/*! 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 180 Gender and Award Preference In E... [FREE SOLUTION] | 91Ó°ÊÓ

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Gender and Award Preference In Example 2.6 on page 51 we consider data from a sample of statistics students that is stored in StudentSurvey. One of the survey questions asked which award students would most like to win from among an Academy Award, Nobel Prize, and Olympic gold medal. Among the 193 male students who responded, 109 chose the Olympic gold medal, while 73 of the 169 females also picked Olympic gold. Use this information to find a \(90 \%\) confidence interval for the difference between the proportions of male and female statistics students who choose Olympic gold.

Short Answer

Expert verified
The 90% confidence interval for the difference in proportions of male and female students who prefer winning an Olympic gold medal is between 0.0407 and 0.2251.

Step by step solution

01

Identify the Sample Proportions

First, calculate the sample proportions. The sample proportion for males, \(p_1\), is the number who chose Olympic gold divided by the total number of males: \(p_1 = 109/193 = 0.5648\). Similarly, find the sample proportion for females, \(p_2\), which is \(p_2 = 73/169 = 0.4319\).
02

Compute the Standard Error

Next, calculate the standard error (SE). Use the formula \(SE = \sqrt{ p_1(1 - p_1) / n_1 + p_2(1 - p_2) / n_2 }\) for the standard error of the difference between two proportions. Substitute the calculated values into the formula: \(SE = \sqrt{0.5648(1 - 0.5648) / 193 + 0.4319(1 - 0.4319) / 169} = 0.0561\).
03

Determine the Z-Score

Since a 90% confidence interval is desired, a z-score corresponding to the significance level should be determined. In a standard normal distribution (z-distribution), a z-score for a two-tailed test that provides 90% confidence is approximately 1.645.
04

Calculate Margin of Error

The margin of error (MOE) can now be calculated by multiplying the standard error by the z-score: \(MOE = SE * z = 0.0561 * 1.645 = 0.0922\). It means the true difference in proportions lies within this range around the calculated difference.
05

Construct the Confidence Interval

The point estimate for the difference of the two proportions is \(p_1 - p_2 = 0.5648 - 0.4319 = 0.1329\). Now, construct the 90% confidence interval for the difference between the two population proportions: \(CI = (p_1—p_2) \pm MOE\), which gives \(CI = 0.1329 \pm 0.0922\). Therefore, the required 90% confidence interval is from \(0.0407\) to \(0.2251\).

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Key Concepts

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

Sample Proportions
Sample proportions serve as estimates of actual population proportions. In the context of the exercise, we calculate the sample proportion for males and females who preferred the Olympic gold medal. For males, the sample proportion, denoted as \( p_1 \), is computed as the number of male students selecting this option divided by the total number of male respondents. This yields a proportion of approximately 0.5648. Similarly, for females, the sample proportion \( p_2 \) is approximately 0.4319.

These proportions are vital because they form the basis for subsequent statistical calculations. They act as a snapshot of the sampled data that helps draw inferences about the broader population.
  • Males: \( p_1 = \frac{109}{193} \approx 0.5648 \)
  • Females: \( p_2 = \frac{73}{169} \approx 0.4319 \)
Understanding sample proportions helps in analyzing how representative the sample is of the total population and in setting up further calculations.
Standard Error
The standard error is a measure that reflects the variability of the sample statistic, providing insight into how much this statistic might fluctuate from sample to sample. In this example, we calculate the standard error for the difference between two sample proportions, which allows us to understand the precision of our estimate.

The formula used is:\[SE = \sqrt{ \frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2} }\]where \( n_1 \) and \( n_2 \) are the sizes of the male and female samples, respectively. After inserting the specific values, the calculated standard error \( SE \) is approximately 0.0561.
  • This calculation is fundamental because it underpins the construction of confidence intervals.
  • It indicates that the larger the sample size, the smaller the standard error, resulting in a more accurate estimation.
Z-Score
The z-score in statistics is a measure of how many standard deviations an element is from the mean. For confidence intervals, it helps in determining how much confidence we might have in our sample's estimate reflecting the true population.

For a 90% confidence interval, which is considered relatively precise, the corresponding z-score is approximately 1.645.
  • This z-score is employed because it represents a balance between a higher confidence that requires a wider interval and the need for practical specificity.
  • The z-score is integral to calculating the margin of error, as it directly impacts the size of the confidence interval.
Margin of Error
Margin of Error (MOE) provides a range above and below the reported statistic that indicates where the true value is likely to lie, given a specific level of confidence. It is calculated by multiplying the standard error by the z-score:

\[MOE = SE \times z\]In this case, \( 0.0561 \times 1.645 \) gives us an MOE of approximately 0.0922.
  • This MOE signifies that the true difference in preferences between male and female students is likely to lie within this range of the calculated difference.
  • A lower MOE is preferred as it shows a narrow interval, indicating higher precision of our sample results.
Statistics Education
Statistics education illuminates the power of understanding how to make inferences about populations from sample data. Through exercises like this, learners appreciate the real-world application of statistical concepts like confidence intervals.

Such exercises teach students critical skills:
  • Understanding how sample data can provide insights into broader population behaviors.
  • The process of constructing confidence intervals and interpreting their meaning.
  • The importance of having accurate estimates and the reliability of these estimates when making data-driven decisions.
Through carefully structured statistical exercises, students enhance their analytical skills, becoming adept at using data for informed decision-making.

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Most popular questions from this chapter

6.8 Percent over 600 on Math SAT In the class of \(2010,25 \%\) of students taking the Mathematics portion of the SAT (Scholastic Aptitude Test) \(^{3}\) scored over a 600 . If we take random samples of 100 members of the class of 2010 and compute the proportion who got over a 600 on the Math SAT for each sample, what will be the mean and standard deviation of the distribution of sample proportions?

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the area in a t-distribution less than -1.4 if the samples have sizes \(n_{1}=30\) and \(n_{2}=40\).

A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) $$ \begin{aligned} &\text { Two-Sample T-Test and Cl }\\\ &\begin{array}{lrrrr} \text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Late } & 32 & 22.56 & 5.13 & 0.91 \\ \text { Early } & 30 & 19.73 & 6.61 & 1.2 \end{array} \end{aligned} $$ Difference \(=\mathrm{mu}\) (Late) \(-\mathrm{mu}\) (Early) Estimate for difference: 2.83 \(95 \%\) Cl for difference: (-0.20,5.86) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=1.87\) P-Value \(=0.066 \quad \mathrm{DF}=54\)

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$

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