X^10 the exponent would be 10
Answer:
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Step-by-step explanation:
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
