1. Ch. 10 Comparing Two Populations or Groups
Ch Comparing Two Means

2. 𝑥 1 =123.8
less decay
𝑥 2 =116.4
more decay
123.8
116.4
7.4
Maybe, but it could be that we got this difference of means purely by chance. We need to perform a significance test.

3. 𝐻 0

5. = # 𝑜𝑓 𝑑𝑜𝑡𝑠 ≥ 7.4 𝑡𝑜𝑡𝑎𝑙 # 𝑜𝑓 𝑑𝑜𝑡𝑠 =
statistically significant?
𝑝-value
Assuming there’s no difference in decay, there is a ____________ probability of obtaining a 𝑥 1 − 𝑥 2 value of 7.4 or more purely by chance.
This provides _____________ evidence, so we __________ conclude that the length of time in the ground does affect the breaking strength of the polyester specimen.

6. 𝑥 1 𝜇 1 𝑡 𝑑𝑓 N 𝜇 1 , 𝜎 𝑥 1 𝜇 1 𝜎 𝑥 1 = 𝜎 1 𝑛 1 𝑠 1 𝑛 1 𝜇 1 𝑧 dist
𝑥 1
𝜇 1
𝑡 𝑑𝑓
N 𝜇 1 , 𝜎 𝑥 1
𝜇 1
𝜎 𝑥 1 =
𝜎 𝑛 1
𝑠 𝑛 1
𝜇 1 OR
𝑧 dist
𝑡 dist
𝑥 2
𝜇 2
𝑡 𝑑𝑓
N 𝜇 2 , 𝜎 𝑥 2
𝜇 2
𝜎 𝑥 2 =
𝜎 𝑛 2
𝑠 𝑛 2 OR
𝜇 2
𝑧 dist
𝑡 dist

7. 𝑥 1 − 𝑥 2 𝜇 1 − 𝜇 2 𝑡 𝑐𝑟𝑎𝑧𝑦 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝜇 1 − 𝜇 2
𝑥 1 − 𝑥 2
𝜇 1 − 𝜇 2
𝑡 𝑐𝑟𝑎𝑧𝑦 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝜇 1 − 𝜇 2
N 𝜇 1 − 𝜇 2 , 𝜎 𝑥 1 − 𝑥 2
𝜎 𝑥 1 − 𝑥 2 =
𝜎 𝑛 𝜎 𝑛 2
𝜇 1 − 𝜇 2
𝑧 dist OR
𝑠 𝑛 𝑠 𝑛 2
𝑡 dist

8. 𝑥 1 − 𝑥 2 𝜇 1 − 𝜇 2 𝑡 𝑐𝑟𝑎𝑧𝑦 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝜇 1 − 𝜇 2
𝑥 1 − 𝑥 2
𝜇 1 − 𝜇 2
𝑡 𝑐𝑟𝑎𝑧𝑦 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝜇 1 − 𝜇 2
N 𝜇 1 − 𝜇 2 , 𝜎 𝑥 1 − 𝑥 2
𝜎 𝑥 1 − 𝑥 2 =
𝜎 𝑛 𝜎 𝑛 2
𝜇 1 − 𝜇 2
𝑧 dist OR
𝑠 𝑛 𝑠 𝑛 2
𝑡 dist

9. 𝑥 1 =96
𝑛 1 =38
𝑠 1 =3.5
𝑥 2 =94
𝑛 2 =40
𝑠 2 =2.5
𝜇 1 → true mean lap time for old tires (in seconds)
𝜇 2 → true mean lap time for new tires (in seconds)
We want to estimate the true difference 𝜇 1 − 𝜇 2 at a 95% confidence level.

10. Two-sample 𝑡 interval for 𝜇 1 − 𝜇 2
Random:
“SRS of 38 laps” and “SRS of 40 laps”
Normal:
𝑛 1 ≥30 and 𝑛 2 ≥30, so by CLT the sampling distribution of 𝑥 1 − 𝑥 2 is approximately normal.
Independent:
Two things to check:
The two samples need to be independent of each other.
Individual observations in each sample have to be independent. When sampling without replacement for both samples, must check 10% condition for both.
We must assume the old lap times are independent from new lap times.
We must also assume that Mr. Hussey has driven more than =380 laps with the old tires and =400 laps with the new tires.

11. 𝑥 1 =96
𝑛 1 =38
𝑠 1 =3.5
𝑥 2 =94
𝑛 2 =40
𝑠 2 =2.5
Estimate ± Margin of Error
𝑠 𝑛 𝑠 𝑛 2
𝑥 1 − 𝑥 2 ± 𝑡 ∗
Standard Error
Calculating degrees of freedom for two sample means:
Option 1: Use calculator (2-SampTInt).
Actual 𝑑𝑓 formula is:
Option 2: (conservative approach)
Use the smaller 𝑑𝑓 of 𝑛 1 −1 and 𝑛 2 −1.
𝑑𝑓= 𝑠 𝑛 𝑠 𝑛 𝑛 1 −1 𝑠 𝑛 𝑛 2 −1 𝑠 𝑛

12. 𝑥 1 =96
𝑛 1 =38
𝑠 1 =3.5
𝑥 2 =94
𝑛 2 =40
𝑠 2 =2.5
Estimate ± Margin of Error
𝑠 𝑛 𝑠 𝑛 2
𝑥 1 − 𝑥 2 ± 𝑡 ∗
smaller 𝑑𝑓
Option 1
Option 2
With calculator:
𝑑𝑓= 𝑛 1 −1=37
STAT  TESTS  2-SampTInt (0)
𝑡 ∗ =invT .975, 37 =2.026
or use 𝑡-table 𝑑𝑓=30, 𝑡 ∗ =2.042
Inpt: Data Stats
x1:
Sx1:
n1:
x2:
Sx2:
n2:
C-Level:
Pooled: No Yes 96
3.5
0.619, 3.381 38
96−94± 94
𝑑𝑓=66.695
2.5
2±1.402 40
0.95
0.598, 3.402
ALWAYS do no pooled for 𝑡 procedures

13. We are 95% confident that the interval from 0. 619 to 3
We are 95% confident that the interval from to captures the true difference in means 𝜇 1 − 𝜇 2 of mean lap times between the old and new tires.
This suggests that the mean lap time with the old tires is between and seconds longer than the mean lap time with the new tires.
Yes, because 0 is NOT captured in the interval.

14. two-sample
No, this is not a matched pairs design.
Use 1 Var-Stats on each list to find 𝑥 and 𝑠 𝑥 .
𝑥 1 =23.2
𝑛 1 =10
𝑠 1 =3.52
𝑥 2 =22.125
𝑛 2 =8
𝑠 2 =3.76
𝜇 1 → true mean ACT score for students who take “smart pill”
𝜇 2 → true mean ACT score for students who take placebo
We want to estimate the true difference 𝜇 1 − 𝜇 2 at a 90% confidence level.

15. Two-sample 𝑡 interval for 𝜇 1 − 𝜇 2
Random:
“10 student chosen at random are given the ‘smart pill’” and “8 students chosen at random are given a placebo”
Normal:
Both population distributions are normal so the sampling distribution of 𝑥 1 − 𝑥 2 is approximately normal.
Independent:
Two things to check:
The two groups need to be independent of each other.
Individual observations in each group have to be independent. When sampling without replacement for both samples, must check 10% condition for both.
Due to random assignment, these two groups can be viewed as independent.
No 10% condition since there was no sampling.
Individual observations in each group should also be independent: knowing one subject’s ACT score gives no information about another subject’s ACT score.

16. Estimate ± Margin of Error
𝑥 1 =23.2
𝑛 1 =10
𝑠 1 =3.52
𝑥 2 =22.125
𝑛 2 =8
𝑠 2 =3.76
Estimate ± Margin of Error
𝑠 𝑛 𝑠 𝑛 2
𝑥 1 − 𝑥 2 ± 𝑡 ∗
smaller 𝑑𝑓
Option 1
Option 2
We can use the “data” option now since our data is in lists.
𝑑𝑓=8−1=7
𝑡 ∗ =invT .95, 7 =1.895
or use 𝑡-table 𝑑𝑓=7, 𝑡 ∗ =1.895
With calculator:
STAT  TESTS  2-SampTInt (0)
Inpt: Data Stats
List1:
List2:
Freq1:
Freq2:
C-Level:
Pooled: No Yes
𝐿 1
23.2−22.125±
𝐿 2
−1.969, 1
1.075±3.286 1
𝑑𝑓=14.66
0.90
−2.211, 4.361
ALWAYS do no pooled for 𝑡 procedures

17. Answer key will use conservative approach for degrees of freedom.
We are 90% confident that the interval from −1.969 to captures the true difference in means 𝜇 1 − 𝜇 2 for ACT scores between kids given a smart pill and kids given a placebo.
This suggests that the “smart pill” may improve scores by up points or the placebo may improve scores by up points.
−0.756, −0.564
Answer key will use conservative approach for degrees of freedom.

18. 𝜇 1 → true mean ACT score for students who take “smart pill”
one sided
𝑥 1 =23.2
𝑛 1 =10
𝑠 1 =3.52
𝑥 2 =22.125
𝑛 2 =8
𝑠 2 =3.76
State:
𝜇 1 → true mean ACT score for students who take “smart pill”
𝜇 2 → true mean ACT score for students who take placebo
𝐻 0 :
𝐻 𝑎 :
𝜇 1 − 𝜇 2 =0
𝜇 1 = 𝜇 2
𝑥 1 − 𝑥 2 =23.2−22.125=1.075 OR
𝜇 1 − 𝜇 2 >0
𝜇 1 > 𝜇 2
𝛼=0.05 OR

19. Plan:
Two sample 𝑡 test for 𝜇 1 − 𝜇 2
Random:
“10 student chosen at random are given the ‘smart pill’” and “8 students chosen at random are given a placebo”
Normal:
Both population distributions are normal so the sampling distribution of 𝑥 1 − 𝑥 2 is approximately normal.
Independent:
Two things to check:
The two groups need to be independent of each other.
Individual observations in each group have to be independent. When sampling without replacement for both samples, must check 10% condition for both.
Due to random assignment, these two groups can be viewed as independent.
No 10% condition since there was no sampling.
Individual observations in each group should also be independent: knowing one subject’s ACT score gives no information about another subject’s ACT score.

20. Do:
𝑥 1 =23.2
𝑛 1 =10
𝑠 1 =3.52
𝑥 2 =22.125
𝑛 2 =8
𝑠 2 =3.76
Sampling Distribution of 𝑥 1 − 𝑥 2
conservative approach:
N 0, ________
1.734
𝑑𝑓=8−1=7
smaller 𝑑𝑓
1.075
𝜎 𝑥 1 − 𝑥 2 = 𝑠 𝑛 𝑠 𝑛 2 = =1.734
𝑡= 𝑥 1 − 𝑥 2 − 𝜇 1 − 𝜇 𝜎 𝑥 1 − 𝑥 2
= 23.2− − =0.62
𝑡cdf 0.62, 99999, 7 =0.277
𝑝-value

21. Conclude:
Assuming 𝐻 0 is true 𝜇 1 = 𝜇 2 , there is a .277 probability of obtaining
a 𝑥 1 − 𝑥 2 value of or more purely by chance.
This provides
weak evidence against 𝐻 0 and is not statistically significant at 𝛼=.05
level .277>.05 .
Therefore, we fail to reject 𝐻 0 and cannot conclude
that the smart pill increases ACT scores.
With calculator:
𝑡=0.62
𝑝=0.272
𝑑𝑓=14.66
𝑥 1 =23.2
𝑥 2 =22.125
𝑆 𝑥 1 =3.52
𝑆 𝑥 2 =3.76
𝑛 1 =10
𝑛 2 =8
STAT  TESTS  2-SampTTest (4)
𝑝-value
Inpt: Data Stats
List1:
List2:
Freq1:
Freq2:
𝜇: ≠𝜇 <𝜇 >𝜇
Pooled: No Yes
𝐿 1
𝐿 2 1 1
ALWAYS do no pooled for 𝑡 procedures

22. What population can we target?
experiment
Yes
Not a random sample, so conclusion only holds for the people in the experiment
What population can we target?

23. matched pairs
Yes, since it’s a matched pairs design.
Very important to know when you use a matched pairs 𝑡 test for 𝜇 𝑑 or a two sample 𝑡 test for 𝜇 1 − 𝜇 2
If groups were formed using a completely randomized design:
⇒ two-sample 𝑡 test for 𝜇 1 − 𝜇 2
If subjects were paired, then split at random into two treatments or if each subject receives both treatments:
⇒ matched pairs 𝑡 test for 𝜇 𝑑

24. We will do (smart – placebo) so that + values represent pill worked
Differences 2 9 −3 −2 10 6 5 4 12
Use 1 Var-Stats
𝑥 𝑑 =4.3
𝑠 𝑑 =5.10
State:
𝐻 0 :
𝐻 𝑎 :
𝜇 𝑑 =0
𝜇 𝑑 → true mean difference (smart pill – placebo)
in test scores
𝜇 𝑑 >0
𝛼=0.05
Plan:
Matched pairs T test for 𝜇 𝑑
Random:
Treatments were randomized
Normal:
Population distribution is normal so the sampling distribution of 𝑥 𝑑 is approximately normal.
Independent:
There are more than =100 students at ECRCHS.

25. Do:
𝑥 𝑑 = 𝑠 𝑑 =5.10
Sampling Distribution of 𝑥 𝑑
𝜎 𝑥 𝑑 = 𝑠 𝑑 𝑛 = =1.613
N 0, ________
1.613
𝑑𝑓=𝑛−1=10−1=9
4.3
𝑡= 𝑥 𝑑 − 𝜇 𝑑 𝜎 𝑥 𝑑
= 4.3− =2.67
area = tcdf 2.67, 99999, 9 =.013
𝑝-value

26. Conclude:
Assuming 𝐻 0 is true 𝜇 𝑑 =0 , there is a .013 probability of obtaining a
𝑥 𝑑 value of 4.3 or higher purely by chance.
This provides strong
evidence against 𝐻 0 and is statistically significant at 𝛼=0.05 level
(.013<.05).
Therefore, we reject 𝐻 0 and can conclude that the smart
pills work.
experiment
Yes
All ECRCHS students

27. Two sample 𝑧 test for 𝑝 1 − 𝑝 2
Matched pairs 𝑡 test for 𝜇 𝑑
more
less
Equal