Question

The annual update of U.S. Overseas Loans and Grants, informally known as the “Greenbook,” contains data on U.S. government monetary economic and military assistance loans. The following table shows military assistance loans, in thousands of dollars, to a sample of 10 countries, as reported by the U.S. Agency for International Development.
102, 69, 280, 180, 33, 89, 1643, 205, 177, 695

The sample mean is
(Round to one decimal place)
The sample standard deviation is
(Round to one decimal place)
The sample size is
The degree of freedom for the Studentized version of the sample mean is
The point estimate for the mean of ALL US Overseas Loans and Grants is
The t-value for a 95% confidence interval is t =
The Margin of error for a 95% confidence interval for this sample is
The 95 % confidence interval estimate of the mean of ALL US Overseas Loans and Grants is
< µ <

(Round to one decimal place)

The sample size is needed in order to obtain a margin of error of no more than 100 is
(Must be a whole number)

Answer 1

To solve this question, we find the sample mean, sample standard deviation, sample size, the number of degrees of freedom, the confidence interval and the required sample size. For each step, the procedures are explained and we find what is asked.

Doing this, we find that the sample mean is 347.3, the sample standard deviation is 467.4, the sample size is 10, the number of degrees of freedom is 9, the t-value for a 95% confidence interval is t = 2.2226, the margin of error for a 95% confidence interval for this sample is 328.5, the 95% CI is $18.8 < µ < $675.8 and the sample size needed in order to obtain a margin of error of no more than 100 is 94.

The sample mean is

Sum of all values divided by the number of values, thus:

$\overline{x} = \frac{102 + 69 + 280 + 180 + 33 + 89 + 1643 + 205 + 177 + 695}{10} = 347.3$

The sample mean is 347.3.

The sample standard deviation is

Square root of the sum of the difference squared between each value and the mean, divided by one less than the sample size. So

$s = \sqrt{\frac{(102-347.3)^2 + (69-347.3)^2 + (280-347.3)^2 + (180-347.3)^2 + (33-347.3)^2 + (89-347.3)^2 + (1643-347.3)^2 + (205-347.3)^2 + (177-347.3)^2 + (695-347.3)^2}{9}} = 467.4$

The sample standard deviation is 467.4.

The sample size is

10 observations, thus the sample size is 10.

The degree of freedom for the Studentized version of the sample mean is

One less than the sample size, so 10 - 1 = 9.

The number of degrees of freedom is 9.

The point estimate for the mean of ALL US Overseas Loans and Grants is

The sample mean, which is 347.3.

The point estimate is 347.3.

The t-value for a 95% confidence interval is t

The t-value is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.2226.

The t-value for a 95% confidence interval is t = 2.2226.

The Margin of error for a 95% confidence interval for this sample is

$M = t\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample.

For this question, $s = 467.4, n = 10$, and thus:

$M = 2.2226\frac{467.4}{\sqrt{10}}$

$M = 328.5$

The margin of error for a 95% confidence interval for this sample is 328.5.

The 95 % confidence interval estimate of the mean of ALL US Overseas Loans and Grants is

The lower end of the interval is the sample mean subtracted by M. So it is 347.3 - 328.5 = 18.8

The upper end of the interval is the sample mean added to M. So it is 347.3 + 328.5 = 675.8

The 95 % confidence interval estimate of the mean of ALL US Overseas Loans and Grants is:

$18.8 < µ < $675.8.

The sample size is needed in order to obtain a margin of error of no more than 100 is

Here, we have to consider the sample standard deviation as the population standard deviation, and use the z-distribution. So

We have to find our 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 Z-table as such z has a p-value of .

That is z with a p-value of , so Z = 1.96.

Now, find the margin of error 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.

In this question, $\sigma = 467.4$

We have to find n for which M = 100. So

$M = z\frac{\sigma}{\sqrt{n}}$

$100 = 1.96\frac{467.4}{\sqrt{n}}$

$100\sqrt{n} = 1.96*467.4$

Dividing both sides by 100:

$\sqrt{n} = 1.96*4.674$

$(\sqrt{n})^2 = (1.96*4.674)^2$

$n = 83.92$

Rounding up:

The sample size needed in order to obtain a margin of error of no more than 100 is 94.

For a problem that you build the confidence interval after finding the sample mean and the sample standard deviation, you can check brainly.com/question/24232455.

For a problem in which we have to find the sample size given the desired margin of error, you can check brainly.com/question/23559442