University:
York University
Course:
AP/ADMS 2320 | Business Statistics
Academic year:

2022
Views:

183

Pages:

4
Author:

Niklas

Two Sample Hypothesis Testing: lecture 7 (6) 2 independent populations example: 30 CEO's, 12 women and 18 men showed average salary for women was 7.6 million and average salary for men was 8.7 million. Means are unknown, but σF^2 = 1^2 million and σM^2 = 1.8^2 million. Test at α=5% if average women's salary differs from men's. X1 ~ N(μ1, σ1^2) X2 ~ N(μ2, σ2^2) difference = X1-X2 ~ N(μ1-μ2, σ1^2/n1 + σ2^2/n2) test hypothesis H0: μM-μF=0 z = (x̄M-x̄F) / sqrt(σM^2/nM + σF^2/nF) = (8.7-7.6) / sqrt(1.8^2/18 + 1^2/12) = 2.14 CV = 1.645 2.14 > 1.645, so reject H0 male CEO are paid more The p-value for this example is P(Z>2.14), and can be computed using z-table as 0.0162 is less than significance of 0.05, we can reject H0 For 2 sample HT populations must be independent, and either normally dist or sample size > 30 [Z=\frac{\overline{X}{1}-\overline{X}{2}}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}] where x1, x2 are sample Means Example σ1^2 = σ2^2 = 64-secs^2 x̄A = 121 x̄B = 112 n = 10 α = 0.05 H0: μA-μB = 0 vs H1: μA-μB ≠ 0 [z=\frac{121-112}{\sqrt{\frac{64}{10}+\frac{64}{10}}}=2.5156] as 2.5156 > 1.645, we can reject the null hypothesis P value = P(Z>2.5156) = 1-0.99413=0.00587 as p<0.05, we confirm rejection when the population variances are unknown we use the sample standard deviations (s) and calculate the mean weight The mean weight is called the pooled variance estimate (sp) [S_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}] We must use the t-score. [t=\frac{\overline{x}{1}-\overline{x}{2}}{\sqrt{s_{P}^{2}(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}] the CV can be found on table where [df=n_{1}+n_{2}-2] Example: independent random samples of checking account balances a 2 branches showed Branch No of sample accounts mean (1) SD (3) A 12 1000 150 B 10 920 120 Экспортировать в Таблицы [\overline{x}{A}=1000 \overline{x}{B}=920 s_{A}=150 s_{B}=120 α=10%] [s_{p}^{2}=\frac{(12-1)150^{2}+(10-1)120^{2}}{12+10-2}=18855] [t=\frac{1000-920}{\sqrt{18855(\frac{1}{12}+\frac{1}{10})}}=1.3606] [df=20] 2 sided test so α/2 = 0.05 as t < tα/2, we can't reject null hypothesis When we are using proportions instead of means, we use the pooled probability of success (p̂) p̂ = n1p1 + n2p2 n1 + n2 and the pooled variance becomes s^2 = p̂(1-p̂)(1/n1 + 1/n2)