Download ENGG2430A-Homework 5

ENGG2430A-Homework 5
Due on Mar 19, 2014 via the assignment box.
1. Ex 9.4 from textbook The heights of a random sample of 50 college students showed a mean of 174.5
centimeters and a standard deviation of 6.9 centimeters.
(a) Construct a 98% confidence interval for the mean height of all college students.
Solution: From the question, the sample size is n = 50, the realization of the sample mean
is x̄ = 174.5, the sample standard deviation is s = 6.9, and the confidence level desired is
0.98 = 1 − α with α = 0.02.
Since n > 30, the sample standard deviation approximates σ reasonably well and so we can use
the large-sample confidence interval (p.276 of the textbook,)
s
6.9
x̄ ± zα/2 √ = 174.5 ± (−Φ(0.02/2)) √
n
50
≈ 174.5 ± 2.27 ≈ [172.2, 176.8]
(b) What can we assert with 98% confidence about the possible size of our error if we estimate the
mean height of all college students to be 174.5 centimeters?
Solution: By the previous part, we can assert with 98% confidence that the error |x̄ − µ| is no
larger than zα/2 √sn = 2.27. This is again because the sample size is sufficiently large so that
the sample standard deviation is a good estimate of the population standard deviation.
2. Ex 9.8 from textbook An efficiency expert wishes to determine the average time that it takes to drill
three holes in a certain metal clamp. How large a sample will she need to be 95% confident that her
sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies
that σ = 40 seconds.
Solution: As stated in the question, the confidence level is 0.95 = 1−α with α = 0.05, the standard
deviation is σ = 40. To satisfy the desired error requirement c = 0.5, we set c ≥ zα/2 √σn , i.e.
σ 2
n ≥ zα/2
≈
c
40 2
1.96 ×
≈ 27.3
15
and so the calculation suggests a sample size of n = 28. However, for the central limit theorem to
give a good approximation, it is desired to have n > 30 when the distribution of the population is
not too skewed.
n.b. if the distribution is skewed, the sample size may need to be larger. If the distribution is
approximately normal to begin with, the sample size may be smaller.
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