Question

sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the population is 15. a. Find the best point estimate of the mean. b. Find the 95% confidence interval of the mean reading scores of all fifth-graders. c. Find the 99% confidence interval of the mean reading scores of all fifth-graders. d. Which interval is larger

Answer 1

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.

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
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (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*(\sigma)/(√(n))`

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

`M = 1.96*(15)/(√(35)) = 5`

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
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (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*(\sigma)/(√(n))`

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

`M = 2.575*(15)/(√(35)) = 6.53`

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.