Question

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

(0.016, 0.171)

Step-by-step solution

To estimate the difference between the proportion of all elementary school and all high school teachers who are very satisfied  

A confidence interval for \({p_1} - {p_2}\)with confidence level of approximately \(100(1 - \alpha )\% \) is

\({\hat p_1} - {\hat p_2} \pm {z_{\alpha /2}} \cdot \sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{m} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{n}} \)

The confidence interval can be used when\(m \cdot {\hat p_1},m \cdot {\hat q_1},n \cdot {\hat p_1}\), and \(n \cdot {\hat q_1}\) are at least 10

There are \(m = 395\) elementary school teachers and \(n = 266\) high school teachers. There are 224 elementary school satisfied teachers with their job, thus

\({\hat p_1} = \frac{{224}}{{395}}\)

is the proportion, and 126 high school teachers satisfied with their job, thus

\({\hat p_2} = \frac{{126}}{{266}}\)

Therefore, the 95 % confidence interval is

\(\begin{array}{l}{{\hat p}_1} - {{\hat p}_2} - {z_{\alpha /2}} \cdot \sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{m} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{n}} ,{{\hat p}_1} - {{\hat p}_2} + {z_{\alpha /2}} \cdot \sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{m} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{n}} \\ = \frac{{224}}{{395}} - \frac{{126}}{{266}} - 1.96 \cdot \sqrt {\frac{{224}}{{395}} \cdot \frac{{171}}{{395}}} \;\;\;395 + \frac{{126}}{{266}} \cdot \frac{{140}}{{266}}\;\;\;266,\\ = \frac{{224}}{{395}} - \frac{{126}}{{266}} + 1.96 \cdot \sqrt {\frac{{224}}{{395}} \cdot \frac{{171}}{{395}}} \;\;\;395 + \frac{{126}}{{266}} \cdot \frac{{140}}{{266}}\;\;\;266\\ = (0.016,0.171).\end{array}\)

Final proof

(0.016, 0.171)