Q:

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.

Accepted Solution

A:
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 : [tex]H_0: p=0.43\\\\ H_a:p\neq0.43[/tex]Confidence interval for population proportion is given by :-[tex]\hat{p}\pm z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where n= sample size [tex]\hat{p}[/tex]= sample proportionand [tex]z_c[/tex] is the two-tailed z-value for confidence level (c).As per given , Sample size of parents : n= 1085Number of parents indicated that they were satisfied= 466Sample proportion :  [tex]\hat{p}=\dfrac{466}{1085}\approx0.429[/tex]Critical value for 90​% confidence interval : [tex]z_c=1.645[/tex]  ( by z-value table)Now, the  90​% confidence interval :[tex]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)[/tex] 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.