Question

Last year, we discovered that 42% of our respondents reported drinking soda on a weekly basis (n=500). This year, we found that 38% reported drinking soda on a weekly basis (n=250). Is this a significant difference? Why or why not? Assume we are looking for 95% significance

Answer 1

Solution:

Here, we have to use z test for difference in population proportions.

H₀: p₁ = p₂ versus H_a: p₁ ≠ p₂

Test statistic is given as below:

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

N1 = 500

N2 = 250

P1 = 0.42

P2 = 0.38

α = 1 – 0.95 = 0.05

P = (N1P1 + N2P2) / (N1 + N2) = (0.42*500 + 0.38*250)/(500+250) = 0.4067

Z = (0.42 – 0.38)/sqrt(0.4067*(1 - 0.4067)*((1/500)+(1/250)))

Z = 1.0513

P-value = 0.2931

(by using z-table)

P-value > 0.05

So, we do not reject the null hypothesis

There is no significant difference.