Question

One-sample proprotion Z

A member of the house of representatives wants to determine the proportion of voters in her district who favor a flat tax income tax. A random sample of 200 voters in her district, 89 was in favor of the flat income tax.

a) find a 95% confidence interval for voters who favor a flat income tax

B) does this confidence interval indicate whether a majority of voters oppose the tax? explain.

(please include all the steps/ work: conditions, hypothesis, calculations, conclusion)

Answer 1

Sample proportion = 89 / 200 = 0.445

a)

95% confidence interval for p is

- Z * sqrt( ( 1 - ) / n) < p < + Z * sqrt( ( 1 - ) / n)

0.445 - 1.96 * sqrt( 0.445 * 0.555 / 200) < p < 0.445 + 1.96 * sqrt( 0.445 * 0.555 / 200)

0.376 < p < 0.514

95% CI is ( 0.376 , 0.514 )

b)

H0: p = 0.50

Ha: p > 0.50

Since claimed value 0.50 contained in confidence interval, we do not have sufficient evidence to reject H0.

We conclude at 0.05 level that we fail to support the claim.