MATH 472 / SPRING 2013
ASSIGNMENT 3: DUE FEBRUARY 20
FINALIZED
Please include a cover sheet that provides a complete sentence answer to each the following
three questions:
(a) In your opinion, what were the main ideas covered in the assignment?
(b) Were there any topics that you believe should have been covered, but were not?
(c) What problems did you have with the assignment, if any?
Write your solutions carefully and neatly, preferably in complete and grammatically correct sentences. Justify all of your arguments.
1. Question 5.3.19, page 311.
2. Question 5.3.28, page 311.
3. Question 6.2.1, pages 359–360.
4. Question 6.2.8, page 360.
5. Question 6.2.11, page 360. Perform a suitable hypothesis test, and then quantify the
result using a P -value.
6. Question 6.3.1, pages 365–366.
7. Question 6.3.9, pages 366.
8. Question 6.4.5, pages 378.
9. Question 6.4.7, pages 378.
Solutions to Selected Problems.
Solution (6.2.11). In the absence of any other information, we assume that the price of a
“food basket” is normally distributed. Set µ0 = $145.75 and σ = $9.50. Given that this is an
inflation study, it seems most natural to test H0 : µ = µ0 against H1 : µ > µ0 at the α = 0.05
ȳ−µ
√ 0 ≥ z0.05 = 1.64.
level of significance. Under these circumstances, we will reject H0 if z = σ/
n
With n = 25, this decision rule becomes
“Reject H0 if ȳ ≥ $148.87.”
The sample we are given is ȳ = $149.75, so we move to reject H0 . That is, the difference
between the expected mean and the observed mean was STATISTICALLY SIGNIFICANT.
Now let’s compute a P-value with our 1-sided hypothesis test. We get
$149.75 − µ0
√
P-value = P Ȳ ≥ $149.75 | µ = µ0 = P Z ≥
= P (Z ≥ 2.105) = 0.0176.
σ/ n
This is a VERY small P-value (relative to α = 0.05).
Note: If instead you decided to use a 2-sided hypothesis test with H1 : µ 6= µ0 , then you
would reject H0 if ȳ ≤ $142.03 or ȳ ≥ $149.47. Your P-value would be
P-value = P (Z ≤ −2.105) + P (Z ≥ 2.105) = 2(0.0176) = 0.035.
Solution (6.3.9). We are going to do a binomial parameter test to of H0 : p = 0.75 against
H1 : p < 0.75 using a sample with n = 7 and the decision rule “Reject H0 if k ≤ 3.”
(a) The level of significance α is the probability of committing a type I error:
α = P (Reject H0 | H0 is true)
= P (k ≤ 3 | p = 0.75)
3 X
7
=
(0.75)k (0.25)7−k ≈ 0.07.
k
k=0
(b) The probability that H0 will be rejected is a function of the true parameter p:
3 X
7 k
f (p) := P ( Reject H0 ) = P (k ≤ 3) =
p (1 − p)7−k .
k
k=0
Plotting y = f (p) in Sage, we find the following curve:
2
As one would expect, the probability of rejecting H0 becomes substantially higher as p gets
closer to zero. Note that if the true parameter satisfies p < 0.75, then f (p) = 1 − β, the
power of the decision rule.
Solution (6.4.5). Set µ0 = 240. Let’s figure out the decision rule for testing H0 : µ = µ0
against H1 : µ < µ0 at the α = 0.01 level of significance with a random sample of n = 25
normally distributed observations. We are given that σ = 50. Our general theory says that
ȳ−µ
√ 0 is a standard normal measurement, and our decision rule will be:
the ratio z = σ/
n
“Reject H0 if z ≤ zα = −2.326.”
Equivalently, writing this in terms of ȳ it becomes
σ
“Reject H0 if ȳ ≤ µ0 + zα √ = 216.74.”
n
Now let’s assume that the true value of the mean satisfies µ ≤ 220. The probability that
our decision rule fails to detect this occurrence is given by
P Ȳ > 216.74|µ ≤ 220 ≥ P Ȳ > 216.74|µ = 220
Ȳ − µ
216.74 − 220
√ >
√
=P
σ/ n
50/ 25
= P (Z > −0.326) ≈ 0.63.
Solution (6.4.7). First, let’s determine the decision for testing H0 : µ = 200 against H1 :
µ < 200 at the α = 0.10 level of significance. Here we write µ0 = 200 and assume that
√ , our decision rule is “Reject H0 if z ≤ −z0.1 = −1.2816.
σ = 15.0. Setting z = ȳ−µ+−
σ/ n
Rewriting this in terms of ȳ, we get
σ
“Reject H0 if ȳ ≤ 200 − 1.2816 √ ”.
n
We want the power of our decision rule to be at least 0.75 when the true mean satisfies
µ = 197. (In fact, this computation will give us the power 0.75 for any true mean ≤ 197.)
To that end, we have
0.75 ≤ 1 − β = P (“Reject H0 ” |µ = 197)
σ
= P ȳ ≤ 200 − 1.2816 √ | µ = 197
n
√
ȳ − 197
3 − 1.2816σ/ n
√ ≤
√
=P
| µ = 197
σ/ n
σ/ n
√
n
=P Z≤
− 1.2816 .
5
√
It follows that 5n − 1.2816 ≥ z0.75 = 0.6745. Solving for n, we obtain n ≥ 95.66. So in order
to achieve this kind of power in our decision rule, we need to take a sample size n ≥ 96.
3