Question 1183963



X2 = {{{sum((O[k] - E[k])^2/E[k], k=1,3) = (5 - 6)^2/6 + (11 - 9)^2/9 + (14 - 15)^2/15  = 1/6 + 4/9 + 1/15 = 61/90 = 0.678 }}} to 3 d.p.


For d.f. = 3 - 1 = 2,  p-value = P(X2 > 0.678) = 0.71 > 0.10, hence do NOT reject the null hypothesis that the three categories have proportion 0.20, 0.30, and 0.50 respectively.

***Chi-square probability was taken from https://stattrek.com/online-calculator/chi-square.aspx.