Question

Suppose a polling agency reported that 44.4​% of registered voters were in favor of raising income taxes to pay down the national debt. The agency states that results are based on telephone interviews with a random sample of 1049 registered voters. Suppose the agency states the margin of error for 95​% confidence is 3.0​%. Determine and interpret the confidence interval for the proportion of registered voters who are in favor of raising income taxes to pay down the national debt.

Answer 1

Answer:

95% of confidence interval for the proportion of registered voters who are in favor of raising income taxes to pay down the national debt.

(0.414 ,0.474)

Step-by-step explanation:

Step(i):-

Given sample proportion

                                    p⁻ = 44.4 % = 0.444

Random sample size 'n' = 1049

Given margin of error for 95% confidence level = 3 % = 0.03

Step(ii):-

95% of confidence interval for the proportion is determined by

$(p^{-} - Z_{\alpha }\sqrt{\frac{p^{-} (1-p^{-} }{n} } , p^{-} + Z_{\alpha }\sqrt{\frac{p^{-} (1-p^{-} }{n} })$

we know that

Margin of error for 95% confidence level is determined by

$M.E = Z_{\alpha }\sqrt{\frac{p^{-} (1-p^{-}) }{n} }$

Step(iii):-

Now

95% of confidence interval for the proportion is determined by

$(p^{-} - M.E, p^{-} + M.E)$

Given Margin of error

                              M.E = 0.03

Now 95% of confidence interval for the proportion

$(0.444 - 0.03, 0.444+ 0.03)$

(0.414 ,0.474)

Conclusion:-

95% of confidence interval for the proportion of registered voters who are in favor of raising income taxes to pay down the national debt.

(0.414 ,0.474)