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insens350 [35]
2 years ago
9

A statistician selected a sample of 16 accounts receivable and determined the mean of the sample to be $5,000 with a standard de

viation of $400. He reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. He neglected to report what confidence level he had used. Based on the above information, determine the confidence level that was used. Assume the population has a normal distribution.
Mathematics
1 answer:
Anon25 [30]2 years ago
4 0

Answer:

The confidence level that was used is 0.25% .

Step-by-step explanation:

We are given that mean of the selected sample of 16 accounts is $5,000 and a standard deviation of $400.

It has also been reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. This represents the Confidence Interval for population mean.

<em>But we have to find that at what confidence level this information about range of population men has been stated.</em>

Since we know that Confidence Interval for population mean is given by :

   <em>C.I. for population mean = Sample mean(xbar) </em>\pm<em> z value * </em>\frac{Standard    deviation}{\sqrt{n} }<em />

i.e.,<em>if we have 95% C.I. = xbar </em>\pm<em> 1.96 * </em>\frac{\sigma}{\sqrt{n} }<em> .</em>

So, our Confidence Interval for population is written as :

 [$4,739.80 , $5,260.20] = $5000 \pm z value * \frac{400}{\sqrt{16} }

 $5000 - z value * 100 = $4739.80    { Solving these we get Z value = 2.602}

 $5000 + z value * 100 = $5260.20

In z table we find that at value of 2.60 the probability is 0.99534 so subtracting this from 1 we get confidence level for one tail i.e.0.5%(approx).

Therefore, for two tail Confidence level will be 0.5%/2 = 0.25% .

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uranmaximum [27]

Answer:

1. The square root of 100

2.c=27.11

d=13.96

e=12.6

<u>Step</u><u> </u><u>by</u><u> </u><u>step</u><u> </u><u>explanation</u><u>:</u><u> </u>

1. 2×50=100 the geometric mean is the square root of 100

2. The diagram is made up of three right angled triangles the big outer one and two triangles inside the big one thus we can use the pythagorean theorem to come up with expressions which will help us in solving the unknown parts as below.

c²=30²-d²

c²=e²+24²

d²=e²+6²

both the following add up to c² meaning they are equal:

c²=30²-d²

c²=e²+24²

thus

30²-d²=e²+24²

I want to remain with d² only thus;

30²-24²-e²=d²

900-576=324 (squareroot of 324=18) so

d²=18²-e² and d²=e²+6²

both the above add up to d² meaning they are equal thus;

18²-e²=e²+6²

18²-6²=e²+e²

324-36=2e²

318=2e²

159=e²

e=12.61

<em>Thus</em><em> </em><em>d</em><em>²</em><em>=</em><em>e</em><em>²</em><em>+</em><em>6</em><em>²</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>d</em><em>²</em><em>=</em><em>1</em><em>2</em><em>.</em><em>6</em><em>1</em><em>²</em><em>+</em><em>6</em><em>²</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>d</em><em>²</em><em>=</em><em>1</em><em>5</em><em>9</em><em>+</em><em>3</em><em>6</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>d</em><em>²</em><em>=</em><em>1</em><em>9</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>d</em><em>=</em><em>1</em><em>3</em><em>.</em><em>9</em><em>6</em>

<em>c</em><em>²</em><em>=</em><em>e</em><em>²</em><em>+</em><em>2</em><em>4</em><em>²</em>

<em>c</em><em>²</em><em>=</em><em>12.61</em><em>²</em><em>+</em><em>2</em><em>4</em><em>²</em>

<em>c</em><em>²</em><em>=</em><em>1</em><em>5</em><em>9</em><em>+</em><em>5</em><em>7</em><em>6</em>

<em>c</em><em>²</em><em>=</em><em>7</em><em>3</em><em>5</em>

<em>c</em><em>=</em><em>27.11</em>

<em> </em>

<em>I</em><em> </em><em>hope</em><em> </em><em>this</em><em> </em><em>helps</em>

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