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:
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
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 . 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
In which s is the standard deviation of the sample and n is the size of the sample.
For this question, , and thus:
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:
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
In which is the standard deviation of the population and n is the size of the sample.
In this question,
We have to find n for which M = 100. So
Dividing both sides by 100:
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