Answer to Question #285397 in Statistics and Probability for anna

Question

In a public opinion survey, 60 out of a sample of 70 high-income voters and 90 out of a sample of 100 low-income voters supported the introduction of a new national security tax.

a/ Estimate, with 95% confidence level, the true proportion of low-income people who will vote for the introduction of the tax.

b/ Can we conclude at the 5% level of significance that the proportion of high-income voters favoring the new security tax is lower than that of low-income voters?

Answer 1

a) For the low-income voters:

$95\%CI=(0.9-1.96\sqrt{\frac{0.9*0.1}{100}},0.9+1.96\sqrt{\frac{0.9*0.1}{100}})=(0.8412,$ 0.9588).

b)For the high-income voters:

$\hat p= \frac{60}{70}=0.8571.$

$95\%CI=(0.8571-1.96\sqrt{\frac{0.8571(1-0.8571)}{70}},0.8571+1.96\sqrt{\frac{0.8571(1-0.8571)}{70}})=$

$=(0.6228,0.9381).$

Since the 95%CI's for the low-income and high-income overlapped, there is no sufficient evidence that the proportion of high-income voters favoring the new security tax is lower than that of low-income voters.