MATH1725 Introduction to Statistics: Worked examples

Worked Example: Lectures 4–6
A firm investigates the length of telephone conversations of their office staff. Ten consecutive
conversations had lengths, in minutes:
10.7, 9.5, 11.1, 7.8, 11.9, 4.1, 10.0, 9.2, 6.5, 9.2.
Derive a 95% confidence interval for the mean conversation length. Test whether the mean length
of a conversation is eight minutes.
Answer:
¯x = 1 n∑
x i = 90
n 10 = 9 minutes.
i=1
{ n∑
}
s 2 = 1 x 2 i
n − 1
− n¯x2 = 5.42.
i=1
Estimate the population variance σ 2 by s 2 with s = √ 5.42 = 2.33. Then
¯X − µ
s/ √ n ∼ t n−1.
95% confidence interval for µ is ¯x ± t 9 (2.5%)s/ √ 10. Here s/ √ 10 = 0.737, t 9 (2.5%) = 2.262.
s
¯x ± t 9 (2.5%) √ = 9 ± (2.262 × 0.737)
10
= 9 ± 1.667 = (7.3,10.7).
Since 8 minutes lies inside the 95% confidence interval we would accept H 0 in testing H 0 : µ =
8 vs. H 1 : µ ≠ 8 at the 5% significance level.
Worked Example: Lectures 5–6
A population has a Poisson distribution but it is not known whether the mean µ is 1 or 4. To
test the hypothesis H 0 : µ = 1 vs. H 1 : µ = 1 on the basis of one observation X the following test
procedure is considered: reject H 0 if X ≥ i.
Type I error is defined to be “rejecting H 0 when H 0 is true”. Find the probability of type I
error for the three cases i = 2, 3, 4.
Answer: If H 0 is true, µ = 1 and
so that pr{Type I error} = pr{X ≥ i}.
If i = 2,
pr{X = x} = e−1
x! , x = 0,1,2,... ,
pr{Type I error} = pr{X ≥ 2} = 1−pr{X < 2} = 1−pr{X = 0}−pr{X = 1} = 1−e −1 −e −1 = 0.264.
Similarly if i = 3,
If i = 4,
pr{Type I error} = pr{X ≥ 3} = 1 − pr{X < 3} = 0.080.
pr{Type I error} = pr{X ≥ 4} = 0.019.
Notice that an exact 5% or 10% significance level test does not exist for this discrete distribution.
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