Respuesta :
Answer:
a) 82
b) The 95% confidence interval of the mean reading scores of all fifth-graders is between 77 and 87.
c) The 99% confidence interval of the mean reading scores of all fifth-graders is between 75.47 and 88.53.
d) The 99% confidence interval is larger, due to the higher critical value of z.
Step-by-step explanation:
a. Find the best point estimate of the mean.
The best point estimate of the mean is the mean of the sample, which is 82.
b. Find the 95% confidence interval of the mean reading scores of all fifth-graders.
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{15}{\sqrt{35}} = 5[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 82 - 5 = 77
The upper end of the interval is the sample mean added to M. So it is 82 + 5 = 87
The 95% confidence interval of the mean reading scores of all fifth-graders is between 77 and 87.
c. Find the 99% confidence interval of the mean reading scores of all fifth-graders.
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{15}{\sqrt{35}} = 6.53[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 82 - 6.53 = 75.47
The upper end of the interval is the sample mean added to M. So it is 82 + 6.53 = 88.53
The 99% confidence interval of the mean reading scores of all fifth-graders is between 75.47 and 88.53.
d. Which interval is larger
The 99% confidence interval is larger, due to the higher critical value of z.
