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andrew11 [14]
3 years ago
6

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consum

ption to be 74 gallons with a standard deviation of 16 gallons. Assume that the population distribution is normal. (Use t Distribution Table.)
a-1. What is the value of the population mean?
16
Unknown
74
a-2. What is the best estimate of this value?
Estimate population mean
c. For a 90% confidence interval, what is the value of t? (Round your answer to 3 decimal places.)
Value of t
d. Develop the 90% confidence interval for the population mean. (Round your answers to 3 decimal places.)
Confidence interval for the population mean is and .
e. Would it be reasonable to conclude that the population mean is 68 gallons?
a) Yes
b) No
c) It is not possible to tell.
Mathematics
1 answer:
PIT_PIT [208]3 years ago
8 0

Correct question is;

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consumption to be 74 gallons with a standard deviation of 16 gallons.

a. What is the value of the population mean? What is the best estimate of this value?

b. Explain why we need to use the t distribution. What assumption do you need to make?

c. For a 90 percent confidence interval, what is the value of t?

d. Develop the 90 percent confidence interval for the population mean.

e. Would it be reasonable to conclude that the population mean is 68 gallons?

Answer:

A) Best estimate = 74 gallons

B) because the population standard deviation is unknown. The assumption we will make is that the population follows the normal distribution.

C) t = 1.725

D) 90% confidence interval for the population mean is (67.9772, 80.0228) gallons

E) Yes

Step-by-step explanation:

We are given;

Sample mean; x' = 74

Sample population; n = 21

Yearly Standard deviation; s = 16

A) We are not given the population mean.

So the closest estimate to the population mean would be the sample mean which is 74.

B) We are not given the population standard deviation and as such we can't use normal distribution. So what is used when population standard deviation is not known is called t - distribution table. The assumption we will make is that the population follows the normal distribution.

C) At confidence interval of 90% and DF = n - 1 = 21 - 1 = 20

From t-tables, the t = 1.725

D) Formula for the confidence interval is;

x' ± t(s/√n) = 74 ± 1.725(16/√21) = 74 ± 6.0228 = 67.9772 or 80.0228

Thus 90% confidence interval for the population mean is (67.9772, 80.0228) gallons

E) 68 gallons lies within the range of the confidence interval, thus we can say that "Yes, it is reasonable"

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Step-by-step explanation:

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HELP PLEASE!!!!!! Find the speed of an athlete who makes 4and3/4 laps in 3mins45 seconds on a 400m field in m/s​
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Given:

An athlete who makes 4\dfrac{3}{4} laps in 3 mins 45 seconds on a 400m field.

To find:

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Solution:

We know that,

Distance covered in 1 lap = 400 m

Distance covered in 4\dfrac{3}{4} laps = 4\dfrac{3}{4}\times 400 m

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We know that,

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The speed of the athlete is:

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Answer:

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Step-by-step explanation:

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We define the random variable D = before-after and we can calculate the inidividual values:

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And the deviation with:

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And the confidence interval for this case would be given by:

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The degrees of freedom are given by:

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For the 95% of confidence the value for the significance is \alpha=0.05 and the critical value would be t_{\alpha/2}= 3.182. And replacing we got:

7-3.182 \frac{13.638}{\sqrt{4}}= -14.698

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