Null hypothesis "H_0:\\;p_1=p_2"
Alternative hypothesis "H_a:\\;p_1< p_2"
Pooled sample proportion: "E=mc^2p=\\frac{p_1n_1+p_2n_2}{n_1+n_1}=\\frac{0.01*200+0.02*300}{200+300}=0.016"
Test statistic: "z=\\frac{p_1-p_2}{\\sqrt{p(1-p)(\\frac{1}{n_1}+\\frac{1}{n_2})}}=-0.87."
P-value: "p=0.191."
Since the P-value is greater than 0.05, reject the null hypothesis.
Comments
Leave a comment