Question

1) Find the 95 % confidence interval for the mean if σ = 11 , using this sample:

43 52 18 20 25 45 43 21 42 32 24 32
19 25 26 44 42 41 53 22 25 23 21 27
33 36 47 19 20 41

Answer 1

The 95 % confidence interval for the mean is between 28.09 and 35.97.

Explanation:

Mean of the sample:

30 values.

So we sum all the values and divide by 30.

The sum of all the values(43 + 52 + 18 + ... + 20 + 41) is 961.

961/30 = 32.03

Confidence interval:

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 the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96(11)/(30) = 3.94`

The lower end of the interval is the sample mean subtracted by M. So it is 32.03 - 3.94 = 28.09

The upper end of the interval is the sample mean added to M. So it is 32.03 + 3.94 = 35.97

The 95 % confidence interval for the mean is between 28.09 and 35.97.