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Week 8: 13/03/2012
CM0128M
1. A manager randomly samples 50 containers of juice. The manager is concerned that
the containers may be filled to an amount different from 240 grams.
A. Develop a suitable null and alternative hypothesis.
Ans: H0: μ=240
H1: μ≠240
B. How can the manager make a Type I error?
Ans:Reject H0 when H0 is correct.
C. How can the manager make a Type II error?
Ans: Accept H0 when H0 is not correct.
D. Will the hypothesis test procedure prove that the containers are filled to an
amount different from 240 grams?
Ans: The hypothesis test procedure could only show that whether there is
sufficient evidence to suggest that the containers are filled to an amount different
from 240 grams.
2. State if an error is made in the following situations and, if so, what type of error.
A. The null hypothesis is true and the calculated value of the test statistic falls in the
rejection region.
Ans: Wrong decision will be made, with Type I error.
B. The null hypothesis is true and the calculated value of the test statistic does not
fall in the rejection region.
Ans: Correct decision will be made.
C. The alternative hypothesis is true and the calculated value of the test statistic falls
in the rejection region.
Ans: Correct decision will be made.
D. The alternative hypothesis is true and the calculated value of the test statistics
does not fall in the rejection region
Ans: Wrong decision will be made, with Type II error.
3. The mean of a normally distributed population is believed to be equal to 50.1. A
sample of 36 observations is taken, and the sample mean is found to be 53.2. The
alternative hypothesis is that the population mean is not equal to 50.1. Complete the
hypothesis test, assuming that the population standard deviation is equal to 4. Use a
0.05 significance level.
Ans:
step 1. state the null and alternative hypotheses
H0: μ=50.1
H1: μ≠50.1
step 2. select a level of significance: α=0.05
step 3. Calculate the test statistic: Z 0 =
̄x −μ 0 53.2−50.1
=
=4.65
σ/ √ n
4 / √ 36
step 4. Find rejection region, make a decision
This is a two-tailed test, critical values are : ±Z α /2 =±Z 0.05/ 2=±1.96
The rejection region is Z <−Z α /2 or Z>Z α/2 → Z<−1.96 or Z>1.96
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Reject H0 since 4.65>1.96
step 5. based on the sample data, there is sufficient evidence to conclude that
μ≠50.1
4. Set up the null hypothesis and the alternative hypothesis for each of the following
situations
A. An automotive analyst believes that the miles per gallon on a new model is less
than what the company is advertising. A random sample of data is taken to
support the analyst's belief.
Ans: H0: μ= μ0
H1: μ< μ0
B. A phone company claims that the average customer pays less than £30 a month. A
random sample of data is taken to verify this claim
Ans: H0: μ= μ0
H1: μ< μ0
C. A marketing research firm believes that the average number of hours per week
that Americans watch TV differs from 15 hours a week. A sample of data is taken
to support this belief.
Ans: H0: μ= μ0
H1: μ≠ μ0
5. A random sample of 49 observations from a normal population yields a sample mean
of 85 with a standard deviation of 14.
A. Test the claim that the population mean differs from 81. Use a 0.05 significance
level.
Ans:
step 1. state the null and alternative hypotheses
H0: μ=81
H1: μ≠81
This is a two-tailed test.
step 2. select a level of significance: α=0.05
step 3. Calculate the test statistic, in this case σ is not given, we use s instead as
the sample size is large (49) : Z 0 =
x −μ 0 85−81
̄
=
=2
s / √n 14 / √ 49
step 4. Find the rejection region.
This is a two-tailed test, so critical values are : ±Z α /2 =±Z 0.05/ 2=±1.96
The rejection region is Z <−Z α /2 or Z>Z α/2 → Z<−1.96 or Z>1.96
Reject H0 since 2>1.96
step 5. based on the sample data, there is sufficient evidence to conclude that
μ≠81
B. Test the claim that the population mean is greater than 85. Use a 0.05 significance
level.
Ans:
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CM0128M
step 1. state the null and alternative hypotheses
H0: μ=85
H1: μ>85
This is a one-tailed test.
step 2. select a level of significance: α=0.05
step 3. Calculate the test statistic: Z 0 =
x −μ 0 85−85
̄
=
=0
s / √ n 14 / √ 49
step 4. Find the rejection region.
This is a right-tailed test, so critical value is : Z α =1.65
The rejection region is Z >1.65
accept H0 since 0<1.65
step 5. based on the sample data, there is sufficient evidence to conclude that
μ=85
6. A factory manager asserts that the mean running time of machines is14 hours a day.
A random sample of 64 machines shows that their mean running time was only 13
hours 20 minutes, with a standard deviation of 3 hours. Test the manager’s assertion
at the 5% significant level.
Ans:
step 1. state the null and alternative hypotheses
H0: μ=14
H1: μ≠14
This is a two-tailed test.
step 2. select a level of significance: α=0.05
step 3. Calculate the test statistic: in this case σ is not given, we use s instead as
the sample size is large (n=64)
Z0=
x̄−μ 0 13.33−14
=
=−1.79
s / √n
3 / √ 64
step 4. Find the rejection region.
This is a two-tailed test, so critical values are : ±Z α /2 =±Z 0.05/ 2=±1.96
The rejection region is Z <−Z α /2 or Z>Z α/2 → Z<−1.96 or Z>1.96
accept H0 since −1.79>−1.96
step 5. based on the sample data, there is sufficient evidence to conclude that
μ=14
7. The owner of a restaurant believes that the average amount spent on a meal per
couple is £50.00. She takes a random sample of 40 couples and finds that the mean
expenditure is £53.70 with standard deviations of £7.80. Test the owner’s claim at
the 5% level of significance.
Ans:
step 1. state the null and alternative hypotheses
H0: μ=50
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H1: μ≠50
This is a two-tailed test.
step 2. select a level of significance: α=0.05
step 3. Calculate the test statistic: in this case σ is not given, we use s instead as
the sample size is large (n=40)
Z0=
x −μ 0 53.70−50
̄
=
=3
s / √n
7.8 / √ 40
step 4. Find the rejection region.
This is a two-tailed test, so critical values are : ±Z α /2 =±Z 0.05/ 2=±1.96
The rejection region is Z <−Z α /2 or Z>Z α/2 → Z<−1.96 or Z>1.96
reject H0 since 3>1.96
step 5. based on the sample data, there is sufficient evidence to conclude that the
average amount spent on a meal per couple is not £50.00.
8. The expected life of a car engine is 90 000miles. A sample of 190 cars gives a mean
life of 96 700 miles with a standard deviation of 37 5000 miles. Test whether the
cars can last more than the expected 90000 miles at the 1% level of significance.
Ans:
step 1. state the null and alternative hypotheses
H0: μ=90000
H1: μ>90000
This is a one-tailed test.
step 2. select a level of significance: α=0.01
step 3. Calculate the test statistic: in this case σ is not given, we use s instead as
the sample size is large (n=190)
Z0=
x̄−μ 0 96700−90000
=
=2.46
s / √n
37500/ √ 190
step 4. Find the rejection region.
This is a right-tailed so critical value is: Z α =Z 0.01=2.33
The rejection region is Z >2.33
Reject H0 since 2.46>2.33
step 5. based on the sample data, there is sufficient evidence to conclude that the
life time of a car engine will be more than 90000 miles.
9. An estate agent advertises that it can sell houses in 40 days or less. A sample of 64
recently sold homes shows a sample mean selling time of 45 days with a standard
deviation of 20 days. Using a 0.05 significance level, test the validity of the agent’s
claim.
Ans:
step 1. state the null and alternative hypotheses
H0: μ≤40
H1: μ>40
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This is a one-tailed test.
step 2. select a level of significance: α=0.05
step 3. Calculate the test statistic: in this case σ is not given, we use s instead as
the sample size is large (n=64)
Z0=
̄x −μ 0 45−40
=
=2
s / √ n 20 / √ 64
step 4. Find the rejection region.
This is a right-tailed test, so critical value is : Z α =Z 0.05=1.65
The rejection region is Z >1.65
Reject H0 since 2>1.65
step 5. based on the sample data, there is sufficient evidence to conclude that the
estate agent will sell a house in more than 40 days.
10. A poll of 121 randomly selected students reveals that the mean number of CD’s they
buy a year is 7.01 with a standard deviation of 3.74. Test the claim that the mean for
all students is less than 7.5 CDs at the 1% level of significance.
Ans:
step 1. state the null and alternative hypotheses
H0: μ ≥ 7.5
H1: μ < 7.5
This is a one-tailed test.
step 2. select a level of significance: α=0.01
step 3. Calculate the test statistic: in this case σ is not given, we use s instead as
the sample size is large (n=121)
Z0=
x̄−μ 0 7.01−7.5
=
=−1.44
s / √ n 3.74 / √ 121
step 4. Find the rejection region.
−Z α=−Z 0.01=−2.33
This is a left-tailed test, so critical value is :
The rejection region is Z <−2.33
Accept H0 since −1.44>−2.33
step 5. based on the sample data, there is sufficient evidence to conclude that all
students will buy 7.5 CDs or more.
11. The breaking strengths of cables produced by a manufacturer have a mean of 1800
pounds (lb) and a standard deviation of 100 lb. By a new technique in the
manufacturing process, it is claimed that the breaking strength can be increased. To
test this claim, a sample of 50 cables s tested and it is found that the mean breaking
strength is 1850lb. Can we support that claim at the 0.01 significance level?
Ans:
step 1. state the null and alternative hypotheses
H0: μ =1800
H1: μ > 1800
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This is a one-tailed test.
step 2. select a level of significance: α=0.01
step 3. Calculate the test statistic: in this case σ is given,
Z0=
x̄−μ 0 1850−1800
=
=3.54
σ/ √ n
100/ √ 50
step 4. Find the rejection region.
Z α =Z 0.01=2.33
This is a right-tailed test, so critical value is :
The rejection region is Z <2.33
Reject H0 since 3.54>2.33
step 5. based on the sample data, there is sufficient evidence to conclude that the
new technique can improve the break strength.
The following two questions are additional exercises, which I will leave for you to
analyse. You are required to test the hypotheses using significance level of both 0.01
and 0.05. Note: to encourage attempts at these questions, answers will not be provided
(i.e. for Q12&13) but you can show me your answers in the tutorial session.
12. Do middle-aged male executives have different average blood pressure then the
general population? The National Centre for Health Statistics reports that the mean
systolic blood pressure for males 35t to 44 years of age is 128 and the standard
deviation in this population is 15. The medical director of a company looks at the
medical records of 72 company executive in this age group and finds that the mean
systolic blood pressure in this sample is 126.06. Is this evidence that executive blood
pressures differ from the national average?
13. Doctors have warned the government that women having their first child when they
are over 40 years old are more likely to have children with chromosome disorders. A
long-term study of women’s pregnant ages in a particular area indicates that the mean
age is 42.1 years. Following a government sponsored campaign to highlight this
issue; a survey of 60 new mothers gives a mean pregnant age of 38.9 years and a
standard deviation of 8.3 years. Has the campaign really reduced the pregnant ages of
new mothers?
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