Question

Please write the full derivation of the answers. Thank you!

Q1) 1 Point for part a, 1 point for part b; total 2 points In a research survey in a certain city, a sample of 400 voters are asked if they favor an additional 4% gasoline sales tax to provide badly needed revenues for street repairs. If more than 220 but fewer than 260 favor the sales tax, we shall conclude that 60% of the voters are for it. a) Find the probability of committing a type I error if 60% of the voters favor the increased tax. b) What is the probability of committing a type II error using this test procedure if actually only 48% of the voters are in favor of the additional gasoline tax?

Answer 1

a) The null and alternative hypothesis

H0: p =0.60

Ha : p $\neq$ 0.60

Accept H0 if , 220 < X < 260

that is , accept H0 if, 220/ 400 < $\hat{p}$ < 260/400

Reject H0 if , $\hat{p}$ < 220/ 400 or  $\hat{p}$ > 260/400

P( type I error ) = P( reject H0 I H0 is true )

= P( $\hat{p}$ < 220/ 400 or $\hat{p}$ > 260/400 I p =0.60 )  

= P( $\hat{p}$   < 0.55 or $\hat{p}$ > 0.65 I p =0.60 )

$=P( z < \frac{0.55-0.60}{\sqrt{0.60*(1-0.6)/400}}or z > \frac{0.65-0.60}{\sqrt{0.60*(1-0.6)/400}})$

= P( z < -2.04 ) +P(z > 2.04 )

= 0.0207 + 0.0207 ( from z table )

= 0.0414

Probability of type I error = 0.0414

b) P( type II error ) = P( accept H0 I H0 false )

= P ( 220/ 400 < $\hat{p}$ < 260/400 I p =0.48 )

= P( 0.55 < $\hat{p}$ < 0.65 I p = 0.48 )

$=P( \frac{0.55-0.48}{\sqrt{0.48*(1-0.48)/400}}< z < \frac{0.65-0.48}{\sqrt{0.48*(1-0.48)/400}})$

= P( 2.80 < z < 6.81 )

= P ( 0 < z < 6.81 ) - P( 0 < z < 2.80)

= 0.5 - 0.4974 ( from z table )

= 0.0026

Probability of type II error = 0.0026