作业帮====各位数学大牛帮解一道小题,关于置信区间的========
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====各位数学大牛帮解一道小题,关于置信区间的========
Two random samples are taken,with each group asked if they support a particular candidate.A summary of the sample sizes and proportions of each group answering ``yes'' are given below:
Pop1:n1=93 p1=0.753
Pop2:n2=93 p2=0.575
Suppose that the data yields (0.0506,0.3054) for a confidence interval for the difference p1-p2 of the population proportions.What is the confidence level?(Give your answer in terms of percentages.)
Question:Confidence Level =_________________________?
在得到了critical value Z的情况下怎么反推出 confidence level 就是%的那个。
Two random samples are taken,with each group asked if they support a particular candidate.A summary of the sample sizes and proportions of each group answering ``yes'' are given below:
Pop1:n1=93 p1=0.753
Pop2:n2=93 p2=0.575
Suppose that the data yields (0.0506,0.3054) for a confidence interval for the difference p1-p2 of the population proportions.What is the confidence level?(Give your answer in terms of percentages.)
Question:Confidence Level =_________________________?
在得到了critical value Z的情况下怎么反推出 confidence level 就是%的那个。
Suppose the true expectations of the two kinds of random variables, denoted as X1 and X2, are p10 and p20 respectively. The null hypothesis is H0: p10 - p20 = 0.
Thus, we can construct the statistic : p1 - p2 / sd(p1 - p2) to test H0, where sd(p1 - p2) signifies the standard deviation of (p1 - p2). Since sd(p1 - p2) is unknown, we can replace it by its consistent estimator se(p1 - p2), i.e., the standard error of (p1 - p2).
Now, note that Var(p1 - p2) = Var(p1) + Var(p2) = Var(X1)/93 + Var(X2)/93. Moreover,
Var(X1) can be estimated by the sample variance of sample 1, which is p1(1-p1). Similarly, we can use p2(1-p2) to estimate Var(X2). Consequently, se(p1 - p2) = {[p1(1-p1) + p2(1-p2)]/93}^(0.5).
Suppose the confidence level is c. Thus, the confidence interval is
-c < (p1 - p2 - (p10 - p20))/se(p1-p2)
再问： 我得到的Z也是1.8728但是confidence level 怎么求呢？
再答： 这个一般按中心极限定理假设正态分布能近似该统计量，所以只需查表或用软件算一下它对应的percentile就行了，我刚算了下，应该是6.109573%。
再问： 大神好像不对诶。。哎郁闷了。。你是用表算的还是软件。用表是用table a吗 然后用1.87去寻找百分比？ 但我这样算出来是0.9693. 我第一次的时候得出的答案就是这个，但不对。。不知道错哪了。
再答： 你算出来的是P(X
Thus, we can construct the statistic : p1 - p2 / sd(p1 - p2) to test H0, where sd(p1 - p2) signifies the standard deviation of (p1 - p2). Since sd(p1 - p2) is unknown, we can replace it by its consistent estimator se(p1 - p2), i.e., the standard error of (p1 - p2).
Now, note that Var(p1 - p2) = Var(p1) + Var(p2) = Var(X1)/93 + Var(X2)/93. Moreover,
Var(X1) can be estimated by the sample variance of sample 1, which is p1(1-p1). Similarly, we can use p2(1-p2) to estimate Var(X2). Consequently, se(p1 - p2) = {[p1(1-p1) + p2(1-p2)]/93}^(0.5).
Suppose the confidence level is c. Thus, the confidence interval is
-c < (p1 - p2 - (p10 - p20))/se(p1-p2)
再问： 我得到的Z也是1.8728但是confidence level 怎么求呢？
再答： 这个一般按中心极限定理假设正态分布能近似该统计量，所以只需查表或用软件算一下它对应的percentile就行了，我刚算了下，应该是6.109573%。
再问： 大神好像不对诶。。哎郁闷了。。你是用表算的还是软件。用表是用table a吗 然后用1.87去寻找百分比？ 但我这样算出来是0.9693. 我第一次的时候得出的答案就是这个，但不对。。不知道错哪了。
再答： 你算出来的是P(X
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