Question

Several years​ ago, 42​% 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 075 parents who have children in grades​ K-12 if they were satisfied with the quality of education the students receive. Of the 1 comma 075 ​surveyed, 462 indicated that they were satisfied. Construct a 90​% confidence interval to assess whether this represents evidence that​ parents' attitudes toward the quality of education have changed.

Answer 1

The 90​% confidence interval for the proportion of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive is (0.405, 0.455). 0.42 = 42% is part of the confidence interval, so we are 90% sure that there is no evidence that​ parents' attitudes toward the quality of education have changed.

Explanation:

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:

`\pi = 462, n = (462)/(1075) = 0.4298`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4298 - 1.645\sqrt{(0.4298*0.5702)/(1075)} = 0.405`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4298 + 1.645\sqrt{(0.4298*0.5702)/(1075)} = 0.455`

The 90​% confidence interval for the proportion of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive is (0.405, 0.455). 0.42 = 42% is part of the confidence interval, so we are 90% sure that there is no evidence that​ parents' attitudes toward the quality of education have changed.