Question

a 2011 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

a) find a 90% confidence interval for the proportion of Americans who would agree with this
b)interpret your interval in this context
c)explain what 90% confidence means
d) do these data refute a pundits claim that 2/3 of Americans believe this statement? Explain

Answer 1

Answer on Question #56150 – Math – Statistics and Probability

A 2011 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

a) find a 90% confidence interval for the proportion of Americans who would agree with this

b) interpret your interval in this context

c) explain what 90% confidence means

d) do these data refute a pundits claim that 2/3 of Americans believe this statement? Explain

Solution

a) A 90% confidence interval for the proportion of Americans who would agree with this is

$$CI = \left(p - z^* \sqrt{\frac{p(1 - p)}{n}}; p + z^* \sqrt{\frac{p(1 - p)}{n}}\right),$$

where $z^* = 1.645$ for 90% confidence level.

$$CI = \left(0.76 - 1.645 \sqrt{\frac{0.76(1 - 0.76)}{1002}}; 0.76 + 1.645 \sqrt{\frac{0.76(1 - 0.76)}{1002}}\right) = (0.738; 0.782),$$

b) We can be 90% confident that true proportion of Americans who would agree with this lies between 0.738 and 0.782.

c) 90% confidence means that we have a 0.90 probability of containing the population mean proportion in this interval.

d) Yes. $\frac{2}{3} \approx 0.667$ is outside 90% confidence interval.

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