase. This suggests the confidence interval becomes wider, offering a higher chance of containing the true population parameter.

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Most popular questions from this chapter

Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Why would the error bound change if the confidence level were lowered to 95%?

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. In words, define the random variable \(X\) .

Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. Identify the following: a. \(\quad \overline{x}=\) b. \(\sigma=\) c. \(n=\)

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why?

Forbes magazine published data on the best small firms in 2012 . These were firms that had been publicly traded for at least a year, have a stock price of at least \(\$ 5\) per share, and have reported annual revenue between \(\$ 5\) million and \(\$ 1\) billion. The Table 8.13 shows the ages of the corporate \(\mathrm{CEOs}\) for a random sample of these firms. $$\begin{array}{|c|c|c|c|c|}\hline 48 & {58} & {51} & {61} & {56} \\ \hline 59 & {74} & {63} & {53} & {50} \\ \hline 59 & {60} & {60} & {57} & {46} \\\ \hline 55 & {63} & {57} & {47} & {55} \\ \hline 57 & {43} & {61} & {62} & {49} \\ \hline 67 & {67} & {55} & {55} & {49} \\ \hline\end{array}$$ Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution.

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