HD EDU
HD EDU
ECON 7300
Project 相关知识点
TUTOR: Arya
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Project 作业要求
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Project 作业提示（重要）
Project- Part I 改编作业题目+知识点+答题
要点
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Project- Part I 知识点
目标：the effect of treatment on Y (例如 不同学历对薪酬的影响)
1. CRD – One Way ANOVA
o Treatment test
▪ Hypothesis statement
▪ Test statistics ()
▪ Critical values ()
▪ Decision & Conclusion
o Tukey-Kramer procedure (基于前面 treatment test 的 Decision 的结论
来判断是否需要做 T-K)
2. RBD- Randomized Block Design
o Block test
▪ Hypothesis statement
▪ Test statistics ()
▪ Critical values ()
▪ Decision & Conclusion
o Treatment test
▪ Hypothesis statement
▪ Test statistics ()
▪ Critical values ()
▪ Decision & Conclusion
o Tukey Multiple Comparisons (基于前面 treatment test 的 Decision 的
结论来判断是否需要做 T-M)
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Project- Part I – Part A 知识点
目标：the effect of treatment on Y (例如 不同学历对薪酬的影响)
• CRD – One Way ANOVA （例题 Tutorial 6_Q14）
o Treatment test
▪ Hypothesis statement
▪ Test statistics ()
=
− 1⁄
− ⁄
=
▪ Critical values ()
= ,(−1),(−)
▪ Decision & Conclusion
• Decision: refer to H0 ( reject H0 or Don’t reject H0)
• Conslusion: refer to H1.
Reject 0, then we say there is sufficient evidence at certain level of significance
to conclude that not all population means are equal (1).
Don’t Reject 0, then we say there is insufficient evidence at certain level of
significance to conclude that not all population means are equal (1).
到目前为止，通过这个 treatment test, 我们得出的结论是：不是所有的 population means
are equal, 但我们是否知道 到底是 哪个 ppl mean 和 哪个 ppl mean 之前不相等呢？
A：知道
B： 不知道
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o Tukey-Kramer procedure (基于前面 treatment test 的 Decision 的结论来判断是
否需要做 T-K)
• Don’t Reject H0 => inappropriate to apply T-K procedure
• Reject H0 => Apply T-K procedure
• Step 1: Calculate Qu:
For c degrees of freedom in the numerator and n-c degrees of freedom in the denominator
• Step2: Calculated # of pairwise comparisons:
c(c-1)/2 pairwise comparisons
• Step3: 算 CR 和 |̅ − ̅′|
Check if the sample size are equal
 how many CR that you need => calculate the absolute mean difference
• Step4: 比较 CR 和 |̅ − ̅′|
Project- Part I – Part A 例题
1. Using Sample 1 dataset test at a 5% level of significance if there is any evidence
of a significant difference in the mean commuting time (in minutes) from home
to UQ Campus for students who live in three areas; St. Lucia coded 1, South
Bank coded 2, and Garden City coded 3.
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2. If your results in (1) indicate that it is appropriate, use the Tukey-Kramer
procedure to determine which groups of students differ in mean commuting
time from their home to UQ. Use a 5% level of significance.
具体步骤：
1. 先看一下数据-excel 表
2. 在 excel表中把数据的样子变成类似 Tutorial6中数据的样子
3. Phstat-Multiple-Sample Test – One way ANOVA
4. 写出完整的 ANOVA hypothesis test 步骤，参考 Tutorial6 的
solution; （写的要简明扼要在点上！多说多错）
5. 看看是否需要做 Tukey-Kramer procedure，参考 Tutorial6
的 solution 做就好
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Q1：
• 0: 1 = 2 = 3
• 1:
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=
=
154.1818
122.2496
= 1.2607
,−1,− = 0.05,2,68 ≈ 3.15 ( 0.05,2,60)；
因为 || < || 所以我们不拒绝 0
There is insufficient evidence at 5% level of significance to conclude H1
Q2：Solutions
Do we still need to apply the Tukey-Kramer Test if we didn‘t reject the null hypothesis?
以下内容是假设，为了讲 Turkey Kramer Procedure。
假如
F test statistic = 5.89 > F critical value
We reject the H0 and there is sufficient evidence at 5% level of significance to conclude
that not all population means are equal.
这时候，我们需要做 Turkey Kramer Test to figure out which group is different from
the rest of groups.
Q = ,,− = 0.05,3,68 ≈ 3.4 ( 0.05,3,60)
3 × 2/2 = 3
Group size are different.
• = 19 ′ = 24 (group 1 and group 2)
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= √
2
(
1

+
1
′
) = 3.4√
122.2946
2
(
1
19
+
1
24
) = 8.1643
|̅1 − ̅2| = |31.0526 − 29.9167| = 1.1260 < 8.1643
∴ ℎ ℎ ℎ 1 2 .
• = 19 ′ = 28 (group 1 and group 3)
= √

2
(
1
+
1
′
) = 3.4√
122.2946
2
(
1
19
+
1
28
) = 7.9023
|̅1 − ̅3| = |31.0526 − 26.25| = 4.8026 < 7.9023
∴ ℎ ℎ ℎ 1 3 .
• = 24 ′ = 28 (group 2 and group 3)
= √

2
(
1
+
1
′
) = 3.4√
122.2946
2
(
1
24
+
1
28
) = 7.3963
|̅2 − ̅3| = |29.9167 − 26.25| = 3.6667 < 7.3963
∴ ℎ ℎ ℎ 2 3 .
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Project- Part I – Part B 知识点
目标：the effect of treatment on Y (例如 不同学历对薪酬的影响)
1. RBD- Randomized Block Design
o Block test
▪ Hypothesis statement
0: 1 = 2 = ⋯ = = 0 (There is no block effects)
1: (There is block effects)
▪ Test statistics ()
=
− 1⁄
( − 1)( − 1)⁄
=
▪ Critical values ()
= ,(−1),(−1)(−1)
▪ Decision & Conclusion
Reject 0, then we say there is sufficient evidence at certain level of significance
to conclude that no block effects (1).
Don’t Reject 0, then we say there is insufficient evidence at certain level of
significance to conclude that no block effects (1).
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o Treatment test
▪ Hypothesis statement
0: 1 = 2 = ⋯ = = 0 (There is no treatment effects)
1: (There is treatment effects)
▪ Test statistics ()
=
( − 1)⁄
( − 1)( − 1)⁄
=
▪ Critical values ()
= ,(−1),(−1)(−1)
▪ Decision & Conclusion
Reject 0, then we say there is sufficient evidence at certain level of significance
to conclude that not all population means are equal (1).
Don’t Reject 0, then we say there is insufficient evidence at certain level of
significance to conclude that not all population means are equal (1).
到目前为止，通过这个 treatment test, 我们得出的结论是：不是所有的 population means
are equal, 但我们是否知道 到底是 哪个 ppl mean 和 哪个 ppl mean 之前不相等呢？
A：知道
B： 不知道
o Tukey Multiple Comparisons (基于前面 treatment test 的 Decision 的
结论来判断是否需要做 T-M)
▪ Don’t Reject H0 => inappropriate to apply T-K procedure
▪ Reject H0 => Apply T-K procedure
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Project- Part I – Part B 例题
1. An Education company has developed four new Programming products and
conducted a pivot test to collect data on Students’ preferences. Five Students were asked
to rate each products on a scale 1-40. The variable Student is the blocking variable.
(a) At the 5% level of significance, was blocking effective?
(b) Using a 5% level of significance, is there a significance difference in the mean rating of
the four Products?
(c) If your results in (b) indicates that it is appropriate, use the Tukey procedure to
determine which groups differ in the mean rating.
具体步骤：
1. 先看一下数据-excel 表
2. 在 excel表中把数据的样子变成类似 Tutorial6的样子
3. Data-Data Analysis-ANOVA-Two Factor Without Replication
(Row – Between Blocks ; Columns – Between Groups)
4. 写出完整的 hypothesis test 步骤，按照 Tutorial6 的 solution。
（写的要简明扼要在点上！多说多错）
5. 看是否需要做 Tukey Procedure，按 Tutorial6 的 solution 做就好
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a)
• 0: 1 = 2 = 3 = 4 = 5
• 1:
=
=
145.625
36.8917
= 3.9474
,−1,(−1)(−1) = 0.05,4,12 = 3.26；
因为 || > ||
所以拒绝ℎ 0
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There is sufficient evidence at 5% level of significance to conclude H1
b)
• 0: 1 = 2 = 3 = 4
• 1:
=
=
143.2667
36.8917
= 3.8834
,−1,(−1)(−1) = 0.05,3,12 = 3.49；
因为 || > ||
所以 ℎ 0
There is sufficient evidence at 5% level of significance to conclude that H1
c)
⸪ we reject the H0 in Question-(b), therefore, we perform the Tukey Procedure.
Q = ,,(−1)(−1) = 0.05,4,12 = 4.2
= √
= 4.2√
36.8917
5
= 11.4085
|̅1 − ̅2| = |28.8 − 33.8| = 5 < 11.4085
|̅1 − ̅3| = |28.8 − 22.2| = 6.6 < 11.4085
|̅1 − ̅4| = |28.8 − 33.2| = 4.4 < 11.4085
|̅2 − ̅3| = |33.8 − 22.2| = 11.6 > 11.4085
|̅2 − ̅4| = |33.8 − 33.2| = 0.6 < 11.4085
|̅3 − ̅4| = |22.2 − 33.2| = 11 < 11.4085
Except for the absolute difference in the sample means between program 2 and
program3 (which is 11.6), the rest absolute mean differences are smaller than the critical
range (11.4085). Thus, the population mean is significantly different between program
2 and program 3