0
$\begingroup$

A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).

What are the values of the sample mean and the variance? (X bar and S)

I got X bar to be 47.3.

I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4

$\endgroup$

1 Answer 1

0
$\begingroup$

Yes, you have gotten the mean right, it is the average of the end points.

$$\bar{x} - k S = 44.7$$

$$\bar{x} + k S = 44.7$$

where $k$ is known to you.

Since you already know $\bar{x}$, you can now solve for $\sigma$ in terms of $k$ and $\bar{x}$.

$\endgroup$
1
  • $\begingroup$ I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4 $\endgroup$
    – kt046172
    Nov 19, 2018 at 1:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .