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

Answer:

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.

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 1095, \pi = \frac{478}{1095} = 0.4365$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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.