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soldier1979 [14.2K]
3 years ago
6

Can anyone help please?

Mathematics
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

Blank 1: 18

Blank 2: 10

Blank 3: 12

Blank 4: 18

Blank 5: 1.5

Blank 6: 155

Blank 7: 224

Blank 8: 280

B & C

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,46
JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

M = \frac{50463-38737}{2} = 5863

Relating the margin of error with the sample size:

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 - 0.9}{2} = 0.05

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.64.

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.

For this problem, we have that \sigma = 14300, M = 5863. So

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

5863 = 1.645\frac{14300}{\sqrt{n}}

5863\sqrt{n} = 1.645*14300

\sqrt{n} = \frac{1.645*14300}{5863}

(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2

n = 16

The sample size was of 16.

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Small point give away! Hope you guys enjoy!
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Answer:

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