MATH SOLVE

2 months ago

Q:
# 1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the field. He finds that the mean pH level of the samples is 5.7 with a standard deviation of 0.26.A. Find the mean and standard deviation of the sampling distribution of all possible soil samples of size n = 32. Round to three decimal places.B. Construct a 95% confidence interval for the mean pH level of soil samples from every 1-foot by 1-foot section. Show your work. Even if you use technology, show the expression that is used to determine the margin of error for the interval. Interpret the interval in a complete sentence. C. The farmer tested soil samples from 34 randomly selected sections from another field. He found the mean pH level of those samples to be 6.1 with a standard deviation of 0.34. Construct a 95% confidence interval for the difference between the mean pH levels of all possible soil samples from both fields. Show your work. Even if you use technology, show the expression that is used to determine the margin of error for the interval. Interpret the interval in a complete sentence.Answer:

Accepted Solution

A:

Part A

Answers:

Mean = 5.7

Standard Deviation = 0.046

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The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32

So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

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Part B

The 95% confidence interval is roughly (3.73, 7.67)

The margin of error expression is z*s/sqrt(n)

The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

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At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error

ME = z*s/sqrt(n)

ME = 1.96*5.7/sqrt(32)

ME = 1.974949

The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:

L = xbar-ME

L = 5.7-1.974949

L = 3.725051

L = 3.73

and

U = xbar+ME

U = 5.7+1.974949

U = 7.674949

U = 7.67

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Part C

Confidence interval is (5.99, 6.21)

Margin of Error expression is z*s/sqrt(n)

If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

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At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error

ME = z*s/sqrt(n)

ME = 1.96*0.34/sqrt(34)

ME = 0.114286657

The margin of error is roughly 0.114286657

L = lower limit

L = xbar-ME

L = 6.1-0.114286657

L = 5.985713343

L = 5.99

U = upper limit

U = xbar+ME

U = 6.1+0.114286657

U = 6.214286657

U = 6.21

Answers:

Mean = 5.7

Standard Deviation = 0.046

-----------------------

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32

So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)

The margin of error expression is z*s/sqrt(n)

The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error

ME = z*s/sqrt(n)

ME = 1.96*5.7/sqrt(32)

ME = 1.974949

The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:

L = xbar-ME

L = 5.7-1.974949

L = 3.725051

L = 3.73

and

U = xbar+ME

U = 5.7+1.974949

U = 7.674949

U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)

Margin of Error expression is z*s/sqrt(n)

If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error

ME = z*s/sqrt(n)

ME = 1.96*0.34/sqrt(34)

ME = 0.114286657

The margin of error is roughly 0.114286657

L = lower limit

L = xbar-ME

L = 6.1-0.114286657

L = 5.985713343

L = 5.99

U = upper limit

U = xbar+ME

U = 6.1+0.114286657

U = 6.214286657

U = 6.21