Confidence Interval. Bernoulli Distribution. I am reviewing the construction of confidence intervals for a random sample with Bernoulli distribution. The book uses the statistics of the central limit theorem that distributes N(0,1) to estimate the interval : Z_n=(X_1+X_2+cdots+X_n-n mu)/(sigma sqrtn)

Spactapsula2l

Spactapsula2l

Answered question

2022-09-17

Confidence Interval. Bernoulli Distribution
I am reviewing the construction of confidence intervals for a random sample with Bernoulli distribution. The book uses the statistics of the central limit theorem that distributes N(0,1) to estimate the interval :
Z n = X 1 + X 2 + + X n n μ σ n
Why are the intervals constructed from these statistics symmetrical around the origin?
The book says: "Since it is desirable that the length of the interval be as small as possible and since the standard normal distribution is symmetrical around the origin, it turns out that the minimum length interval must also be symmetric around the origin", but I don't understand this.

Answer & Explanation

Yadira Mcdowell

Yadira Mcdowell

Beginner2022-09-18Added 13 answers

Step 1
First, the confidence interval your book is calculating is not an "exact interval" but it is only an asymptotic one. (you can use this approx if both np and n ( 1 p ) are 5). For smaller samples you must use the histogram of the binomial r.v.
Step 2
Second, the statistic used (note that this is a CSS Statistics) is not symmetrical around the origin but it is symmetrical around n μ. In fact
T n = i = 1 n X i
has an asymptotic gaussian distribution N ( n μ ; n σ 2 )
It is symmetrical around its mean because if not, the asymmetrical queue becomes zero.
The Statistic symmetrical around zero is the "Pivotal Quantity" Z n = T n μ σ n = T n p n p ( 1 p )

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