# Do teachers find their work rewarding and satisfying? The article presents the results of a survey of 399 primary school teachers and 264 senior teach

Do teachers find their work rewarding and satisfying? The article presents the results of a survey of 399 primary school teachers and 264 senior teachers. Of the elementary school teachers, 223 said they were very satisfied with their jobs, whereas 127 of the high school teachers were very satisfied with their work. Estimate the difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied by calculating a $95\mathrm{%}CI.\left(\text{Use}{P}_{\text{elementary}}-{P}_{\text{high}}$ school. Round your answers to four decimal places.)
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Step 1
Let ${P}_{\text{elementary}}$ denotes the population proportion of all elementary school teachers who are satisfied, and ${P}_{\text{high school}}$ denotes the population proportion of all high school teachers who are satisfied. The test assertion is that there is a difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied. The hypothesis is,
Null hypothesis:
${H}_{0}:{P}_{\text{elementary}}-{P}_{\text{high school}}=0$
Alternative hypothesis:
${H}_{1}:{P}_{\text{elementary}}-{P}_{\text{high school}}\ne 0$
The sample proportion of all elementary school teatcher who are satisfied is,
${\stackrel{^}{P}}_{\text{elementary}}=\frac{{x}_{1}}{{n}_{1}}$
$=\frac{223}{399}$
$=0.5589$
The sample proportion of all elementary school teachers who are satisfied is 0.5589.
The sample proportion of all high school teachers who are satisfied is,
${\stackrel{^}{P}}_{\text{high school}}=\frac{{x}_{1}}{{n}_{1}}$
$=\frac{127}{264}$
$=0.4811$
The sample proportion of all high school teachers who are satisfied is 0.4811.
Step 2
Computation of critical value:
$\left(1-\alpha \right)=0.95$
$\alpha =0.05$
$\frac{\alpha }{2}=0.025$
$1-\frac{\alpha }{2}=0.0975$
The critical value of z-distribution can be obtained using the excel formula $“=NORM.S.INV\left(0.975\right)\text{"}.$ The critical value is 1.96.
The $95\mathrm{%}$ confidence interval for the difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied is, $CI={\stackrel{^}{P}}_{\text{elementary}}-{\stackrel{^}{P}}_{\text{high school}}±{z}_{\frac{\alpha }{2}}\sqrt{\frac{{\stackrel{^}{P}}_{\text{elementary}}\left(1-{\stackrel{^}{P}}_{\text{elementary}}\right)}{{n}_{\text{elementary}}}+\frac{{\stackrel{^}{P}}_{\text{high school}}\left(1-{\stackrel{^}{P}}_{\text{high school}}\right)}{n\left({n}_{\text{high school}}\right)}}$
$=\left(0.5589-0.4811\right)±1.96\sqrt{\frac{0.5589\left(1-0.5589\right)}{399}+\frac{0.4811\left(1-0.4811\right)}{264}}$
$=0.0778±0.0775$
$=\left(0.0778-0.0775,0.0778+0.0775\right)$
$=\left(0.0003,0.1553\right)$
Hence, the $95\mathrm{%}$ confidence interval for the difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied is (0.0003, 0.1553).
Decision rule:
If the confidence interval does not contains value zero, then reject the null hypothesis.
Conclusion:
The confidence interval does not contain a zero value.
Based on the decision rule, reject the null hypothesis.
Thus, there is difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied.