Download Hypothesis Test Practice

Hypothesis Test Practice
1) Eleven percent of the products produced by an industrial process over the past several months have
failed to conform to specifications. The company modifies the process in an attempt to reduce the rate
of nonconformities. In a random sample of 300 items from a trial run, the modified process produces 16
nonconforming items. Do these results provide convincing evidence that the modification of the process
has been effective?
2) The germination rate of seeds is defined as the proportion of seeds that, when properly planted and
watered, sprout and grow. A certain variety of grass seed usually has a germination rate of 0.80, and a
company wants to see if spraying the seeds with a chemical that is known to change germination rates
in other species will change the germination rate of this grass species. The company researchers spray
400 seeds with the chemical, and 307 of the seeds germinate. Does this data provide enough evidence
that the chemical has changed the germination rate?
3) Nationally, the proportion of red cars on the road is 0.12. A statistically-minded fan of the
Philadelphia Phillies (whose team color is red) wonders if fans who park at Citizens Bank Park (the
Phillies home field) are more likely to drive red cars. One day during a home game, he takes an SRS of
210 cars parked in the lot while a game is being played, and counts 35 red cars. (There are 21,000
parking spaces.) Is this convincing evidence that Phillies fans prefer red cars more than the general
population? Support your conclusion with a test of significance.
4) Do political “attack ads” work? A congressional candidate who currently has the support of only 44%
of the voters runs a television spot that aggressively attacks the character of his opponent. He wants to
know if the television spot has changed his support level among voters. The pollsters survey an SRS of
450 voters and find that 186 support the candidate. This produces a 95% confidence interval for the
proportion of voters who support him of (0.368, 0.459). Is this convincing evidence that the attack ad
has changed the support for this candidate?
5) LeRoy, a starting player for a major college basketball team, made only 40% of his free throws last
season. During the summer, he worked on developing a softer shot in hopes of improving his free throw
accuracy. In the first eight games of this season, LeRoy made 25 free throws in 40 attempts. You want to
investigate whether LeRoy’s work over the summer will result in a higher proportion of free-throw
successes this season. What conclusion would you draw at the α = 0.01 level about LeRoy’s free throw
shooting? Justify your answer with a complete significance test.
6) The Environmental Protection Agency has determined that safe drinking water should contain no
more than 1.3 mg/liter of copper. You are testing water from a new source, and take 30 water samples.
The mean copper content in your samples is 1.36 mg/l and the standard deviation is 0.18 mg/l. There do
not appear to be any outliers in your data. Do these samples provide convincing evidence at the α = 0.05
level that the water from this source contains unsafe levels of copper? Would your conclusion change if
we instead used α = 0.01?
7) Sweet corn of a certain variety is known to produce individual ears of corn with a mean weight of 8
ounces. A farmer is testing a new fertilizer designed to produce larger ears of corn, as measured by their
weight. He finds that 32 randomly-selected ears of corn grown with this fertilizer have a mean weight of
8.23 ounces and a standard deviation of 0.8 ounces. There are no outliers in the data. Do these samples
provide convincing evidence at the α = 0.05 level that the fertilizer had a positive impact on the weight
of the corn ears? Justify your answer.
8) The label on bottles of a certain brand of grapefruit juice say that they contain 180 ml of liquid. Your
friend Jerry suspects that the true mean is less than that, so he takes an SRS of 40 bottles, and finds that
the mean contents of his sample is 178.8 ml with standard deviation 5.4 ml. Does this provide
convincing evidence that the mean contents of all bottles of this brand of grapefruit juice is less than
180 ml?
Solutions
1) Eleven percent of the products produced by an industrial process over the past several months have
failed to conform to specifications. The company modifies the process in an attempt to reduce the rate
of nonconformities. In a random sample of 300 items from a trial run, the modified process produces 16
nonconforming items. Do these results provide convincing evidence that the modification of the process
has been effective?
a)
H0: p = 0.11
Ha: p < 0.11
b)
16
𝑝̂ = 300 = 0.05333
0.11×0.89
300
𝜇𝑝̂ = 𝑝 = 0.11, 𝜎𝑝̂ = √
𝑧=
= 0.01806
0.05333 − 0.11
= −3.14
0.01806
c)
P(z < -3.14) = 0.0008
d)
Since the p-value is so small (0.0008 < 0.05) we reject H0, there is strong evidence that the
proportion of nonconforming parts is now less than 0.11.
2) The germination rate of seeds is defined as the proportion of seeds that, when properly planted and
watered, sprout and grow. A certain variety of grass seed usually has a germination rate of 0.80, and a
company wants to see if spraying the seeds with a chemical that is known to change germination rates
in other species will change the germination rate of this grass species. The company researchers spray
400 seeds with the chemical, and 307 of the seeds germinate. Does this data provide enough evidence
that the chemical has changed the germination rate?
a)
H0: p = 0.80
Ha: p ≠ 0.80
b)
307
𝑝̂ = 400 = 0.7675
0.80×0.20
400
𝜇𝑝̂ = 𝑝 = 0.80, 𝜎𝑝̂ = √
𝑧=
= 0.02
0.7675 − 0.80
= −1.625
0.02
c)
P(z < -1.63) = 0.0516, p-value = 0.0516x2 = 0.1032
d)
Since the p-value is so large (0.1032 > 0.05) we fail to reject H0, there is not enough evidence to
say that the germination rate is different from 0.80 after spraying with the chemical.
3) Nationally, the proportion of red cars on the road is 0.12. A statistically-minded fan of the
Philadelphia Phillies (whose team color is red) wonders if fans who park at Citizens Bank Park (the
Phillies home field) are more likely to drive red cars. One day during a home game, he takes an SRS of
210 cars parked in the lot while a game is being played, and counts 35 red cars. (There are 21,000
parking spaces.) Is this convincing evidence that Phillies fans prefer red cars more than the general
population? Support your conclusion with a test of significance.
a)
H0: p = 0.12
Ha: p > 0.12
b)
35
𝑝̂ = 210 = 0.16667
0.12×0.88
210
𝜇𝑝̂ = 𝑝 = 0.12, 𝜎𝑝̂ = √
𝑧=
= 0.02242
0.16667 − 0.12
= 2.08
0.02242
c)
P(z > 2.08) = 1 - 0.9812 = 0.0188
d)
Since the p-value is so small (0.0188 < 0.05) we reject H0, we have strong evidence that the
proportion of red cars in the entire Phillies parking lot is greater than 0.12.
4) Do political “attack ads” work? A congressional candidate who currently has the support of only 44%
of the voters runs a television spot that aggressively attacks the character of his opponent. He wants to
know if the television spot has changed his support level among voters. The pollsters survey an SRS of
450 voters and find that 186 support the candidate. This produces a 95% confidence interval for the
proportion of voters who support him of (0.368, 0.459). Is this convincing evidence that the attack ad
has changed the support for this candidate?
a)
H0: p = 0.44
Ha: p ≠ 0.44
b)
𝑝̂ =
186
450
= 0.41333
0.44×0.56
450
𝜇𝑝̂ = 𝑝 = 0.44, 𝜎𝑝̂ = √
𝑧=
= 0.0234
0.41333 − 0.44
= −1.14
0.0234
c)
P(z < -1.14) = 0.1271, p-value = 0.1271x2 = 0.2542
d)
Since the p-value is so large (0.2542 > 0.05) we fail to reject H0, there is not enough evidence to
say that the attack ad has made support for this candidate different than 44%.
NOTE: Sorry about throwing the CI in this problem. The connection is that since 0.44 IS IN the
confidence interval, you fail to reject H0.
5) LeRoy, a starting player for a major college basketball team, made only 40% of his free throws last
season. During the summer, he worked on developing a softer shot in hopes of improving his free throw
accuracy. In the first eight games of this season, LeRoy made 25 free throws in 40 attempts. You want to
investigate whether LeRoy’s work over the summer will result in a higher proportion of free-throw
successes this season. What conclusion would you draw at the α = 0.01 level about LeRoy’s free throw
shooting? Justify your answer with a complete significance test.
a)
H0: p = 0.40
Ha: p > 0.40
b)
25
𝑝̂ = 40 = 0.625
0.40×0.60
40
𝜇𝑝̂ = 𝑝 = 0.40, 𝜎𝑝̂ = √
𝑧=
= 0.07746
0.625 − 0.40
= 2.90
0.07746
c)
P(z > 2.90) = 1 - 0.9981 = 0.0019
d)
Since the p-value is so small (0.0019 < 0.05) we reject H0, there is strong evidence that the
proportion of all free throws made after the summer training is more than 44%
6) The Environmental Protection Agency has determined that safe drinking water should contain no
more than 1.3 mg/liter of copper. You are testing water from a new source, and take 30 water samples.
The mean copper content in your samples is 1.36 mg/l and the standard deviation is 0.18 mg/l. There do
not appear to be any outliers in your data. Do these samples provide convincing evidence at the α = 0.05
level that the water from this source contains unsafe levels of copper? Would your conclusion change if
we instead used α = 0.01?
a)
H0: µ = 1.3
Ha: µ > 1.3
b)
𝑥̅ = 1.36, n = 30, df = 29
𝑠
0.18
𝜇𝑥̅ = 𝜇 = 1.3, 𝑠𝑥̅ = 𝑛 = 30 = 0.03286
√
√
1.36 − 1.3
𝑡=
= 1.83
0.03286
c)
P(t > 1.83) = 0.0388 by calculator, p-value between 0.025 and 0.05 by table
d)
at α = 0.05: Since the p-value is so small (< 0.05) we reject H0, there is strong evidence that the
mean copper content in this water source is greater than 1.3 mg/liter
at α = 0.01: Since the p-value is so large (> 0.01) we fail to reject H0, there is no enough evidence
to say that the mean copper content in this water source is greater than 1.3 mg/liter
7) Sweet corn of a certain variety is known to produce individual ears of corn with a mean weight of 8
ounces. A farmer is testing a new fertilizer designed to produce larger ears of corn, as measured by their
weight. He finds that 32 randomly-selected ears of corn grown with this fertilizer have a mean weight of
8.23 ounces and a standard deviation of 0.8 ounces. There are no outliers in the data. Do these samples
provide convincing evidence at the α = 0.05 level that the fertilizer had a positive impact on the weight
of the corn ears? Justify your answer.
a)
H0: µ = 8
Ha: µ > 8
b)
𝑥̅ = 8.23, n = 32, df = 31
𝑠
0.8
𝜇𝑥̅ = 𝜇 = 8, 𝑠𝑥̅ = 𝑛 = 32 = 0.14142
√
√
8.23 − 8
𝑡=
= 1.626
0.14142
c)
P(t > 1.626) = 0.057 by calculator, p-value between 0.05 and 0.10 by table
d)
Since the p-value is so large (> 0.05) we fail to reject H0, there is not enough evidence that the
average weight for ears of corn using this new fertilizer is greater than 8 ounces
8) The label on bottles of a certain brand of grapefruit juice say that they contain 180 ml of liquid. Your
friend Jerry suspects that the true mean is less than that, so he takes an SRS of 40 bottles, and finds that
the mean contents of his sample is 178.8 ml with standard deviation 5.4 ml. Does this provide
convincing evidence that the mean contents of all bottles of this brand of grapefruit juice is less than
180 ml?
a)
H0: µ = 180
Ha: µ < 180
b)
𝑥̅ = 178.8, n = 40, df = 39
𝑠
5.4
𝜇𝑥̅ = 𝜇 = 180, 𝑠𝑥̅ = 𝑛 = 40 = 0.8538
√
√
178.8 − 180
𝑡=
= −1.405
0.8538
c)
P(t < -1.405) = 0.084 by calculator, p-value between 0.05 and 0.10 by table
d)
Since the p-value is so large (> 0.05) we fail to reject H0, there is not enough evidence that the
average content in all bottles of this brand of grapefruit juice is less than 180 ml