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)