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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?

Answer :

sonaa
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.
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.

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