The concentration of copper in the sediments in a particular stream in the Yukon was studied...

The concentration of copper in the sediments in a particular stream in the Yukon was studied using 40 samples. The mean of the samples was found to be 40.0 ppm, with a standard deviation of 16.0 ppm.

a) Construct a 95% confidence interval for the true mean level of copper in the stream sediments.

b) Suppose that 70 samples were collected (instead of 40), would the 95% confidence interval widen or tighten? Explain how you know.

c) Suppose that 40 samples were collected you wanted to be 99% confident (instead of 95%), would the confidence interval widen or tighten compared to a)? Explain.

d) Suppose that 20 samples were collected; what additional piece of information would you need in order to construct a 95% confidence interval?

Expert Solution

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 40- 1 ) = 2.023
40 ± t(0.05/2, 40 -1) * 16/√(40)
Lower Limit = 40 - t(0.05/2, 40 -1) 16/√(40)
Lower Limit = 34.8822
Upper Limit = 40 + t(0.05/2, 40 -1) 16/√(40)
Upper Limit = 45.1178
95% Confidence interval is ( 34.8822 , 45.1178 )
Width = 45.1178 - 34.8822 = 10.236

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 70- 1 ) = 1.995
40 ± t(0.05/2, 70 -1) * 16/√(70)
Lower Limit = 40 - t(0.05/2, 70 -1) 16/√(70)
Lower Limit = 36.1848
Upper Limit = 40 + t(0.05/2, 70 -1) 16/√(70)
Upper Limit = 43.8152
95% Confidence interval is ( 36.1848 , 43.8152 )
Width = 43.8152 - 36.1848 = 7.630

The confidence interval is tighten

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 40- 1 ) = 2.708
40 ± t(0.01/2, 40 -1) * 16/√(40)
Lower Limit = 40 - t(0.01/2, 40 -1) 16/√(40)
Lower Limit = 33.1492
Upper Limit = 40 + t(0.01/2, 40 -1) 16/√(40)
Upper Limit = 46.8508
99% Confidence interval is ( 33.1492 , 46.8508 )
Width = 46.8508 - 33.1492 = 13.702

The confidence interval is widen

Part d)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 20- 1 ) = 2.093
40 ± t(0.05/2, 20 -1) * 16/√(20)
Lower Limit = 40 - t(0.05/2, 20 -1) 16/√(20)
Lower Limit = 32.5119
Upper Limit = 40 + t(0.05/2, 20 -1) 16/√(20)
Upper Limit = 47.4881
95% Confidence interval is ( 32.5119 , 47.4881 )

orchestra answered 1 year ago

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