Respuesta :
Part A
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
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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.
-----------------------
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
A) The mean and standard deviation of the sampling distribution of all possible soil samples of size n = 32 are; Mean= 5.7 and standard deviation = 0.046
B) The 95% confidence interval for the mean pH level of soil samples from every 1-foot by 1-foot section is; CI = (3.694, 7.706)
C) The 95% confidence interval for the mean pH level of soil samples from every 1-foot by 1-foot section is; CI = (5.986, 6.214)
A) Sample size; n = 32
Mean; x' = 5.7
Original standard deviation; s = 0.26
Formula for standard deviation of the mean which is the standard error of the mean is;
s_m = (s/√n)
s_m = 0.26/√32
s_m = 0.046
B) Formula for confidence interval;
CI = x' ± z(s/√n)
where z is critical value at confidence level
We are given CL = 95%
From online sources, the Critical value at CL of 95% is z = 1.96
z(s/√n) is the margin of error
Thus;
ME = 1.96(0.26/√32)
ME = 2.006
CI = 5.7 ± 2.006
CI = (3.694, 7.706)
Interpretation is that we are 95% confident that the pH level of soil samples from every 1-foot by 1-foot section is between 3.694 and 7.706.
C) Sample size; n = 34
Mean; x' = 6.1
standard deviation; s = 0.34
Confidence interval is;
CI = x' ± z(s/√n)
where z is critical value at confidence level
We are given CL = 95%
From online sources, the Critical value at CL of 95% is z = 1.96
z(s/√n) is the margin of error
Thus;
ME = 1.96(0.34/√34)
ME = 0.114
Thus;
CI = 6.1 ± 0.114
CI = (5.986, 6.214)
The interpretation of the confidence interval is that;
We are 95% confident that the pH level of soil samples from every 1-foot by 1-foot section is between 3.694 and 7.706.
Read more about confidence intervals at; https://brainly.com/question/23837916