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Do teachers find their work rewarding and satisfying? The article "Work- Related Attitudes" (Psychological Reports, 1991: 443-450) reports the results of a survey of 395 elementary school teachers and 266 high school teachers. Of the elementary school teachers, 224 said they were very satisfied with their jobs, whereas 126 of the high school teachers were very satisfied with their work. Estimate the difference between the proportion of all elementary school teachers who are very satisfied and all high school teachers who are very satisfied by calculating and interpreting a CI.

Short Answer

Expert verified
The 95% CI for the difference in satisfaction is (0.0074, 0.1794).

Step by step solution

01

Identify Given Information

The problem provides the following data:- Number of elementary school teachers surveyed, \( n_1 = 395 \).- Number of very satisfied elementary school teachers, \( x_1 = 224 \).- Number of high school teachers surveyed, \( n_2 = 266 \).- Number of very satisfied high school teachers, \( x_2 = 126 \).
02

Calculate Sample Proportions

The sample proportion of very satisfied elementary school teachers is calculated as:\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{224}{395} \approx 0.5671 \]The sample proportion of very satisfied high school teachers is:\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{126}{266} \approx 0.4737 \]
03

Find the Point Estimate of the Difference

The point estimate of the difference between the two proportions is:\[ \hat{p}_1 - \hat{p}_2 = 0.5671 - 0.4737 = 0.0934 \]
04

Calculate the Standard Error

The standard error (SE) for the difference between two proportions is given by the formula:\[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]Substituting the known values:\[ SE = \sqrt{\frac{0.5671 (1-0.5671)}{395} + \frac{0.4737 (1-0.4737)}{266}} \approx 0.0439 \]
05

Determine the Critical Value

For a 95% confidence interval, the critical value (z) from the standard normal distribution is approximately 1.96.
06

Calculate the Confidence Interval

The confidence interval is calculated as:\[ (\hat{p}_1 - \hat{p}_2) \pm z \times SE \]Substituting the known values:\[ 0.0934 \pm 1.96 \times 0.0439 \]The interval is:\[ 0.0934 \pm 0.086 \]Hence, the 95% confidence interval is approximately (0.0074, 0.1794).
07

Interpret the Confidence Interval

The 95% confidence interval for the difference between the proportions of elementary and high school teachers who are very satisfied is (0.0074, 0.1794), suggesting that we are 95% confident that the proportion of very satisfied elementary teachers is between 0.74% to 17.94% higher than that of high school teachers.

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Key Concepts

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

Sample Proportion
In any survey, one of the key calculations involves figuring out the sample proportion. This is a straightforward method to determine the relative presence of a specific characteristic within a sample group. To calculate a sample proportion, you divide the number of successes (i.e., responses that meet your criteria) by the total number of observations in the sample.
In our exercise, for instance, we learned that among 395 elementary teachers, 224 were very satisfied; thus, the sample proportion for this group is computed as \( \hat{p}_1 = \frac{224}{395} \approx 0.5671 \). Similarly, for 266 high school teachers with 126 very satisfied, the sample proportion is \( \hat{p}_2 = \frac{126}{266} \approx 0.4737 \).
These proportions tell us how big a slice of each group is 'very satisfied' compared to the whole group. They serve as essential building blocks for further statistical analysis, such as comparing group differences.
Standard Error
The standard error (SE) is an important concept that quantifies the accuracy of a sample statistic, like a sample proportion. In the context of our teacher satisfaction survey, the standard error helps us measure how much variation we might expect between the surveyed sample and the larger population.
Calculating the standard error for the difference between two proportions involves a specific formula: \[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \] Where \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions, and \( n_1 \) and \( n_2 \) are the respective sample sizes.
In our case, substituting the sample proportions and sizes gives us a SE of approximately 0.0439. This calculation tells us about the variability expected in the difference between the two group's proportions if we were to repeat this survey numerous times.
Elementary and High School Teachers Survey
This insightful survey aimed to gauge job satisfaction levels among elementary and high school teachers. The survey results provided a clear snapshot: out of 395 elementary school teachers surveyed, 224 expressed they were very satisfied with their jobs. Conversely, out of 266 surveyed high school teachers, only 126 conveyed the same satisfaction level.
These surveys are crucial tools in educational research, helping stakeholders understand and compare how different teaching environments can impact job satisfaction. It paints a picture of where one teaching group may be thriving more than the other, guiding enhancements in work conditions and policies.
Difference in Proportions
The difference in proportions is a vital metric to recognize disparities between two groups in research data. When applied to our teacher satisfaction survey, it quantifies the gap between the proportion of 'very satisfied' teachers in elementary versus high schools.
This is calculated by simply taking the difference between the two sample proportions: \[ \hat{p}_1 - \hat{p}_2 = 0.5671 - 0.4737 = 0.0934 \] This indicates that the proportion of very satisfied elementary teachers exceeds that of their high school counterparts by about 9.34%.
Further, when interpreted with a confidence interval, it suggests that the true difference might lie within a specific range, providing a more nuanced interpretation of satisfaction levels across educational settings.

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