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Chapter 11: Problem 24

An educator wants to determine the difference between the proportion of males and females who have completed 4 or more years of college. What sample size should be obtained if she wishes the estimate to be within two percentage points with \(90 \%\) confidence, assuming that (a) she uses the 1999 estimates of \(27.5 \%\) male and \(23.1 \%\) female from the U.S. Census Bureau? (b) she does not use any prior estimates?

Short Answer

Expert verified
For part (a), the required sample size is approximately 1242. For part (b), the required sample size is approximately 1693.

Step by step solution

01

- Identify the Given Information for Part (a)

We are given the following: - Confidence Level: 90%- Margin of Error (E): 0.02 (2 percentage points)- Proportion of Males (\(\bar{p}_1\)): 0.275- Proportion of Females (\(\bar{p}_2\)): 0.231- The combined proportion (\(\bar{p}\)) is not provided directly, but can be estimated as \(\bar{p} = \frac{\bar{p}_1 + \bar{p}_2}{2}\ = \frac{0.275 + 0.231}{2} = 0.253\)
02

- Find the Z-value for 90% Confidence Level

Use the Z-table to find the critical value for a 90% confidence level. The critical value (Z) for 90% confidence is approximately 1.645.
03

- Calculate the Combined Proportion

Given \(\bar{p} = 0.253\), note that \(\bar{q} = 1 - \bar{p} = 0.747\).This combined proportion will be used to find the required sample size.
04

- Apply the Sample Size Formula for Part (a)

Use the formula: \[ n = \left( \frac{Z^2 \times \bar{p} \times \bar{q}}{E^2} \right) \times 2 \]where \( n \) is the sample size for each group. Substituting the values: \[ n = \left( \frac{(1.645)^2 \times 0.253 \times 0.747}{(0.02)^2} \right) \times 2 \] = 1242 Calculating this value yields n 鈮� 1242.
05

- Identify the Given Information for Part (b)

For part (b):- Confidence Level: 90%- Margin of Error (E): 0.02 (2 percentage points)- No prior estimates are given, so assume \(\bar{p}\) = 0.5 and \(\bar{q}\) = 0.5.
06

- Apply the Sample Size Formula for Part (b)

Use the formula: \[ n = \left( \frac{Z^2 \times \bar{p} \times \bar{q}}{E^2} \right) \times 2 \]where \(\bar{p} = 0.5\), \(\bar{q} = 0.5\), and \(Z = 1.645\). Substituting the values: \[ n = \left( \frac{(1.645)^2 \times 0.5 \times 0.5}{(0.02)^2} \right) \times 2 \] 鈮� 1693.Calculating this value yields n 鈮� 1693.

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Key Concepts

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

confidence level
When we talk about the confidence level in statistics, we are referring to the degree of certainty that our estimate falls within a certain range. It's usually expressed as a percentage. For example, a 90% confidence level means we are 90% sure that our sample accurately represents the population. Confidence levels commonly used are 90%, 95%, and 99%.

The higher the confidence level, the more certain we can be about the accuracy of our results, but the trade-off is usually a larger required sample size. In our exercise, a 90% confidence level means we need to be fairly confident, but not excessively so, which balances out our need for precision and practicality in sample size.
margin of error
The margin of error is a measure of the range within which we expect our true population parameter to lie, based on our sample statistic. It is usually denoted by E and is expressed as a percentage. For instance, a 2% margin of error means our estimate is within 卤2% of the actual value.

A smaller margin of error requires a larger sample size. In our example, the educator wants a margin of error of 2 percentage points. This specific need affects the sample size needed to ensure that our confidence interval is narrow enough to be useful.
proportion estimation
Proportion estimation involves calculating the ratio of individuals in a sample that possess a particular characteristic, and then using that sample proportion to estimate the population proportion. In our problem, we are estimating the proportion of males and females who have completed 4 or more years of college.

For part (a) of our problem, previous estimates of 27.5% for males and 23.1% for females were given. By averaging these, we get an estimate (0.253) to use in our calculations. For part (b), we don't have prior estimates, so we assume proportions of 0.5 for a more conservative approach.
z-value
The Z-value, also known as the Z-score, is a statistical measure that describes a value's relation to the mean of a group of values. In the context of our problem, the Z-value is used to translate a confidence level into a critical value, which then feeds into our sample size formula.

For a 90% confidence level, the Z-value is approximately 1.645. This means that 90% of the population falls within 1.645 standard deviations from the mean. This value plays a crucial role in determining the number of samples needed. For example, our formula for sample size requires squaring the Z-value as one of its components.

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Most popular questions from this chapter

Kids and Leisure Young children require a lot of time. This time commitment cuts into a parent's leisure time. A sociologist wanted to estimate the difference in the amount of daily leisure time (in hours) of adults who do not have children under the age of 18 years and the amount of daily leisure time (in hours) of adults who have children under the age of 18 years. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.62 hours with a standard deviation of 2.43 hours. A random sample of 40 adults with children under the age of 18 years results in a mean daily leisure time of 4.10 hours with a standard deviation of 1.82 hours. Construct and interpret a \(90 \%\) confidence interval for the mean difference in leisure time between adults with no children and adults with children. (Source: American Time Use Survey)

The drug Prevnar is a vaccine meant to prevent certain types of bacterial meningitis. It is typically administered to infants starting around 2 months of age. In randomized, double-blind clinical trials of Prevnar, infants were randomly divided into two groups. Subjects in Group 1 received Prevnar, while subjects in Group 2 received a control vaccine. After the second dose, 137 of 452 subjects in the experimental group (Group 1) experienced drowsiness as a side effect. After the second dose, 31 of 99 subjects in the control group (Group 2 ) experienced drowsiness as a side effect. (a) Does the evidence suggest that a different proportion of subjects in Group 1 experienced drowsiness as a side effect than subjects in Group 2 at the \(\alpha=0.05\) level of significance? (b) Construct a \(99 \%\) confidence interval for the difference between the two population proportions, \(p_{1}-p_{2}\)

Visual versus Textual Learners Researchers wanted to know whether there was a difference in comprehension among students learning a computer program based on the style of the text. They randomly divided 36 students into two groups of 18 each. The researchers verified that the 36 students were similar in terms of educational level, age, and so on. Group 1 individuals learned the software using a visual manual (multimodal instruction), while Group 2 individuals learned the software using a textual manual (unimodal instruction). The following data represent scores the students received on an exam given to them after they studied from the manuals. (a) What type of experimental design is this? (b) What are the treatments? (c) A normal probability plot and boxplot indicate it is reasonable to use Welch's \(t\) -test. Is there a difference in test scores at the \(\alpha=0.05\) level of significance? (d) Construct a \(95 \%\) confidence interval about \(\mu_{\text {visual }}-\mu_{\text {textual }}\) and interpret the results.

Suppose a researcher believes the mean from population 1 is less than the mean from population 2 in matchedpairs data. How would you define \(\mu_{d} ?\) How would you determine \(d_{i} ?\)

Explain why we determine a pooled estimate of the population proportion when testing hypotheses regarding the difference of two proportions, but do not pool when constructing confidence intervals about the difference of two proportions.
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