Derive 100(1−α) percent lower and upper confidence intervals for p, when the data consist of the values of n independent Bernoulli random variables wi

Expert Answers (1)

2021-05-30

Lower confidence interval: \(\displaystyle{\left(-\infty,\overbrace{{{p}}}+{z}_{{\alpha}}\sqrt{{\overbrace{{{p}}}{\left({\frac{{{1}-\overbrace{{{p}}}}}{{{n}}}}\right)}}}\right.}\)
Upper confidence interval: \(\displaystyle{\left(\overbrace{{{p}}}-{z}_{{\alpha}}\sqrt{{\overbrace{{{p}}}{\left({\frac{{{1}-\overbrace{{{p}}}}}{{{n}}}}\right)},\infty}}\right\rbrace}\)

Relevant Questions

asked 2021-05-14

When σ is unknown and the sample size is \(\displaystyle{n}\geq{30}\), there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(\displaystyle{n}\geq{30}\), use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?

See answers (1)

asked 2021-05-23

Random variables \(X_{1},X_{2},...,X_{n}\) are independent and identically distributed. 0 is a parameter of their distribution.
If \(q(X,0)\sim N(0,1)\) is a pivotal function for 0, explain how you would use this result to obtain a symmetrical 95% confidence interval for 0.

See answers (1)

asked 2021-09-03

Let \(\displaystyle{X}_{{1}},{X}_{{2}},\dot{{s}},{X}_{{n}}\) be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables. Find n such that Pr\(\displaystyle{\left({X}{>}{2000}\right)}\geq{0.95}\)

See answers (1)

asked 2021-06-04

Let \(X_{1}, X_{2},...,X_{n}\) be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables.
Find n such that \(Pr(X>2000)\geq 0.95\).

See answers (1)

...