Download Class 19 Exam 2 Scores summary and comments

Exam2
A learning experience….
Scores
•
•
•
•
Raw Scores went from 68 to 147
As percentage of total….40% to 86%
Scaled scores went from 60.5 to 100
Some still left to be graded…
90s
80s
70s
60s
TOTAL
8
23
23
5
59
Question by Question
Question
count
min
max
avg
s
avail
1
2
3
4
5
6 7
59 59 59 59 59 59 59
4
2
2
4
4 10 1
20 20 20 34 18 20 15
13.7 16.4 12.9 23.2 10.5 16.0 8.9
4.5 4.1 4.4 7.5 4.1 1.7 3.0
20 20 20 40 20 20 15
8
59
2
10
4.3
2.7
10
9
59
0
2
0.8
0.6
5
raw
59
68
147
106.7
17.3
170
scaled
59
60.5
100
79.8
8.7
100
Data for Q1 to Q3
Categorical
n=60
Numerical
Numerical
Categorical
Car
Class
Displacement Fuel Type Hwy MPG
1
Midsize
3.5
R
28
2
Midsize
3
R
26
3
Large
3
P
26
4
Large
3.5
P
25
.
.
.
.
.
.
.
.
.
.
58
Compact
6
P
20
59
Midsize
2.5
R
30
60
Midsize
2
R
32
Q1
• expect that the size of the car engine
(measured by displacement) would change
based on car class (compact, midsize, large)
• H0: MU(compact)=MU(mid)=MU(large)
• Ha: not all equal
• ANOVA single factor (3 samples)
• Unstack the data, excel data analysis
Q2
• expect to see a relationship between car class
and recommended fuel type
• Relationship between two categorical
variables (car class and fuel type)
• Chi-sq independence test
– 3x2 contingency table of counts…summing to 60
Compact
Large
Midsize
P
16
11
9
36
R
3
5
16
24
19
16
25
60
Q3. Fuel type and mpg
• expect that because premium gasoline is higher quality, cars for which it is
recommended will get higher gas mileage (on average) than cars for which
regular fuel is recommended
R got higher
• Ho: MU(prem) = MU(reg)
sample mean
• Ha: MU(prem) > MU(reg)
t-Test: Two-Sample Assuming Equal
Variances
• Unstack, T-test two sample
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
The wrong
P(T<=t) one-tail
t Critical one-tail
value
P(T<=t) two-tail
t Critical two-tail
• NOTE: We guessed the wrong tail.
•
Do not reject HO in favor of THIS Ha.
p
P
R
24.33333 27.70833
12.4 9.519928
36
24
11.2579
0
58
-3.81704
0.000165 0.999835
1.671553
0.000331
2.001717
The correct
p value
Q4a Aspirin and Heart Attack
• Relationship between two 0/1 variable.
• 2x2 contingency table from the facts in the
question (like lights and myopia).
• Chi-sq independence test for 2T alternative.
• Half the pvalue if you want a 1T alternative
(Paspirin < Pplacebo)
Aspirin
Placebo
Heart
Attack
104
189
293
No Heart Attack
10896
10811
21707
11000
11000
22000
Q4b. How many heart attacks using
new design (given Ps)
• It is easy to calculate the mean (most likely) of
250.5.
• Tell me that the actual number is a random
variable
• Provide a probability distribution for that random
variable
𝑛 ∗ 𝑝 ∗ (1 − 𝑝)
Group
aspirin
placebo
TOTAL
Number
16500
5500
22000
Probability
0.009454545
0.017181818
Number of Heart Attacks
mean
variance std dev
156
154.53
12.4
94.5
92.88
9.6
250.5
247.40
15.7
Normal
approx to
binomial
Q4c. Will new design affect p-value?
• Yes. We will be more certain about Aspirin’s
effect and LESS certain about Placebo’s effect.
• The test is focused on the difference.
• The gain in accuracy for aspirin is not as great as
the loss in accuracy for placebo (diminishing
returns)
• Our test will be less powerful.
• P-value will go up.
• 50/50 v 75/25 v 100/0
Best design
Worst design
Q5. Is Di significantly better than El?
• Not about whether P=0.5
• About whether P(di)=P(el)
• 2x2 chi-squared
independence test
2 tailed pvalue
1 tailed pvalue
Di
El
Expected
In
10
5
15
Out
10
17
27
7.1
7.9
12.9
14.1
Distances 1.142857 0.634921
1.038961 0.577201
calculated chisuared
Pvalue
3.393939
0.065436
Pvlaue/2
0.032718
20
22
42
Q6. Rportfolio
• Rportfolio = (R1+R2+R3)/3
R1, R2, R3
Will not be
Independent.
Stock
Mean Return
Variance
Standard Deviation
1
0.1
0.01
0.1
2
0.05
0.0016
0.04
3
0.2
0.16
0.4
TOTAL
0.35
0.1716
0.414
Rportfolio
0.117
0.019066667
0.138
Sum of
variances
(independent)
.414/3
Q7. Total (Avg) weight of n=20
• Mean = 20*μ
• Variance = 20*σ2
• Normal (sum of normals)
One guest
Total of 20
guests
Mean
150
3000
Family hotel means…..
Weights in elevator not
independent.
More likely to be under 3500.
Variance Std Dev
1600
40
32000
178.8854
Pr(total<3500) =
NORMDIST(3500,3000,178.9,true)
= 0.9974
Q8. Al and Bo
• Neither knows σ
• Both get the same
𝑋−𝜇
𝑠
• Al uses t.dist, Bo uses normdist
• The t correctly reflects extra
uncertainty…giving Al a higher p-value
• Bo’s cheating is rewarded with a lower pvalue.
Q9
• If students don’t cheat, then their IQs are
independent identically distributed N(100,15)
• The null hypothesis (mean men = mean women)
IS TRUE!!!
• When H0 is true, and we do any test correctly, we
reject with probability 0.05.
• We will reject H0 with probability 0.05 and fail to
reject with probability 0.95
• What will happen under H0 is “easy”
• What will happen under Ha is very difficult…