Question

September 20 is coming up! In Redditville, 15,000 out of the 20,000 residents plan on storming Area 51. Meanwhile, 36,000 out of the 50,000 residents of Facebooktown plan to storm Area 51. At the .02 level of significance, is there a difference in the proportion of Redditville and Facebooktown residents who plan to take part in the Area 51 part?

Answer 1

Let p₁ = The proportion of residents from Redittville = 15000 / 20000 = 0.75

Let p₂ = The proportion of residents from Facebooktown = 36000 / 50000 = 0.72

Let p = Overall proportion = (15000+36000)/(20000+70000) = 51000 / 70000 = 0.7286

1 - p = 0.2714

   = 0.02

(a) The Hypothesis:

H0: p₁ = p₂ : The proportion of residents from Redittville is equal to the proportion of residents from Facebooktown.

Ha: p₁ p₂ : The proportion of residents from Redittville is different from the proportion of residents from Facebooktown.

This is a 2 Tailed Test.

The Test Statistic:

The p Value: The p value (2 Tail) for Z = 8.06, is; p value = 0.0000

The Critical Value:   The critical value (2 tail) at = 0.02, Zcritical = +2.326 and -2.326

The Decision Rule:    If Zobserved is > Zcritical or if Zobserved is < -Zcritical, Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision: Since Z observed (8.06) is > Zcritical (2.326), We Reject H0.

Also since P value (0.0000) is < (0.02), We Reject H0.

The Conclusion: There is sufficient evidence at the 98% significance level to conclude that the proportion of residents from Redittville is different from the proportion of residents from Facebooktown.