578 Assignment-4 (Chs. 9, 10 and 12): solutions due by midnight of Sunday, April 5, 2013: drop box 4): 70 points (show work when possible) True/False (1 point each) Chapter 9 1.The manager of the...

578 Assignment-4 (Chs. 9, 10 and 12): solutions due by midnight of Sunday, April 5, 2013: drop box 4): 70 points (show work when possible)


True/False (1 point each)
Chapter 9

1.The manager of the quality department for a tire manufacturing company wants to know the average tensile strength of rubber used in making a certain brand of radial tire. The population is normally distributed and the population standard deviation is known. She uses a Z test to test the null hypothesis that the mean tensile strength is 800 pounds per square inch. The calculated Z test statistic is a positive value that leads to a p-value of .045 for the test. If the significance level (a) is .01, the null hypothesis would be rejected.
2.You cannot make a Type I error when the null hypothesis is true.
3.The power of a statistical test is the probability of rejecting the null hypothesis when it is false.

4.The level of significance indicates the probability of rejecting a false null hypothesis.
5. When conducting a hypothesis test about a single mean, other relevant factors held constant, increasing the level of significance from .05 to .10 will decrease the probability of a Type II error.


Chapter10

6.In testing the difference between two means from two independent populations, the sample sizes do not have to be equal to be able to use the Z statistic.
7.In testing the difference between the means of two independent populations, if neither population is normally distributed, then the sampling distribution of the difference in means will be approximately normal provided that the sum of the sample sizes obtained from the two populations is at least 30.
8.If the limits of the confidence interval of the difference between the means of two normally distributed populations were 0.5 and 2.5 at the 95% confidence level, then we can conclude that we are 95% certain that there is a significant difference between the two population means.

Chapter 12

9.In a contingency table, if all of the expected frequencies equal the observed frequencies, then we can conclude that there is a perfect dependence between rows and columns.
10.In performing a chi-square test of independence, as the difference between the respective observed and expected frequencies increase, the probability of concluding that the row variable is independent of the column variable increases.
11.When we carry out a chi-square test of independence, the expected frequencies are based on the alternative hypothesis.

Multiple Choices (2 points each)


Chapter 9

1. If a null hypothesis is rejected at a significance level of .10, it will ______ be rejected at a significance level of .05
A.Always
B.Sometimes
C.Never
2. If a null hypothesis is not rejected at a significance level of .05, it will ______ be rejected at a significance level of .01
A.Always
B.Sometimes
C.Never
3.When carrying out a large sample test of H₀:
m
=10 vs. H_a:
m
¹ 10 by using a p-value, we reject H₀
at level of significance
a
when the p-value is:
A.Greater than
a
/2
B.Greater than
a

C.Less than
a

D.Less than
a
/2
E.Less than Z
_a


4. If you live in California, the decision to buy earthquake insurance is an important one. A survey revealed that only 133 of 350 randomly selected residences in one California county were protected by earthquake insurance.
Calculate the appropriate test statistic (Z value) to test the hypotheses that less than 40% buy the insurance. (Rounded)
A. 0.76
B. 0.38
C. -0.76

D. -0.40
E. -0.38

Chapter 10

5.In order to test the effectiveness of a drug called XZR designed to reduce cholesterol levels, 20 heart patients' cholesterol levels are measured before they are given the drug. The same 20 patients use XZR for two continuous months. After two months of continuous use the 20 patients' cholesterol levels are measured again. The comparison of cholesterol levels before vs. after administering the drug is an example of testing the difference between:
A.Two means from independent populations
B.Matched pairs from two dependent populations
C. Two means from independent populations with unknown variances
D.Two means from independent populations with equal variances
6. If the t statistic (critical value) is used in lieu of the Z statistic when comparing two means from independent populations using small samples, the probability of rejecting the null hypothesis __________.
A. Decreases
B.Increases
C.Remains the same
D. Depends on whether the population is normal
7.Given the following information about a hypothesis test of the difference between two means based on independent random samples, which one of the following is the correct rejection region at a significance level of .01? Assume that the samples are obtained from normally distributed populations having equal variances.

H
₀:
m
_A

=
m
_B
, H
₁:
m
_A
?
m
<sub>B


1</sub>
=
13,
₂
=9,
s
₁

=
5,
s
₂

=
4 ,
n
₁
=
15,
n
₂

=
13.
A.Reject H₀
if Z >2.576
B.Reject H₀
if Z >2.33
C.Reject H₀
if t >2.479
D.Reject H₀
if t >2.779
E.Reject H₀if t >2.763
8. Given the following information about a hypothesis test of the difference between two means based on independent random samples, what is the calculated value of the test statistic? Assume that the samples are obtained from normally distributed populations having equal variances.

H
₀:
m
_A

=
m
_B
, H
₁:
m
_A
?m
<sub>B


1</sub>
=
13,
₂
=
9,
s
₁

=
5,
s
₂

=
4,
n
₁
=15,
n
₂

=
13.
A. 1.730

B. 2.312

C. 4.566
D. 4.000

E. 2.479

Chapter 12

9.Which if any of the following statements about the chi-square test of independence is false?
A.If r_i
is row total for row i and c_j
is the column total for column j, then the estimated expected cell frequency corresponding to row i and column j equals (r_i) (c_j)/n
B.The test is valid if all of the estimated cell frequencies are at least five
C.The chi-square statistic is based on (r-1)(c-1) degrees of freedom where r a nd c denote, respectively the number of rows and columns n the contingency table
D.The alternative hypothesis states that the two classifications are statistically independent.
10. A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: The Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items and Process 3 manufactured 9 defective units in 150 items. At a significance level of .05, the management wants to perform a hypothesis test to determine whether the quality of items produced appears to be independent of the production process used. What is the rejection point condition?
A. Reject H₀
if
c

²

> 0.10257
B. Reject H₀
if
c

²

> 5.99147
C. Reject H₀
if
c

²

> 9.3484
D. Reject H₀
if
c

²

> 7.37776
E. Reject H₀
if
c

²

> 7.81473
11.A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: The Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items and Process 3 manufactured 9 defective units in 150 items. At a significance level of .05, the management wants to perform a hypothesis test to determine whether the quality of items produced appears to be independent of the production process used. Calculate the expected number of defective units produced by Process 2.
A.29
B.21
C. 30
D.16
E.15
12. A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: The Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items and Process 3 manufactured 9 defective units in 150 items.
At a significance level of .05, the management wants to perform a hypothesis test to determine if the quality of the items produced appears to be independent of the production process used. Based on Chi-square test, we:
A.Reject H₀
and conclude that the quality of the product is not the same for all processes
B.Reject H₀
and conclude that the quality of the product is dependent on the manufacturing process
C.Failed to reject H₀
and conclude that the quality of the product does not significantly differ among the three processes
D.Failed to reject H₀
and conclude that the quality of the product is not the same for all processes
E.Reject H₀
and conclude that the quality of the product is independent of the manufacturing process used.

Essay Type (Five points each)


Chapter 9

1.The average waiting time per customer at a fast food restaurant has been 7.5 minutes. The customer waiting time has a normal distribution. The manager claims that the use of a new cashier system will decrease the average customer waiting time in the store. Based on a random sample of 16 customer transactions the mean waiting time is 6.3 minutes and the standard deviation is 2 minutes per customer. Test the manager’s claim at 5% and 1% significance level tests.
2.In an early study, researchers at an Ivy University found that 33% of the freshmen had received at least one A in their first semester. Administrators are concerned that grade inflation has caused this percentage to increase. In a more recent study, of a random sample of 500 freshmen, 185 had at least one A in their first semester Calculate the appropriate test statistic to test the hypotheses related to the concern and test at 5% and 1%.

Chapter 10

3.At a=.05 and .10, test the hypothesis that the proportion of Consumer (CON) industry companies winter quarter profit growth is
more than 2 percentage points
greater than
the proportion of Banking (BKG) companies winter quarter profit growth, given that p_CON
= 0.20, p_BKG
= 0.13,
n
_CON
= 300,
n
_BKG

=
400.
4.The mid-distance running coach, Zdravko Popovich, for the Olympic team of an eastern European country claims that his six-month training program significantly reduces the average time to complete a 1500-meter run. Five mid-distance runners were randomly selected before they were trained with coach Popovich's six-month training program and their completion time of 1500-meter run was recorded (in minutes). After six months of training under coach Popovich, the same five runners' 1500 meter run time was recorded again. The results are given below.

Runner	1	2	3	4	5
Completion time before training	5.9	7.5	6.1	6.8	8.1
Completion time after training	5.5	7.0	6.3	6.7	7.9

At alpha levels of .05 and 0.10, test whether there has been a significant decrease in the mean time per mile?
5. Test H₀: µ₁

=
µ₂; H₁: µ₁
> µ₂
at a = 0.05, when
₁
= 75,
₂
= 72, s₁
= 3.3, s₂
= 2.1, n₁= 6, n₂= 6. Assume equal variances. Indicate which test you are performing; show the hypotheses, the test statistic and the critical values and mention whether one-tailed or two-tailed.

Chapter 12

6.Consider the 3X2 contingency table below

Factor A	Factor B
B1	B2
A1	15	15
A2	8	22
A3	7	33

At a significance levels of 0.05 and 0.01, test H₀: the factors A and B are independent.
7.In the past, of all the students enrolled in "Basic Business Statistics" 10% earned A's 20% earned B's, 30% earned C's, 20% earned D's and the rest either failed or withdrew from the course. Dr Johnson is a new professor teaching "Basic Business Statistics" for the first time this semester. At the conclusion of the semester, in Dr. Johnson's class of 60 students, there were 10 A's, 20 B's, 20 C's, 5 D's and 5 W's or F's. Assume that Dr. Johnson's class constitutes a random sample. Dr Johnson wants to know if there is sufficient evidence to conclude that the grade distribution of his class is different than the historical grade distribution.
Use a =.05 and .01 and perform a goodness of fit test.