Download μ μ π μ δ μ μ μ μ

Solutions
Stat652
Homework 3
[Optional] 6.1.1 In an experiment to measure the lifetimes of parts manufactured from a
certain aluminum alloy, 73 parts were loaded cyclically until failure. The mean number
of kilocycles to failure was 783, and the standard deviation was 120. Let μ represent the
mean number of kilocycles to failure for parts of this type. A test is made of
H 0 = μ ≤ 750 versus H1 = μ > 750.
a. Find the P-value
b. Either the mean number of kilocycles to failure is greater than 750, or the sample
is in the most extreme ____% of its distribution.
Sln:
a. By definition, P-value is
sup π ( μ | δ t ) = sup Pr (T ≥ t | μ )
μ∈Ω0
:
μ∈Ω0
= Pr
(
X − μ0
σ
n
≥
783− 750
120
73
| μ0 ≤ 750
)
= Pr ( Z ≥ 2.35 | μ0 ≤ 750 )
Therefore, p = 1 − Pr ( Z ≤ 2.35 | μ0 ≤ 750 ) = 1 − 0.9906 = 0.0094.
b. If H0 is true, then the sample is in the most extreme 0.94% of its distribution when
reject H0.
[Optional] 6.2.1 For which P-value is the null hypothesis more plausible: P = 0.5 or
P = 0.05?
Sln: P = 0.5. Because larger P-value will have less chance that our test statistics fall into
the critical region, thus, the null hypothesis is more plausible.
[Optional] 6.2.2 True or False:
a.
b.
c.
d.
If we reject H0, then we conclude that H0 is false.
If we do not reject H0, then we conclude that H0 is true.
If we reject H0, then we conclude that H1 is true.
If we do not reject H0, then we conclude that H1 is false.
[Optional] 6.2.3 If P = 0.01, which is the best conclusion?
i. H0 is definitely false.
ii. H0 is definitely true.
iii. There is a 1% probability that H0 is true.
iv. H0 might be true, but it’s unlikely.
v. H0 might be false, but it’s unlikely.
vi. H0 is plausible.
Sln: (iv) is the best conclusion.
1 TRUE
FALSE
TRUE
FALSE
Solutions
Stat652
Homework 3
[Optional] 6.2.5 True or false: If P = 0.02, then
a. The result is statistically significant at the 5% level.
b. The result is statistically significant at the 1% level.
c. The null hypothesis is rejected at the 5% level.
d. The null hypothesis is rejected at the 1% level.
TRUE
FALSE
TRUE
FALSE
6.10.1 Rivets are manufactured for a certain purpose. The length specification is
1.20~1.30 cm. It is thought that 90% of the rivets manufactured meet the specification,
while 5% are too short and 5% are too long. In a random sample of 1000 rivets, 860 met
th specification, 60 were too short, and 80 were too long. Can you conclude that the true
percentages differ from 90%, 5%, and 5%.
a. State the appropriate null hypothesis.
b. Compute the expected values under the null hypothesis.
c. Compute the value of the chi-square statistic.
d. Find the P-value. What do you conclude?
Sln:
a. Let p1 be the probability of randomly chosen sample meet the specification, p2 be
the probability that is is too short, and p3 be the probability that it is too long.
Therefore, H0: p1 = 0.9, p2 = 0.05, p3 = 0.05
b. E1 = np1 = 900, E2 = np2 = 50, E3 = np3 = 50
∑
c.
21.7778
, Thus,
2
d. For χ distribution with degree of freedom of 2, we calculate
p = Pr (T ≥ 21.7778 ) = 1 − Pr (T ≤ 21.7778 ) = 1.866 ×10−5
Therefore, we reject H0 for significant level 5%.
6.10.4 The article “Analysis of Time Headways on Urban Roads: Case Study from
Riyadh” (A. Al-Ghamdi, Journal of Transportation Engineering, 2001:289-294) presents
a model for the time elapsed between the arrival of consecutive vehicles on urban roads.
Following are 137 arrival times (in seconds) along with the values expected from a
theoretical model.
Time
0 2
2 4
4 6
6 8
8 10
10 12
12 18
18 22
>22
Observed
18
28
14
7
11
11
10
8
30
Expected
23
18
16
13
11
9
20
8
19
Sln:
∑
21.3785. For χ2 distribution with degree of freedom of 8, we
calculate
p = Pr (T ≥ 21.7778 ) = 1 − Pr (T ≤ 21.3785) = 6.21×10−3
Therefore, we reject H0 for significant level 5%.
2