Question

Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 397 elementary school teachers and 268 high school teachers. Of the elementary school teachers, 226 said they were very satisfied with their jobs, whereas 129 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% CI. (Use pelementary − phigh school. Round your answers to four decimal places.)

Answer 1

Answer:

The 95% confidence interval would be given (0.0109;0.1651).  

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_A$ represent the real population proportion for elementary school  

$\hat p_A =\frac{226}{397}=0.569$ represent the estimated proportion for elementary school

$n_A=397$ is the sample size required for Brand A

$p_B$ represent the real population proportion for high school teachers  

$\hat p_B =\frac{129}{268}=0.481$ represent the estimated proportion for high school teachers

$n_B=268$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.569-0.481) - 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.0109$  

$(0.569-0.481) + 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.1651$  

And the 95% confidence interval would be given (0.0109;0.1651).  

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$