6. If the decision is to reject the null hypothesis of no difference between two population
parameters, at the 5% level of significance, what are examples of the alternate hypothesis
and rejection region?
A) ?1 ??2 ; z > 1.65 and z < negative 1.65
B) ?1 ??2; z > 1.96 and z < negative 1.96
C) ?1 > ?2; z < negative 1.65
D) ?1 > ?2; z < negative 1.96
Answer: B
Since at the 5% level of significance the critical value is 1.96. Thus for the alternative
hypothesis of the form “not equal to” the rejection region is z > 1.96 and z < negative
1.96
7. A poll of 400 people from Dobbs Ferry showed 250 preferred chocolate raspberry
coffee while 170 out of 350 in Irvington preferred the same flavor. To test the hypothesis
that there is no difference in preferences in the two villages, what is the alternate
hypothesis?
A) ???1 < ?2
B) ?1 > ?2
C) ?1 = ?2
D) ???1 ??2
Answer: D
Here we want to test the hypothesis that there is no difference in preferences in the two
villages and nothing is specified about the alternative hypothesis. So the alternative
hypothesis will be p1≠p2.
8. Suppose we are testing the difference between two proportions at the 0.05 level of
significance. If the computed z is ¿1.07, what is our decision?
A) Reject the null hypothesis
B) Do not reject the null hypothesis
C) Take a larger sample
D) Reserve judgment
Answer: B) Do not reject the null hypothesis
At the 0.05 level of significance, the critical value is 1.96. So the rejection region is z >
1.96 or z < negative 1.96. Here the computed z =1.07 < 1.96. So we do not reject the null
hypothesis.
9. The net weights of a sample of bottles filled by a machine manufactured by Edne, and
the net weights of a sample filled by a similar machine manufactured by Orno, Inc., are
(in grams):
Edne: 5, 8, 7, 6, 9 and 7
Orno: 8, 10, 7, 11, 9, 12, 14 and 9
Testing the claim at the 0.05 level that the mean weight of the bottles filled by the
Orno machine is greater than the mean weight of the bottles filled by the Edne machine,
what is the critical value?
A) 2.179
B) 2.145
C) 1.782
D) 1.761
Answer: C) 1.782
Here the test statistic for testing the equality of means is the t-statistic, which follows a
Student’s t distribution with (n1+n2-2) =(6 + 8 -2) =12 degrees of freedom. Since the
alternative hypothesis is a one sided test, at the 0.05 level of significance from student’s t
able with 12 degrees of freedom, the critical value is 1.782
10. Using two independent samples, two population means are compared to determine if
a difference exists. The number in the first sample is fifteen and the number in the second
sample is twelve. How many degrees of freedom are associated with the critical value?
A) 24
B) 25
C) 26
D) 27
Answer: B) 25
The degrees of freedom of the test statistic is (n1+n2-2) =(15 + 12 -2) =25
T F 11. The F distribution is positively skewed and its values may range from 0 to plus
infinity.
Answer: True
T F 12. For the hypothesis test, Ho: ?21 = ?22, with n1 = 10 and n2 = 10, the F-test
statistic is 2.56. At the 0.01 level of significance, we would reject the null hypothesis.
Answer: False
At the 0.01 level of significance, from F tables with (n1-1, n2-1) =(9,9) degrees of
freedom the critical value is 5.35 and the rejection region is F >5.35. Here F =2.56<5.35.
So we do not reject the null hypothesis
T F 13. For the hypothesis test, Ho: ?21 = ?22, with n1 = 9 and n2 = 9, the F-test
statistic is 4.53. At the 0.05 level of significance, we would reject the null hypothesis.
Answer: True
At the 0.05 level of significance, from F tables with (n1-1, n2-1) =(8,8) degrees of
freedom the critical value is 3.44 and the rejection region is F >3.44. Here F =4.53>3.44.
So we reject the null hypothesis.
T F 14. In an ANOVA table, k represents the number of treatments, b represents the
number of blocks, and n represents the total number of sample observations.
Answer: True
T F 15. If a confidence interval for the difference between a pair of treatment means
includes 0, then there is no difference in the pair of treatment means.
Answer: True
T F 16. In a two-way ANOVA, the sum of the treatment, block, and error degrees of
freedom equal the total degrees of freedom.
Answer: True
17. An F statistic is:
A) a ratio of two means.
B) a ratio of two variances.
C) the difference between three means.
D) a population parameter.
Answer: B) a ratio of two variances.
18. A large department store examined a sample of the 18 credit card sales and recorded
the amounts charged for each of three types of credit cards: MasterCard, Visa and
Discover. Six MasterCard sales, seven Visa and five Discover sales were recorded. The
store used ANOVA to test if the mean sales for each credit card were equal. What are the
degrees of freedom for the F statistic?
A) 18 in the numerator, 3 in the denominator
B) 3 in the numerator, 18 in the denominator
C) 2 in the numerator, 15 in the denominator
D) 6 in the numerator, 15 in the denominator
Answer: C) 2 in the numerator, 15 in the denominator
The degrees of freedom of the F statistic is (k-1, n-k) = (3-1, 18 -3) = (2, 15)
19. Suppose that an automobile manufacturer designed a radically new lightweight
engine and wants to recommend the grade of gasoline that will have the best fuel
economy. The four grades are: regular, below regular, premium, and super premium.
The test car made three trial runs on the test track using each of the four grades and the
miles per gallon recorded. At the 0.05 level, what is the critical value of F used to test the
hypothesis that the miles per gallon for each fuel is the same.
A) 1.96
B) 4.07
C) 2.33
D) 12.00
Answer: B)
4.07
The F test statistic follows an F distribution with (k-1, n-k) = (4-1, 12 -4) = (3, 8) degrees
of freedom. So at the 0.05 level, the critical value is 4.07.
20. Suppose a package delivery company purchased 14 trucks at the same time. Five
trucks were purchased from manufacturer A, four from B and five from manufacturer C.
The cost of maintaining each truck was recorded. The company used ANOVA to test if
the mean maintenance cost of the trucks from each manufacturer were equal. To apply
the F test, how many degrees of freedom are in the denominator?
A) 2
B) 3
C) 11
D) 14
Answer: C) 11
The degrees of freedom in the denominator = n –k = 14 – 3 = 11
BONUS
21. A preliminary study of hourly wages paid to unskilled employees in three
metropolitan areas was conducted. Seven employees were included from Area A, 9 from
Area B and 12 from Area C. The test statistic was computed to be 4.91. What can we
conclude at the 0.05 level?
A)
B)
C)
D)
Mean hourly wages of unskilled employees all areas are equal
Mean hourly wages in at least 2 metropolitan areas are different
More degrees of freedom are needed
None of these is correct
Answer: B) Mean hourly wages in at least 2 metropolitan areas are different
Here k = 3 and n = 7+9+12 =28 The test statistic follows an F distribution with (k-1, n-k)
= (3-1, 28 -3) = (2, 25) degrees of freedom. So at the 0.05 level, the critical value is 4.24.
Thus the rejection region is F > 4.24. Here F = 4.91>4.24. So we reject Ho.