Question

Several years​ ago, 43​% 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 085 parents who have children in grades​ K-12 if they were satisfied with the quality of education the students receive. Of the 1 comma 085 ​surveyed, 466 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

Answer with explanation:

Let p be the population proportion of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive.

Set of hypothesis :

$H_0: p=0.43\\\\ H_a:p\neq0.43$

Confidence interval for population proportion is given by :-

$\hat{p}\pm z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ , where

n= sample size

$\hat{p}$= sample proportion

and $z_c$ is the two-tailed z-value for confidence level (c).

As per given ,

Sample size of parents : n= 1085

Number of parents indicated that they were satisfied= 466

Sample proportion :  $\hat{p}=\dfrac{466}{1085}\approx0.429$

Critical value for 90​% confidence interval : $z_c=1.645$  ( by z-value table)

Now, the  90​% confidence interval :

$0.429\pm (1.645)\sqrt{\dfrac{0.429(1-0.429)}{1085}}\\\\=0.429\pm 0.0247\\\\=(0.429-0.0247,\ 0.429+0.0247)\\\\=(0.4043,\ 0.4537)$

Thus , the 90​% confidence interval: (0.4043, 0.4537).

Since 0.43 lies in 90​% confidence interval , it means we do not  have enough evidence to reject the null hypothesis .

i.e. We are have no evidence that parents' attitudes toward the quality of education have changed.