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A survey of 1000 students concluded that 274 students chose a professional baseball team. \(A\), as his or her favorite team. In 1991 , the same survey was conducted involving 760 students. It concluded that 240 of them also chose team \(A\) as their favorite. Compute a \(95 \%\) confidence interval for the difference between the proportion of students favoring team .4 between the two surveys. Is there a significant difference?

Short Answer

Expert verified
The 95% confidence interval for the difference in proportions is \(-0.086, 0.002\). There isn't a significant difference since the interval contains zero.

Step by step solution

01

Calculate Proportions

Firstly, we need to calculate the proportions of students favoring team A in both surveys. In the first survey with 1000 students, 274 chose team A, thus the proportion, denoted as \( p_1 \), for the first survey is \( \frac{274}{1000} = 0.274 \). Similarly, in the second survey with 760 students, 240 chose team A. So, the proportion, denoted as \( p_2 \), for the second survey is \( \frac{240}{760} = 0.316 \).
02

Calculate the Difference in Proportions

Next, calculate the difference in proportions \( d = p_1 - p_2 = 0.274 - 0.316 = -0.042 \). Negative value suggests more students chose team A in survey 2.
03

Compute Standard Error

The standard error for the difference in proportions is computed using the formula: \( SE = \sqrt{ \frac{{p_1 \times (1 - p_1)}}{n1} + \frac{{p_2 \times (1 - p_2)}}{n2} } = \sqrt{ \frac{{0.274 \times (1 - 0.274)}}{1000} + \frac{{0.316 \times (1 - 0.316)}}{760} } = 0.022 \).
04

Compute Confidence Interval

Now, the confidence interval can be calculated. For a 95% confidence interval, the z-score is approximately 1.96. The confidence interval is \( d \pm z \times SE \), so it is \( -0.042 \pm 1.96 \times 0.022 \), which results in an interval of \(-0.086, 0.002 \).
05

Determine Significance

Because the confidence interval contains zero, it suggests that there is not a significant difference between the proportions in the two surveys.

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Key Concepts

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

Proportion Difference
Proportion difference is an essential concept when comparing two groups to understand relative preference or choice. In the context of the exercise, the surveys aim to determine how students' preferences for a professional baseball team changed over time. To find the proportion of students who favored team A, we divide the number of students choosing team A by the total number of surveyed students for each survey.
  • **First survey:** 274 students out of 1000 favored team A. Therefore, the proportion (\( p_1 \) ) is 0.274.
  • **Second survey:** 240 students out of 760 preferred team A. Thus, the proportion (\( p_2 \) ) is 0.316.
The difference in these proportions is the focal point: \( d = p_1 - p_2 = -0.042 \). This negative value indicates that team A was more popular in the second survey than the first.
Standard Error
The standard error (SE) plays a significant role in understanding the variability or uncertainty of the proportion difference between two samples. It helps estimate how much the sample statistic might differ from the actual population parameter. In simple terms, the smaller the standard error, the more confident we are in the estimated difference.For this exercise, the standard error of the difference is calculated using both proportions:\[ SE = \sqrt{ \frac{p_1 \times (1 - p_1)}{n_1} + \frac{p_2 \times (1 - p_2)}{n_2} } \]Plugging in the values, we find the SE is approximately 0.022.By computing the standard error, we can then construct a confidence interval to assess if the difference in proportions is statistically significant.
Significance Testing
Significance testing helps determine if the observed difference in data is likely due to chance. It commonly involves calculating a confidence interval and checking if it contains a certain value, like zero, which indicates no effect.In this exercise, a 95% confidence interval means we're 95% sure that the true difference in proportions lies within this interval. Using a z-score of 1.96, the confidence interval is calculated as:\[ d \pm 1.96 \times SE \]Substituting the values:\[ -0.042 \pm 1.96 \times 0.022 \]This results in the interval \([-0.086, 0.002]\).Because this interval includes zero, it signifies that there is no statistically significant difference in the proportions of students favoring team A between the two surveys.
Survey Analysis
Survey analysis entails examining survey data to draw meaningful insights and conclusions. For the exercise at hand, two surveys provide a comparison of students' preferences at different times. Several steps make up effective survey analysis: - **Identify the Objective**: Understand what you aim to discover. Here, it's the difference in preference for team A between two surveys. - **Calculate Proportions**: Determine the portion of students favoring team A in both surveys. - **Examine Differences**: See if there’s a noticeable change in preferences, reflected by the difference in proportions. - **Assess Significance**: Use statistical methods, like confidence intervals, to see if observed changes are statistically meaningful. By analyzing these survey results, educators and analysts can infer if factors between the two survey periods warranted changes in student preferences, thus guiding further studies or decisions.

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Most popular questions from this chapter

A manufacturer turns out a product item that is labeled either "defective" or "not defective." In order to estimate the proportion defective, a random sample of 100 items is taken from production and 10 are found to be defective. Following the implementation of a quality improvement program, the experiment was conducted again. A new sample of 100 is taken and this time only 6 are found defective. (a) Give a \(95 \%\) confidence interval on \(p \mid-p_{2},\) where \(p_{1}\) is the population proportion defective before improvement, and \(p_{2}\) is the proportion defective after improvement. (b) Is there information in the confidence interval found in (a) that would suggest that \(p_{1}>p_{2} ?\) Explain.

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Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 chemical engineers, 40 being women. Compute a \(90 \%\) confidence interval for the difference between the proportion of women in these two fields of engineering. Is there a significant difference between the two proportions?

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