3) a) A 90% Confidence Interval would be narrower than a ...



Chapter 7 Homework Answers

3) a) A 90% Confidence Interval would be narrower than a 95% Confidence Interval. This occurs because the as the precision of the confidence interval increases (ie CI width decreasing), the reliability of an interval containing the actual mean decreases (less of a range to possibly cover the mean). Look at 100% confidence; in order to ensure the mean is captured with 100% certainty, the interval must contain every possible value.

b) The statement is incorrect. The moment a confidence interval has been set (ie [7.8, 9.4]), it is no longer a random interval. Since the interval is set and the mean (μ) is also a set value, the mean either falls into the interval or it doesn’t, and thus cannot be spoken of in terms of a 95% probability.

c) This statement is incorrect. The interval [7.8, 9.4] is an attempt to capture the actual population mean, and is not representative of the alcohol contents of individual bottles.

d) This statement is incorrect. Although we expect that 95% of the confidence intervals to contain the population mean, we cannot say that in 100 trials exactly 95 intervals will contain the mean.

5. a. 95% confidence interval

x = 4.85, n=20, σ=.75

CI = x ± 1.96 * σ/√(n)

= (4.52, 5.18)

b. 98% confidence interval

x = 4.56, n=16, σ=.75, za/2=2.33

CI = x ± 2.33* σ/√(n)

=(4.12, 5.00)

c) 95% confidence interval, width = .4

n= (2*za/2*σ/w)^2

n=54.02 = 55 (Rounded up to next integer value)

d) 99% confidence interval, estimate average to within .2 of actual value (width =.4)

n= (2* za/2*σ/w)^2

n= 93.24 = 94 (Again rounded up)

37. a) 95% Confidence Interval

x = 0.9255, n=20, s=0.0809, ta/2, n-1=2.093

CI = x ± ta/2, n-1* s/√(n)

= (0.888, 0.964)

b) 95% Prediction Interval

x = 0.9255, n=20, s=0.0809, ta/2, n-1=2.093

PI = x ± ta/2, n-1* s*√(1/n + 1)

= (0.752, 1.099)

48 98% Confidence Interval

x = 188.0, n=9, s=7.2, ta/2, n-1=2.896

CI = x ± ta/2, n-1* s/√(n)

= (181.05, 194.95)

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