Respuesta :
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
[tex]\mu_{x} = \frac{8+9+10+13+14+16+17+20+21}{9} = 14.22[/tex]
95% confidence interval:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96*\frac{2.8}{\sqrt{9}} = 1.83[/tex]
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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575*\frac{2.8}{\sqrt{9}} = 2.40[/tex]
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.
