Question

We wish to test H0:1p = 0.30 vs. H1: p > 0.30, where p is the proportion of students who want to attend the game. Let X be the number of individuals in a random sample of n = 25 students who want to attend the game. a) As the sample size n increases, the power of the test also increases. Consider n = 150. For the rejection region "Reject Ho if X >= 53", find ... (i) the significance level (ii) power at p = 0.40 and at p = 0.50

Answer 1

Но
P = 0.30   vs $1596189469973_blob.png$ P > 0.30

Given : x = 53, n = 150 , $1596190023712_blob.png$ = x /n = 0.3533 , p = 0.30 and q = 1- 0.30 = 0.7

Rejection region : Reject H0 if x  ≥ 53

i) Significance level ( α ) = P( Reject H0 , when H0 is true )

P ( x  ≥ 53 Given P = 0.30 )

= P( $1596190023712_blob.png$ ≥ 0.3533 , P = 0.30 )

$=P\left (\frac{\hat{p}-p}{\sqrt{\frac{p*q}{n}}} \geq \frac{0.3533-0.30}{\sqrt{\frac{0.3*0.7}{150}}} \right )$

=P( z ≥ 1.42 )

= 1 - P( z ≤ 1.42 )

= 1 - 0.9222 --- ( from z score table )

α = 0.0778

ii ) Power = P( Reject H0, when H0 is false )

#P ( x  ≥ 53 Given P = 0.40 )

= P( $1596190023712_blob.png$ ≥ 0.3533 , P = 0.40 )

$=P\left (\frac{\hat{p}-p}{\sqrt{\frac{p*q}{n}}} \geq \frac{0.3533-0.40}{\sqrt{\frac{0.4*0.6}{150}}} \right )$

=P( z ≥ -1.17 )

= 1 - P( z ≤ -1.17 )

= 1 - 0.1210 --- ( from z score table )

Power at 0.40 = 0.8790

# P ( x  ≥ 53 Given P = 0.50 )

= P( $1596190023712_blob.png$ ≥ 0.3533 , P = 0.50 )

$=P\left (\frac{\hat{p}-p}{\sqrt{\frac{p*q}{n}}} \geq \frac{0.3533-0.50}{\sqrt{\frac{0.5*0.5}{150}}} \right )$

= P( z ≥ -3.59 )

= 1 - P( z ≤ -3.59 )

= 1 - 0.0002 --- ( from z score table )

Power at 0.50 = 0.9998