Question

Suppose we are sampling from a normally distributed population, and we wish to use the t procedure to find a confidence interval for the population mean. Consider the following 3 statements:

I. All else being equal, the greater the sample size, the smaller the margin of error.

II. All else being equal, the greater the confidence level, the greater the margin of error.

III. All else being equal, the greater the sample variance, the smaller the margin of error.

Which of these statements are true?

A. Just I

B. Just II

C. I and II

D. II and III

E. All of them

Answer 1

Solution:

Margin of error=(critical value)*(S.E(x-bar))

In case of t procedure to find out confidence interval

S.E(x-bar)=s/√n

So

Margin of error=t-score*(s/√n)

So margin of error is inversaly proportional to sample size so as sample size increases margin of error decrease for fixed sample variance and confidence interval so statement'1'is correct

And if we increase confidence lavel then t- score increses (for fixed sample size) and margin of error is directly proportional to t-score so as confidence lavel increases margin of error increases .so stament '2'is also correct

Margin of error is directly proportional to sample variance (s^2) so as sample variance increase margin of error increases for fixed sample size and fixed confidence lavel.so statement '3' is incorrect

So statement'1'and statement'2'are correct

So option'C' is correct