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Alenkasestr [34]
3 years ago
10

Please help I’m stuck.

Mathematics
1 answer:
Alika [10]3 years ago
8 0

Answer: bro Idk imma try to figure it out

Step-by-step explanation:

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The annual update of U.S. Overseas Loans and Grants, informally known as the “Greenbook,” contains data on U.S. government monet
vivado [14]

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

3 0
2 years ago
Can you help me with this
densk [106]

Answer:

158 is greater than 150

For #1 :

Student B can type 8 more words per minute than student A

For #2 :

You divide the last y value by the last x value : 700/5

700/5=140

For # 3 :

1. Student B - 158 words per minute

2. Student A - 150 words per minute

3. Student C - 140 words per minute


8 0
3 years ago
What percent of 12 is 9?<br> A. 25%<br> B. 67%<br> C. 75%<br> D. 108%<br> E. 133%
Sergeu [11.5K]

Answer:

75%

Step-by-step explanation:

75% = 3/4

3/4 * 12 = 9

6 0
3 years ago
A car can travel 476 miles on 14 gallons of gas how many gallons of gas does this car need to travel 578 miles
mr_godi [17]
The car needs 17 gallons 476/14=34 then 578-476=102 then 102/34=3 then 14+3=17
7 0
3 years ago
At the city museum, child admission is $6.20 and adult admission is $9.90 . On Monday, three times as many adult tickets as chil
murzikaleks [220]

Can you please delete this. I have answered it wrong. Thanks

7 0
3 years ago
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