Question

A population is normally distributed with a standard deviation of 2.8. A random sample is obtained from this population and the observations are: 8, 9, 10, 13, 14, 16, 17, 20, 21. Construct the 95% confidence interval for the mean of this population. Construct the 99% confidence interval for the mean of this population.

Answer 1

Answer:

The 95% confidence interval for the mean of this population is between 12.39 and 16.05.

The 99% confidence interval for the mean of this population is between 11.82 and 16.62.

Step-by-step explanation:

The first step is finding the mean of the sample:

There are 9 observations. So

$\mu_{x} = \frac{8+9+10+13+14+16+17+20+21}{9} = 14.22$

95% 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 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.

$M = 1.96*\frac{2.8}{\sqrt{9}} = 1.83$

The lower end of the interval is the sample mean subtracted by M. So it is 14.22 - 1.83 = 12.39

The upper end of the interval is the sample mean added to M. So it is 14.22 + 1.83 = 16.05

The 95% confidence interval for the mean of this population is between 12.39 and 16.05.

99% confidence interval:

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

$\alpha = \frac{1-0.99}{2} = 0.005$

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 $1-0.005 = 0.995$, so $z = 2.575$

Now, find 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.

$M = 2.575*\frac{2.8}{\sqrt{9}} = 2.40$

The lower end of the interval is the sample mean subtracted by M. So it is 14.22 - 2.40 = 11.82

The upper end of the interval is the sample mean added to M. So it is 14.22 + 2.40 = 16.62

The 99% confidence interval for the mean of this population is between 11.82 and 16.62.