Respuesta :
Answer:
The 90% confidence interval would be between 73.80 and 82.20.
Step-by-step explanation:
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 the margin of errorM 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.
We have that M = 5, n = 64. We have to find the standard deviation of the population. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]5 = 1.96*\frac{\sigma}{\sqrt{64}}[/tex]
[tex]1.96\sigma = 5*8[/tex]
[tex]\sigma = \frac{5*8}{1.96}[/tex]
[tex]\sigma = 20.41[/tex]
90% 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{1-0.9}{2} = 0.05[/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]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/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 = 1.645*\frac{20.41}{\sqrt{64}} = 4.20[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 78 - 4.20 = 73.80.
The upper end of the interval is the sample mean added to M. So it is 78 + 4.20 = 82.20
The 90% confidence interval would be between 73.80 and 82.20.
