Question

Statistical Methods Question 14: A researcher collected data from random samples of 100 cats and 100 dogs. The researcher found that 50 of the cats and 60 of the digs in tbe sample had pet licenses. Test at the 5% significance level that the proportion of cats that have licenses is less than the proportion of dogs that have licenses. What is the p-value for this set, what is your conclusion?

Answer 1

n₁ = n₂ = 100

p₁ = 50/100 = 0.5

p₂ = 60/100 = 0.6

H_o: P₁ = P₂
H_a: P₁ < P₂

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂) = ((0.5*100)+(0.6*100))/(100+100) = 0.55

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }

= sqrt((0.55*0.45)*(1/100+1/100))

= 0.07

Test statistic: z = (p₁ - p₂) / SE = (0.5-0.6)/0.07 = -1.429

p-value (Using Excel function NORM.S.DIST(z, cumulative)) = NORM.S.DIST(-1.429,TRUE) = -0.077

Since p-value is less than 0.05, we reject the ull hypothesis and conclude that P₁ < P₂.

Proportion of cats that have licenses is less than the proportion of dogs that have licenses.