# The supporters of a party claim that their party has 60% following in Delhi. To test those claim, a city paper undertakes a test sample of 400 persons on the internet, 200 people supported these claim test at 1% level of significance. Whether the supporters claim is correct or not?

2
by sumitsatpaty

2014-03-05T12:40:22+05:30
Let P be the proportion following the party
H0: P = 0.60
H1: P ≠ 0.60

Estimated p = 200 / 400 = 0.5
Variance of proportion = p*(1-p)/n
= 0.6(0.4)/400 =0.0006
S.D. of p^ is sqrt[0.0006] = 0.0245
z = ( 0.5 - 0.6 ) / 0.0245 = -4.0825

P-value = P( |z| > 4.0825) = 0.0000
Since the p-value < 0.01 (1% level), reject H0.

No evidence to support the claim.
• Brainly User
2014-03-05T13:31:18+05:30
Let P be the proportion following the party
H0: P = 0.60
H1: P ≠ 0.60

Estimated p = 200 / 400 = 0.5
Variance of proportion = p*(1-p)/n
= 0.6(0.4)/400 =0.0006
S.D. of p^ is sqrt[0.0006] = 0.0245
z = ( 0.5 - 0.6 ) / 0.0245 = -4.0825

P-value = P( |z| > 4.0825) = 0.0000
Since the p-value < 0.01 (1% level), reject H0.

No evidence to support the claim.