3. On the day before his final examination on ENME ABC,
John was reviewing his exam scores during the past hour exams
and quizzes. Can you guess what score John received on the final
examination of ENME ABC.
Solution:
From the t-distribution table, the critical t-values for 7 degrees
of freedom (v=8-1=7) and a significance level of 10% (=1-0.90=0.1)
is tcrit=1.90.
He will have a score between 62.3 and 92.7 with 90% confidence
interval.
4
Mr. John Smith is testing the noise level of two types of
sanders. 2- experiments have been performed for each of the tow types of
sanders. The recorded data in terms of DB are listed as follows. Let us
use two approaches fro the comparison.
(1) Calculate the mean and standard deviation of the 20 measurements for
each of the two types of sanders and then make a comparison. Based on the
results, interpret the observed difference. Comment on your work? Which
type is better in terms of the noise level?
(2) Divide the 20 measurements of Sander 1 into tow groups. Calculate the
mean and standard deviation of the 10 measurements for each of the tow
groups and then make a comparison between the two groups. Based on the
results, interpret the difference observed in (1). Comment on your work.
Which type is better in terms of the noise level?
Sand. 1
Sand. 2
1)
Unpaired t-test is used to compare the two groups.
a. State the null (H0) and alternative (H1) hypotheses.
H0: 1=2 ( No difference in noises between the two sanders)
H1: 1<2 ( The noise level of Sand1 is
lower than that of Sand 2)
b. Select the significance level for the test: =0.10
c. Calculate the critical and observed t-values.
For this problem, a one sided t-distribution table is used (H1 includes "<")
tcrit=t(v=38, =0.1)=1.304
The observed t-value is computed below.
d. Since tobs<tcrit, the null hypothesis
(H0) can not be rejected. Therefore, there is no difference in
the noise level.
(2)
Sand. 1
Paired t-test is used to compare the two groups.
State the null(H0) and alternative (H1) hypotheses:
H0: D=0 ( No difference in noise between the two groups)
H1: D0 ( Difference in noise level between the two groups)
b. Select the significance level for the test: =0.10
c. Calculate the critical and observed t-values.
For this problem, a two sided t-distribution table is used .
tcrit=t(v=9, =0.1)=1.833
The observed t-value is computed below.
d. Since tobs<tcrit, the null hypothesis
(H0) can not be rejected. Therefore, there is no difference in
the two grou ps.
5 Ten students were given a standardized achievement test at the beginning of the school year(February 1, 1995). They were re-tested last week (February 8, 1995). The following table listed the data of their scores. Make a fudgement whether the ten students made an improvement in taking the standardized achievement test after two month study at school. (two side test and significance level:10%)
Paired t-test is used to compare the two groups.
H0: 
( the students did not make an improvement
in taking the test)
H1:
( the students made an improvement
in taking the test)
From the t-distribution table, tcrit=t(v=9,=0.10)=1.383
6
In the early 1960s, the average weight of males at the age of
22 was found to be 160 lbs with a standard deviation of 10.5 lbs. To
determine whether on the average weights of males at the age of 22 are
higher today, a graduate student on the College Park campus selects a
random sample of 64 male students( approximately at 22) and records their
weights.
(1) Use the hypothesis test outline to set up statistical procedure to
help decide if today's males are heavier on the average than they used to
be (significance level=0.001)
(2) Suppose the random sample yields a mean weights of 165.4 lbs. Based
on the above procedure, what conclusion should be drawn concerning average
weight of males at the age of 22?
(1) Test procedure
H0: w1=w2 (No difference in the weight of males between 1960 and today)
H1: w1>w2 ( Today's males are heavier on the average than they used to be)
(2) vW1=nW1-1=64-1=63
tcrit=t(v=, =0.001)=
3.09 (one-side distribution)
Since tobs > tcrit , the null hypothesis
should be rejected, i.e., Today's males are heavier on the average
than they used to be in 1960)
7.
An electrical circuit manufactured by XYZ Company is
supposed to have a current output of 2.5 amps. To confirm this
specification, 11 circuits without coating are selected randomly and
tested. The following data is the record of the current output.
(1) Does the data show that the current output meets the
specification?
(2) An engineer has proposed to use coating for increasing the current
output. he took each circuit listed on the left column, coated with
polyester, and re-tested it. The recorded data is listed in the right
column. Do you believe that the coating method is more effective in terms
of keeping the current output at 2.5 amps.
In the table above, D1 stands for the difference between the specification
and before coating. D2 stands for the difference between the specification
and after coating.
a. State the null(H0) and alternative(H1) hypotheses:
H0: 
( No difference in the current output)
H1:
(Difference from the specification
in current output)
b. Select the significance level for the test: =0.001
c. From the t-distribution table, tcrit=t(v=10,=0.10)=1.812
d. Since tobs<tcrit, the null hypothesis
can not be rejected. Therefore, the current output of the circuit
before coating can meet the specificatio n.
(2) A paired t-test is used to make the comparison between D1
and D2
a. State the null(H0) and alternative(H1) hypotheses:
H0: 
(No difference in the current output
between before and after coating)
H1:
(the current output for after coating
is worse than that of before coating)
b. Select the significance level for the test: =0.1
c. From the t-distribution table, tcrit=t(v=10,=0.10)=1.372
(one side t-distribution)
d. Since tobs>tcrit , the null hypothesis
should be rejected. Therefore, the current output of the after
coating circuit is less effective in terms of keeping the current
output at 2.5 amps.
8. The following data represent the trade deficits recorded in 1991 and 1992. The unit is in millions. In the data, you may notice that the trade deficit in January, 1992 was extremely high because of the Gulf War. The reason for a low trade deficit in July, 1992 was due to the payment from Japan to share the cost of the U.S. troops in Kuwait. The Republican Party claimed that the Bush Administration achieved, at certain degree, a trade deficit reduction in 1992. The Democratic Party complained that the Bush Administration worsened the trade deficit in 1992. What do you think? Provide your evidence to support your viewpoint. you may select a significance level.
H0: 
(No difference in trade deficits)
H1: 
( Difference in trade deficit)
Since tobs<tcrit, based on the t-test, the null
hypothesis(H0) is accepted which means that there is no significant
difference in the deficits from 1991 to 1992.
