A2011 gallup poll found that 76% of americans believe that high achieving high school students should be recruited to become teachers. this poll was based on a random sample of 1002 americans. find a 90% confidence interval for the proportion of americans who would agree with this. interpret your interval in this context. explain what “90% confidence” means. do these data refute a pundit’s claim that 2 / 3 of americans believe this statement? explain.

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=

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.

right

step-by-step explanation:

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !