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Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
HOMEWORK: STATISTICAL INFERENCE
The One-Sample t DISTRIBUTION
PROBLEM 1:
A company manufactures bars of soap that are supposed to weigh exactly (on average) 10
ounces. A sample is taken:
n=20
X = 10.20 oz
s = 0.80 oz
(a) Test at α = .05
H 0 : m = 10
H1 : m ¹ 10
a = 0.05 (two-tailed, df = 20-1 = 19)
Critical value T_crit = 2.093
Test statistic T =
x - m 10.2 - 10
=
= 1.118
s n 0.8 20
Since 1.118 < 2.093 we fail to reject H0 and conclude that the company’s claim is true.
(b) Construct a 95% CIE of μ
95% C.I:
m= x ± t
s
0.8
= 10.2 ± 2.093
= (9.826, 10.574)
n
20
Topic 7: The Students’ T-Distribution
p. 1
Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
PROBLEM 2:
The Momoland Springs Company claims that at most there are 1 ppm (parts per million) of
benzene in their water. Test at =.05
Data:
n=25
X = 1.16 ppm
s = .20 ppm
H0 : m£ 1
H1 : m > 1
a = 0.05 (one-tailed, df = 25-1 = 24)
Critical value T_crit = 1.711
Test statistic T =
x - m 1.16 - 1
=
=4
s n 0.2 25
Since 4 > 1.711 we reject H0 and conclude that the company’s claim is not true.
Topic 7: The Students’ T-Distribution
p. 2
Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
PROBLEM 3:
A company claims that cancer patients using drug X will live at least 10 more years.
n=16
X = 8.8 year
s = 3.4 year
Test at =.05
H 0 : m ³ 10
H1 : m < 10
a = 0.05 (one-tailed, df = 16-1 = 15)
Critical value T_crit = -1.753
Test statistic T =
x - m 8.8 - 10
=
= - 1.412
s n 3.4 16
Since -1.412 > -1.753 we fail to reject H0 and conclude that the company’s claim is true.
Topic 7: The Students’ T-Distribution
p. 3
Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
PROBLEM 4:
A company manufactures wind turbines. A random sample of 25 turbines is taken and the
sample mean life is 20.00 years with a standard deviation of 2.50 years. If you were
constructing a 95% two-sided confidence interval estimate, the upper limit would be:
Df = 25-1 = 24.
U.L = x ± t
s
2.5
= 20 + 1.711
= 20.86
n
25
PROBLEM 5:
You want to estimate the average life of a new kind of computer chip. You take a sample of
16 chips and find the sample mean to be 12.50 years with a sample standard deviation of .80
years. You construct a 99% two-sided confidence interval estimate. It is:
Df = 16-1 = 15
m= x ± t
s
0.8
= 12.5 ± 2.947
= (11.911, 13.089)
n
16
Topic 7: The Students’ T-Distribution
p. 4
Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
PROBLEM 6:
A yogurt company claims that its yogurt ice cream has no more than 2.0 mgs. of fat. You
sample 25 cups of yogurt and find that the sample mean is 2.10 milligrams and the sample
standard deviation is .40 milligrams
(a) Test the claim at the .05 significance level
H 0 : m£ 2
H1 : m> 2
a = 0.05 (one-tailed, df = 25-1 = 24)
Critical value T_crit = 1.711
Test statistic T =
x- m
2.1- 2
=
= 1.25
s n 0.4 25
Since 1.25 < 1.711 we fail to reject H0 and conclude that the company’s claim is true.
(b) No claim was made by the company. They took a sample with the above results and
want to construct a two-sided 95% confidence interval for the population mean. The
upper limit of the confidence interval is:
UL = x ± t
s
0.4
= 2.1 + 1.711
= 2.24
n
25
Topic 7: The Students’ T-Distribution
p. 5
Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
PROBLEM 7:
A school claims that the average English regents score of its students is at least 80. You
sample 16 students and find that the sample mean is 76 and the sample standard deviation is
12
(a) Test the claim at the .05 significance level
H 0 : m ³ 80
H1 : m < 80
a = 0.05 (one-tailed, df = 16-1 = 15)
Critical value T_crit = -1.753
Test statistic T =
x - m 76 - 80
=
= - 1.333
s n 12 16
Since -1.333 > -1.753 we fail to reject H0 and conclude that the company’s claim is true.
(b) No claim was made by the school. They took a sample with the above results and
want to construct a two-sided 95% confidence interval for the population mean. The
upper limit of the confidence interval is:
UL = x + t
s
12
= 76 + 1.753
= 81.26
n
16
Topic 7: The Students’ T-Distribution
p. 6
Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
PROBLEM 8:
An LSAT preparation school claims that its review course will add at least 50 points to the
score of a student retaking the LSAT exam. You sample 25 students and find that average
improvement was 40 points with a standard deviation of 15 points
(a) Test the claim at the .05 significance level
H 0 : m ³ 50
H1 : m < 50
a = 0.05 (one-tailed, df = 25-1 = 24)
Critical value T_crit = -1.711
Test statistic T =
x - m 40 - 50
=
= - 3.333
s n 15 25
Since -3.333 < -1.711 we reject H0 and conclude that the school’s claim is false.
(b) No claim was made by the school. They took a sample with the above results and ask
you to construct a two-sided 95% confidence interval for the population mean.
m= x ± t
s
15
= 40 ± 1.711
= (45.13, 34.87)
n
25
Topic 7: The Students’ T-Distribution
p. 7
Foundations of Business Statistics – Professor Troudt
Materials from Professor Friedman
PROBLEM 9:
A drug company that manufactures a diet drug claims that those using the drug for 30 days
will lose at least 15 pounds. You sample 30 people who have used the drug and find that the
average weight loss was 12 pounds with a standard deviation of 5 pounds.
(a) Test the claim at the .05 significance level.
H 0 : m ³ 15
H1 : m < 15
a = 0.05 (one-tailed, df = 30-1 = 29)
Critical value T_crit = -1.699
Test statistic T =
x - m 12 - 15
=
= - 3.286
s n 5 30
Since -3.286 < -1.699 we reject H0 and conclude that the company’s claim is not true.
(b) No claim was made by the company. They took a sample with the above results and ask
you to construct a two-sided 95% confidence interval for the population mean.
m= x ± t
s
5
= 12 ± 1.699
= (13.55, 10.45)
n
30
Topic 7: The Students’ T-Distribution
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