Hypothesis testing
March 24, 2017
1. According to Wikipedia, Great Danes have an average height of 32 inches with a standard deviation of 1 inch. A
random sample of 5 Great Danes living in California was found to have an average height of 3 feet. Can we conclude
that California Great Danes are larger than average?
Solution:
Let µ be the average height of Great Danes.
H0 : µ = 32
HA : µ > 32
N is small, observations are independent, 1 sample t test for a mean.
t
1
<( 36
pt (t,
-
4
32
)
)
/
(
1
/
sqrt
(
5
))
## [1] 0.0004321055
There is strong evidence to believe that CA great danes are large and in charge.
2. Since they started over 2 decades ago, a kennel that breeds German Shepard dogs has produced pups that were 65kg
on average by 1 year old with a standard deviation of 3kg. This past year the breeder decided to try a new formulation
of muscle building supplements. The next 30 GSD pups from this kennel have an average 1 year weight of 66.5kg. Can
the breeder conclude that the new supplements helped increase the weight of their GSD?
Solution:
Let µ be the average weight of 1 year old GSD.
H0 : µ = 65
HA : µ > 65
N is large, observations are independent, 1 sample z test for a mean.
z
1
<( 66.5
pnorm (z)
-
65
)
/
(
3
/
sqrt
(
30
))
-
## [1] 0.00308495
There is strong evidence to believe that the supplement helped increase the weight of 1 year old GSD’s.
1
3. Previous studies of wild wolf populations have clocked their top speed at 22.7mph. A new study looks to compare the
top speed of domestic wolves (defined as at least 1/4 wolf) to their wild brethren. A sample of 30 domestic wolves were
clocked and an average of 21.3mph with a standard deviation of 3.6 were calculated. Is there reason to believe that
domesticating wolves has an impact on top speed?
Solution:
Let µ be the average top speed of a wild wolf.
H0 : µ = 22.7
HA : µ 6= 22.7
N is large, observations are independent, 1 sample Z test for a mean.
z
2
<(
*
( 21.3
pnorm (z))
22.7
)
/
(
3.6
/
sqrt
(
30
))
## [1] 0.03316896
There is moderate evidence to believe that domestic wolves have a different top speed than their wild counterparts.
4. The average number of mitochondria present in normal adults is 830 mitochondria per muscle cell (mpc) with a
σ = 100. Mitochondria concentration was measured from a sample of 50 college athletes and calculated to have an
average concentration of 1356 mpc. Is there reason to believe that athletes have more mitochondria than average?
Solution:
Let µ be the average mitochondria concentration per muscle cell of adults.
H0 : µ = 830
HA : µ > 830
N is large, observations are independent, 1 sample Z test for a mean.
z
1
<( 1365
pnorm (z)
-
830
)
/
(
100
/
sqrt
(
50
))
-
## [1] 0
There is very strong evidence to believe that athletes have a higher concentration of mitochondria per muscle cell
compared to the normal adult.
2
5. Western Sycamore trees have a population mean height of 120’ with a σ = 10.5. If a random sample of 25 trees by
a lakeside has a mean height of 118, can we conclude that Western Sycamore trees by a lakeside are shorter than
normal?
Solution:
Let µ be the average height of Western Sycamore trees.
H0 : µ = 120
HA : µ < 120
N is small, observations are independent, 1 sample t test for a mean.
t
pt
<(t,
(
115
)
-
120
)
/
(
10.5
/
sqrt
(
25
))
24
## [1] 0.01277591
There is strong evidence to believe that Western Sycamore trees living by a lakeside are shorter than what is normal
for this type of tree.
6. An acre of corn gives off on average 3500 gallons of water vapor each day. A plot of 30 Western Sycamore trees were
planted right next to the corn. That plot is measured 50 times throughout the day and found to have an average
transpiration rate of 3602 gallons of water vapor each day with a standard deviation of 250. Does the plot of Western
Sycamores produce more water vapor per day than an acre of corn?
Solution:
Let µ be the average amount of water vapor given off by an acre of corn.
H0 : µ = 3500
HA : µ > 3500
N is large, observations are independent, 1 sample z test for a mean.
z
1
<( 3602
pnorm (z)
-
3500
)
/
(
250
/
sqrt
(
50
))
-
## [1] 0.001957095
There is strong evidence that the acre of Western Sycamore trees produce on average more water vapor per day
compared to an acre of corn.
3
7. For the past 100 years scientists have been keeping track of the winter bird migration to the south. They have concluded
that the population of migratory birds numbers around 1 million birds (σ = 50, 000) migrate from CA to Mexico for
the winter. Last year 40 researchers independently counted the amount of migratory birds and came up with a group
average of 990,750 birds flying south for the winter. Is there reason to believe that the migratory bird population is
smaller than expected?
Solution:
Let µ be the average amount of migratory birds on the west coast.
H0 : µ = 1, 000, 000
HA : µ < 1, 000, 000
N is large, observations are independent, 1 sample z test for a mean.
z
<(
pnorm (z)
990750
-
1000000
)
/
(
50000
/
sqrt
(
40
))
## [1] 0.1209919
There not enough evidence that the size of the migratory bird population is smaller than previous years.
8. A current regimen of chemotherapy has shown that it can reduce tumor sizes to 10mm (σ = .5mm). A researcher
thinks that a combination of exercise and chemo can reduce the tumor size even further. They conduct a study and
find that the 49 patients who participate in an exercise regimen along with their chemo have an average tumor size of
9.8mm. Is there reason to believe that exercising in addition to chemo helps reduce tumor size?
Solution:
Let µ be the reduced tumor size under the current regimen of chemotherapy.
H0 : µ = 10
HA : µ < 10
N is large, observations are independent, 1 sample z test for a mean.
z
<(
pnorm (z)
9.8
-
10
)
/
(
.5
/
sqrt
(
49
))
## [1] 0.00255513
There is sufficient evidence to indicate that exercising in addition to chemo may help to reduce tumor size.
4
9. According to Wikipedia, adult Chinook salmon tend to be 30 inches long. A sample of 50 chinook salmon carcasses
were gathered from a particular river during the 2013 season and found to have an average length of 29.8 inches with
a sd of 3. Is there reason to believe that the salmon were smaller than average this season?
Solution:
Let µ be the average length in inches of Chinook salmon.
H0 : µ = 30
HA : µ < 30
N is large, observations are independent, 1 sample z test for a mean.
z
<(
pnorm (z)
29.8
-
30
)
/
(
3
/
sqrt
(
50
))
## [1] 0.3186759
There is not enough reason to believe that the average length of Chinook salmon in this particular river during 2013
was smaller than average.
10. A biologist wishes to estimate the effects that water quality has on the reproductive rate of aquatic insects in Big Chico
Creek. Previous experiments have shown that good quality water results in an abundance rate of 13 insects per 10cm2
with a standard deviation of 3 insects per cm2 . You go down to One Mile (Sycamore Pool) where the water quality is
less than spectacular (especially in the summer) and sample the insect abundance in 10 different locations. You find
that the average insect density is 14.8 insects per 10cm2 . What can you conclude about the effect of water quality on
the reproductive rate of aquatic insects in Big Chico Creek?
Solution:
Let µ be the average abundance rate of aquatic insects in BCC.
H0 : µ = 13
HA : µ 6= 13
N is small, observations are independent, 1 sample t test for a mean.
t
t
<-
(
14.8
-
13
)
/
9
))
(
3
/
sqrt
(
10
))
## [1] 1.897367
2
*
(
1
-
pt
(t,
## [1] 0.09026733
There is not reason to believe that lower quality water increases the abundance rate of these aquatic insects in BCC.
5
11. Legend has it that 1 in 5 forest gnomes have red beards. You and your team were able to catch 150 forest gnomes in
Siskyou County and noted that only 16% had red beards. What can you conclude about the proportion of red bearded
gnomes in the legend?
Solution:
Let p be the true proportion of forest gnomes with red beards.
H0 : p = .2
HA : p < .2
150
*
.16
## [1] 24
150
*
(
1
-
.16
)
## [1] 126
The success-failure condition is met, so the sample proportion p̂ can be assumed to have a normal distribution. Thus
a 1-sample Z test for proportions can be conducted.
z
<(
pnorm (z)
.16
-
.2
)
/
(
sqrt
(
.16
*
(
1
-
.16
)
/
150
))
## [1] 0.0907246
There is not enough to believe that the population of forest gnomes in Siskyou County has a smaller proportion of
gnomes with red beards compared what legend claims.
12. A certain disinfectant claims to kill 99% of the bacteria it comes in contact with. As a proper biologist you test this
claim by exposing 50 colonies of bacteria to equal amounts of this disinfectant. After the proper incubation time you
note that 96% of the colonies did not grow. What conclusion can you make about the manufacturers claim about the
effectiveness of this disinfectant?
Solution:
Let p be the proportion of bacteria killed by this disinfectant.
H0 : p = .99
HA : p < .99
Check the success-failure condition
50
*
.96
## [1] 48
50
*
(
1
-
.96
)
## [1] 2
The success-failure condition is not satisfied (both np̂ > 10 and nq̂ > 10), so the sample proportion p̂ cannot be assumed
to have a normal distribution. Thus we cannot use the normal model to test this claim.
6