Question

Several years​ ago, 50​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive. A recent poll asked 1 comma 095 parents who have children in grades​ K-12 if they were satisfied with the quality of education the students receive. Of the 1 comma 095 ​surveyed, 478 indicated that they were satisfied. Construct a 95​% confidence interval to assess whether this represents evidence that​ parents' attitudes toward the quality of education have changed.

Answer 1

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that​ parents' attitudes toward the quality of education have changed.

Explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 1095, \pi = (478)/(1095) = 0.4365`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4365 - 1.96\sqrt{(0.4365*0.5635)/(1095)} = 0.4118`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4365 + 1.96\sqrt{(0.4365*0.5635)/(1095)} = 0.4612`

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that​ parents' attitudes toward the quality of education have changed.