Statistical Analysis

Z Test Comparison of Two Outcome Proportions
by Thomas Cobb, MFT and Phyllis Grannis, MFT

The Z Test is used to test the difference between the two outcome proportions.

H:₀ p₁ =< p₂	The Null Hypothesis is that the on-line course is less effective or as effective as the mass media course.
H:₁ p₁ > p₂	The Alternative Hypothesis is that the on-line headache course is more effective than the mass media course

Comparative Data for each condition, X1 is the On-line Internet Course and X2 is the Television/Books/Cassettes/Radio Course.

X₁ = 18 X₂ = 74

n₁ = 21 n₂ = 164

p^₁ = .85 p^₂ = .45

Calculation of p bar, a weighted estimate of p.

p(bar) = X₁ + X₂ / n₁ + n₂

p(bar) = 18 + 74 / 21 + 164 = 92/185 = .497

q(bar) = .503

Calculate the C.V. for the 0.005 level of Significance for a One Tailed Test using the z score.

At the .005 level of Significance C.V. = + 2.576

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