Question

In a study of 265 subjects, the average score on the examination was 63.8 and s = 3.08. What is a 95 percent confidence for μ?

Answer 1

The 95 percent confidence for μ is (63.4292, 64.1708).

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find M as such

`M = z*(s)/(√(n))`

In which s is the standard deviation of the population and n is the size of the sample.

`M = 1.96*(3.08)/(√(265)) = 0.3708`

The lower end of the interval is the mean subtracted by M. So it is 63.8 - 0.3708 = 63.4292

The upper end of the interval is the mean added to M. So it is 63.8 + 0.3708 = 64.1708

The 95 percent confidence for μ is (63.4292, 64.1708).