Mathematics, 26.03.2020 20:37 marcusajns

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

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Answer from: hazle9586

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 $\alpha$ 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.

$M = 1.96*\frac{15}{\sqrt{35}} = 5$

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:

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

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.

$M = 2.575*\frac{15}{\sqrt{35}} = 6.53$

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.

Answer from: Quest

the direct variation says that:

$y \propto x$

then the equation is of the form is given by:

$y = kx$ where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

$3 = 2k$

divide both sides by 2 we get;

$k =\frac{3}{2} = 1.5$

⇒ $m(x) = 1.5x$

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

$m(x) = 1.5x$

$-1.5 = 1.5(-1)$

-1.5 = -1.5 true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

Answer from: Quest

Is there a picture of a graph or whatever that goes along with this?

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Sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the pop...