Answer:
Step-by-step explanation:
Let p be the proportion of Americans who believe that high achieving high school students should be recruited to become teachers
Given that p = 0.76 and q = 1-p =0.24
Std error for proportion= [tex]\sqrt{\frac{pq}{n} } \\=0.0135[/tex]
For 90% confidence interval we have
0.76±Margin of error
= 0.76±1.645(0.0135)
=0.76±0.0222
thus confidence interval
= (0.7378, 0.7822)
This means we are 90% confidence for randomly selected samples of large size, proportion will fall within this interval
Since proportion lower bound is 73.78% we can say that pundit claim of 2/3 is correct.