题目:
We’re trying to test whether a new,low-fat diet actually helps obese people lose weight.100 randomly assigned obese people are assigned to group 1 and put on the low fat diet.Another 100 randomly assigned obese people are assigned to group 2 and put on a diet of approximately the same amount of food,but not as low in fat.After 4 months,the mean weight loss was 9.31 lbs. for group 1(s=4.67) and 7.40 lbs.(s=4.04) for group 2.
low-fat: x1¯=9.31 s1=4.67
control: x2¯=7.4 s2=4.04
x1¯−x2¯=1.91
对于“sampling distribution of difference of means”
期望:μx1¯−x2¯
方差:σ2x1¯−x2¯
又有公式
其中,σ2x1−x2 为population的variance of difference
置信区间:
以95%的confidence interval为例:
因为是two-tails combine一共占5%,因此,需要在z-table中找到值为2.5%的,此时,z=1.96
此时,置信区间的意义就有了:
我们能说:
95% chance that μx1¯−x2¯ is within 1.96σ2x1¯−x2¯ of 1.91.
接下来的事就是计算σ2x1¯−x2¯,
上面提到的计算这个未知量的公式,此处并不能用,因为并没有一个sample是关于x1−x2的,因此,无法用单个sample的s2来作为population的σ2估计,再使用公式推出sampling distribution的σ2x¯
而由已知量入手,
最后,可以说成:
95% confidence interval for μx1¯−x2¯ 1.91 ± 1.21
假设检验:
先设定significance level=5%
null hypothesis: low-fat diet dose nothing.
H0:μ1−μ2=0=>μx1¯−x2¯=0
H1:μ1−μ2>0=>μx1¯−x2¯>0
首先,根据z-table找出5%的z值,为1.65
z∗σx1¯−x2¯+0=1.01805<μx1¯−x2¯=1.91
因此拒绝假设
(其实分母原始写法为:
)