Farmer Myrtle Field grew two types of pumpkins to see which type yielded bigger pumpkins. She grew 30 of the first type and she grew 31 of the second type. If Ms. Field wished to test whether or not the two pumpkin types yielded, on average, pumpkins of the same weight then the appropriate null hypothesis would be:
A. Mean weight of pumpkin type one = mean weight of pumpkin type two
B. Mean weight of pumpkin type one > mean weight of pumpkin type two
C. Mean weight of pumpkin type one < mean weight of pumpkin type two
D. Mean weight of pumpkin type one ≠ mean weight of pumpkin type two
E. None of the above

Farmer Myrtle Field grew two types of pumpkins to see which type yielded bigger pumpkins. She grew 30 of the first type and she grew
31 of the second type. If Ms. Field wished to test whether or not the two pumpkin types yielded, on average, pumpkins of the same weight then the appropriate alternative hypothesis would be:
A. Mean weight of pumpkin type one = mean weight of pumpkin type two
B. Mean weight of pumpkin type one > mean weight of pumpkin type two
C. Mean weight of pumpkin type one < mean weight of pumpkin type two
D. Mean weight of pumpkin type one ≠ mean weight of pumpkin type two
E. None of the above

Farmer Myrtle Field grew two types of pumpkins to see which type yielded bigger pumpkins. She grew 30 of the first type and she grew 31 of the second type. Ms. Field wished to test whether or not the two pumpkin types yielded, on average, pumpkins of the same weight. The appropriate hypotheses can be stated as:
𝐻 𝑜
: 𝜇
1
𝐻 𝑎
: 𝜇
1
= 𝜇
2
≠ 𝜇
2

The appropriate hypothesis test is called a twosample t test. There are three types of alternative hypotheses just as before:
𝐻 𝑜
: 𝜇
1
𝐻 𝑎
: 𝜇
1
= 𝜇
2
≠ 𝜇
2
𝐻 𝑜
: 𝜇
1
𝐻 𝑎
: 𝜇
1
= 𝜇
2
< 𝜇
2
𝐻 𝑜
: 𝜇
1
𝐻 𝑎
: 𝜇
1
= 𝜇
2
> 𝜇
2

There are two types of two-sample t tests:
• Pooled or equal variances t test
• Non-pooled or unequal variances t test

The pooled or equal variances t test assumes the standard deviations of the two populations being compared are equal:
1
2
If we don’t know all the values in each population then it would be rather unlikely we would know the standard deviation of each population. This implies we would have no idea if the standard deviations are equal.

The non-pooled or unequal variances t test assumes the standard deviations of the two populations being compared are not equal:
1
2
This is a much more reasonable assumption.

In general, I recommend practitioners choose the non-pooled or unequal variances twosample t test unless, for some unusual reason, the practitioner believes the two population standard deviations are equal.
Take a look at formulas in book

Farmer Myrtle Field grew two types of pumpkins to see which type yielded bigger pumpkins. She grew 30 of the first type and 𝑥 = 110 pounds with 𝑠 = 13. She grew 31 of the second type and 𝑥 = 104 pounds with 𝑠 = 12 pounds. Since the sample means are 110 lbs and 104 lbs, Farmer
Field correctly concluded the first type produces bigger pumpkins
A. True
B. False

Farmer Myrtle Field grew two types of pumpkins to see which type yielded bigger pumpkins. She grew 30 of the first type and 𝑥 = 110 pounds with 𝑠 = 13. She grew 31 of the second type and 𝑥 = 104 pounds with 𝑠 = 12 pounds. Conduct a non-pooled two-sample t test. The p-value is?
A. 0.0332
B. 0.0613
C. 0.0659
D. 0.0663

Farmer Myrtle Field grew two types of pumpkins to see which type yielded bigger pumpkins. She grew
30 of the first type and 𝑥 = 110 pounds with 𝑠 = 13. She grew 31 of the second type and 𝑥 =
104 pounds with 𝑠 = 12 pounds. The p-value is
0.0663. If 𝛼 = 0.05
then Myrtle concludes:
A. Mean weight of pumpkin type 1 = mean weight of pumpkin type 2
B. Mean weight of pumpkin type 1 > mean weight of pumpkin type 2
C.
Mean weight of pumpkin type 1 < mean weight of pumpkin type 2
D. Mean weight of pumpkin type 1 ≠ mean weight of pumpkin type 2

Researchers are interested in comparing the mean age of buyers of new domestic cars to the mean age of buyers of new imported cars.
Suppose that they want to perform a hypothesis test to decide whether the mean age of buyers of new domestic cars ( 𝜇
1
) is greater than the mean age of buyers of new imported cars ( 𝜇
2
) . The correct hypotheses are:
A.
𝐻 𝑜
B.
𝐻 𝑜
: 𝜇
1
: 𝜇
1
C.
𝐻 𝑜
: 𝜇
1
D.
𝐻 𝑜
E.
𝐻 𝑜
: 𝜇
1
: 𝜇
1
= 𝜇
2
> 𝜇
2
= 𝜇
2
= 𝜇
2
< 𝜇
2
𝐻 𝑎
𝐻 𝑎
: 𝜇
1
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
𝐻 𝑎
: 𝜇
1
: 𝜇
1
≠ 𝜇
2
= 𝜇
2
> 𝜇
2
< 𝜇
2
≠ 𝜇
2

Researchers studied two subspecies of dark-eyed juncos.
One of the subspecies migrates each year and the other does not migrate. A hypothesis test is to be performed to decide whether the mean wing lengths for the two subspecies are different. The correct hypotheses are:
A.
𝐻 𝑜
: 𝜇
1
B.
𝐻 𝑜
: 𝜇
1
C.
𝐻 𝑜
: 𝜇
1
D.
𝐻 𝑜
: 𝜇
1
E.
𝐻 𝑜
: 𝜇
1
= 𝜇
2
> 𝜇
2
= 𝜇
2
= 𝜇
2
< 𝜇
2
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
≠ 𝜇
2
= 𝜇
2
> 𝜇
2
< 𝜇
2
≠ 𝜇
2

An issue of USA Today discussed the amounts spent by teens and adults at shopping malls. Suppose that we want to perform a hypothesis test to decide whether the mean amount spent by teens ( 𝜇
1
) is less than the mean amount spent by adults ( 𝜇
2
). The correct hypotheses are:
A.
𝐻 𝑜
: 𝜇
1
B.
𝐻 𝑜
: 𝜇
1
C.
𝐻 𝑜
: 𝜇
1
D.
𝐻 𝑜
: 𝜇
1
E.
𝐻 𝑜
: 𝜇
1
= 𝜇
2
> 𝜇
2
= 𝜇
2
= 𝜇
2
< 𝜇
2
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
≠ 𝜇
2
= 𝜇
2
> 𝜇
2
< 𝜇
2
≠ 𝜇
2

The U.S. Bureau of Prisons publishes data on the times served by prisoners released from federal institutions for the first time. Researchers want to see if there is evidence to conclude that the mean time served for fraud ( 𝜇
1
) is less than that for firearms offenses ( 𝜇
2
). The correct hypotheses are:
A.
𝐻 𝑜
: 𝜇
1
B.
𝐻 𝑜
: 𝜇
1
C.
𝐻 𝑜
: 𝜇
1
D.
𝐻 𝑜
E.
𝐻 𝑜
: 𝜇
1
: 𝜇
1
= 𝜇
2
> 𝜇
2
= 𝜇
2
= 𝜇
2
< 𝜇
2
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
𝐻 𝑎
: 𝜇
1
: 𝜇
1
≠ 𝜇
2
= 𝜇
2
> 𝜇
2
< 𝜇
2
≠ 𝜇
2

Researchers want to see if there is evidence to conclude that the mean time served for fraud ( 𝜇
1
) is less than that for firearms offenses ( 𝜇
2
): 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
< 𝜇
2
. The sample information is:
2
= 4.64
, 𝑛
1 𝑥
1
= 10 , and 𝑛
= 10.12
2
,
1
= 4.90
, 𝑥
2
= 18.78
,
= 10. Conduct the appropriate test. The p-value is:
A. 0.000371
B. 0.000369
C. 0.000742
D. 0.000738
E. None of the above

Researchers want to see if there is evidence to conclude that the mean time served for fraud ( 𝜇
1
) is less than that for firearms offenses ( 𝜇
2
): 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
< 𝜇
2
. The sample information is:
2
= 4.64
, 𝑛
1 𝑥
1
= 10 , and 𝑛
= 10.12
2
,
1
= 4.90
, 𝑥
2
= 18.78
,
= 10. Compute the appropriate test. The p-value is 0.000371. The correct decision is:
A. Mean time served for fraud is greater than that for firearms
B. Mean time served for fraud is less than that for firearms
C. Mean time served for fraud is the same as that for firearms

Sex and Direction – Problem in book

Intervention program implemented to improve work conditions of bus drivers in Stockholm. Intervention group of 10 drivers drove improved route (intervention group) and 31 drivers drove normal route (control group). Drivers were assigned to each group randomly.
Heart rates of drivers during work were recorded. Researchers want to see if there is evidence to conclude that the mean heart rate for intervention group ( 𝜇
1
) is less than that for control group ( 𝜇
2
). The correct hypotheses are:
A.
𝐻 𝑜
: 𝜇
1
B.
𝐻 𝑜
: 𝜇
1
C.
𝐻 𝑜
: 𝜇
1
D.
𝐻 𝑜
: 𝜇
1
E.
𝐻 𝑜
: 𝜇
1
= 𝜇
2
> 𝜇
2
= 𝜇
2
= 𝜇
2
< 𝜇
2
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
≠ 𝜇
2
= 𝜇
2
> 𝜇
2
< 𝜇
2
≠ 𝜇
2

Researchers want to see if there is evidence to conclude that the mean heart rate for intervention group ( 𝜇
1
) is less than that for control group ( 𝜇
2
). 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
< 𝜇
2
.
The sample information is:
66.81
,
2
= 9.04, 𝑛
1
= 𝑥
1
10, and
= 67.90
𝑛
2
,
1
= 5.49
, 𝑥
= 31. Conduct the
2
= appropriate test.
Before we compute the p-value, do you notice anything about the data?

Researchers want to see if there is evidence to conclude that the mean heart rate for intervention group ( 𝜇
1
) is less than that for control group ( 𝜇
2
). 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
< 𝜇
2
.
The sample information is:
66.81
,
2
= 9.04, 𝑛
1
= 𝑥
1
10, and
= 67.90
𝑛
2
,
1
= 5.49
, 𝑥
= 31. Conduct the
2
= appropriate test. The p-value is:
A. 0.6748
B. 0.4586
C. 0.6504
D. 0.7218
E. 5

Researchers want to see if there is evidence to conclude that the mean heart rate for intervention group ( 𝜇
1
) is less than that for control group ( 𝜇
2
). 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
< 𝜇
2
.
The sample information is:
66.81
,
2
= 9.04, 𝑛
1
= 𝑥
1
10, and
= 67.90
𝑛
2
,
1
= 5.49
, 𝑥
= 31. Conduct the
2
= appropriate test. The p-value is 0.6748. The correct decision is:
A. The intervention program worked
B. The intervention program did not work

Assumption of the two-sample t hypothesis test in order for the decision to be valid:
• Random samples are independent of each other
• The random samples each came from a normal distribution OR each random sample size is 𝑛 ≥ 30

A researcher has two independent random samples with sample sizes 𝑛
1
= 10 and 𝑛
2
= 10 . Below are the normal probability plots for each sample. The t distribution assumption for a two-sample t test is met.
A. True
B. False

A researcher has two independent random samples with sample sizes 𝑛
1
= 31 and 𝑛
2
=
The normal probability plots for each sample
38. imply the samples are not from normal distributions. The t distribution assumption for a two-sample t test is met.
A. True
B. False

Thirty married couples are randomly selected. The amount of TV the men view and the amount of TV the women view in a week is recorded. Researchers want to test the hypothesis that the mean amount of
TV married men watch is less than the mean amount of TV married women watch. The t distribution assumption for a two-sample t test is met.
A. True
B. False

Thirty married couples are randomly selected.
The amount of TV the men view and the amount of TV the women view in a week is recorded.
Researchers want to test the hypothesis that the mean amount of TV married men watch is less than the mean amount of TV married women watch.
This is called paired data. A two-sample t test is not appropriate. What can we do?

Thirty married couples are randomly selected.
The amount of TV the men view and the amount of TV the women view in a week is recorded.
Researchers want to test the hypothesis that the mean amount of TV married men watch is less than the mean amount of TV married women watch.
𝐻 𝑜
: 𝜇 𝑚𝑒𝑛
𝐻 𝑎
: 𝜇 𝑚𝑒𝑛
= 𝜇 𝑤𝑜𝑚𝑒𝑛
< 𝜇 𝑤𝑜𝑚𝑒𝑛

Researchers want to test the hypothesis that the mean amount of TV married men watch is less than the mean amount of TV married women watch.
𝐻 𝑜
𝐻 𝑎
: 𝜇 𝑚𝑒𝑛
: 𝜇 𝑚𝑒𝑛
= 𝜇 𝑤𝑜𝑚𝑒𝑛
< 𝜇 𝑤𝑜𝑚𝑒𝑛
Examine the difference:
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝜇 𝑚𝑒𝑛
− 𝜇 𝑤𝑜𝑚𝑒𝑛

𝐻 𝑜
𝐻 𝑎
: 𝜇
: 𝜇 𝑚𝑒𝑛 𝑚𝑒𝑛
= 𝜇 𝑤𝑜𝑚𝑒𝑛
< 𝜇 𝑤𝑜𝑚𝑒𝑛
Examine the difference:
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝜇 𝑚𝑒𝑛
− 𝜇 𝑤𝑜𝑚𝑒𝑛
This implies the hypotheses can be written as:

𝐻 𝑜
𝐻 𝑎
: 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0
: 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 < 0
Conduct a one-sample t test on the differences. This is called a paired t test.

Two strains of tobacco leaf virus were studied by randomly selecting 8 tobacco leaves and putting virus
1 on half of each leaf and virus 2 on the other half of each leaf. Researchers then counted the number of lesions each virus caused. Researchers wanted to test the hypothesis that one virus produces more lesions, on average, than the other.
𝐻 𝑜
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
𝐻 𝑎
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
= 𝜇 𝑣𝑖𝑟𝑢𝑠 2
≠ 𝜇 𝑣𝑖𝑟𝑢𝑠 2
Where 𝜇 is the mean number of lesions produced for each virus.

Two strains of tobacco leaf virus were studied by randomly selecting 8 tobacco leaves and putting virus
1 on half of each leaf and virus 2 on the other half of each leaf. Researchers then counted the number of lesions each virus caused. Researchers wanted to test the hypothesis that one virus produces more lesions, on average, than the other. The data follow:
Leaf 1 2 3 4 5 6 7 8
Virus 1 31 20 18 17 9 8 10 7
Virus 2 18 17 14 11 10 7 5 6

Two strains of tobacco leaf virus were studied by randomly selecting 8 tobacco leaves and putting virus 1 on half of each leaf and virus 2 on the other half of each leaf. Researchers then counted the number of lesions each virus caused.
Now, compute the differences.
Leaf 1 2 3 4 5 6 7 8
Virus 1 31 20 18 17 9 8 10 7
Virus 2 18 17 14 11 10 7 5 6
Differences 13 3 4 6 -1 1 5 1

Conduct a one-sample t test on the differences:
𝐻 𝑜
𝐻 𝑎
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
= 𝜇 𝑣𝑖𝑟𝑢𝑠 2
< 𝜇 𝑣𝑖𝑟𝑢𝑠 2
Leaf 1 2 3 4 5 6 7 8
Virus 1 31 20 18 17 9 8 10 7
Virus 2 18 17 14 11 10 7 5 6
Differences 13 3 4 6 -1 1 5 1

Conduct a one-sample t test on the differences making sure to use correct null and alternative hypotheses with difference = 0 vs. difference < 0.
𝐻 𝑜
𝐻 𝑎
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
= 𝜇 𝑣𝑖𝑟𝑢𝑠 2
< 𝜇 𝑣𝑖𝑟𝑢𝑠 2
Differences 13 3 4 6 -1 1 5 1
P-value = 0.034

P-value = 0.034 so data supports alternative hypothesis which means virus 1 produces more lesions, on average, than virus 2.
𝐻 𝑜
𝐻 𝑎
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
: 𝜇 𝑣𝑖𝑟𝑢𝑠 1
= 𝜇 𝑣𝑖𝑟𝑢𝑠 2
< 𝜇 𝑣𝑖𝑟𝑢𝑠 2

Researchers measured the corneal thickness of eight patients who had glaucoma in one eye but not in the other. They want to test the hypothesis that mean corneal thickness is greater in normal eyes ( 𝜇
1
) than in eyes with glaucoma ( 𝜇
2
) . The correct null and alternative hypotheses are:
A.
𝐻 𝑜
: 𝜇
1
B.
𝐻 𝑜
: 𝜇
1
C.
𝐻 𝑜
: 𝜇
1
D.
𝐻 𝑜
: 𝜇
1
E.
𝐻 𝑜
: 𝜇
1
= 𝜇
2
> 𝜇
2
= 𝜇
2
= 𝜇
2
< 𝜇
2
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
𝐻 𝑎
: 𝜇
1
≠ 𝜇
2
= 𝜇
2
> 𝜇
2
< 𝜇
2
≠ 𝜇
2

Researchers measured the corneal thickness of eight patients who had glaucoma in one eye but not in the other. They want to test the hypothesis that mean corneal thickness is greater in normal eyes ( 𝜇
1
) than in eyes with glaucoma
( 𝜇
2
) . The correct null and alternative hypotheses are 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
> 𝜇
2
. The differences are as follows. Conduct the appropriate test and compute the p-value
-4 0 12 18 -4 -12 6 16

Researchers want to test the hypothesis that mean corneal thickness is greater in normal eyes
( 𝜇
1
) than in eyes with glaucoma ( 𝜇
2
) . The correct null and alternative hypotheses are 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
> 𝜇
2
. The differences are as follows. The p-value is?
-4 0 12 18 -4 -12 6 16
A. 0.312
B. 0.129
C. 0.164
D. 0.258

Researchers want to test the hypothesis that mean corneal thickness is greater in normal eyes
( 𝜇
1
) than in eyes with glaucoma ( 𝜇
2
) . The correct null and alternative hypotheses are 𝐻 𝑜
: 𝜇
1
= 𝜇
2
𝐻 𝑎
: 𝜇
1
> 𝜇
2
. The p-value is 0.164. The researchers should conclude:
A. Mean corneal thickness is greater in normal eyes than in eyes with glaucoma
B. Mean corneal thickness is the same in normal eyes and eyes with glaucoma
C. Mean corneal thickness is less in normal eyes than in eyes with glaucoma

Assumption of the paired t hypothesis test in order for the decision to be valid. Same as assumption of the one-sample t test:
• The random sample came from a normal distribution
OR
• The random sample size is 𝑛 ≥ 30 .