Question

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

Answer 1

Answer:

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

Step-by-step 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:

$\alpha = \frac{1-0.95}{2} = 0.025$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.96\frac{11}{30} = 3.94$

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.