Bus Stat Test 3 Practice Test
Answer Section
MULTIPLE CHOICE
1.ANS:CPTS:1
2.ANS:BPTS:1
3.ANS:BPTS:1
4.ANS:DPTS:1
5.ANS:BPTS:1
6.ANS:CPTS:1
7.ANS:BPTS:1
8.ANS:DPTS:1
9.ANS:BPTS:1
10.ANS:APTS:1
11.ANS:APTS:1
12.ANS:CPTS:1
13.ANS:BPTS:1
14.ANS:DPTS:1
15.ANS:CPTS:1
16.ANS:BPTS:1
17.ANS:APTS:1
18.ANS:CPTS:1
19.ANS:BPTS:1
20.ANS:DPTS:1
21.ANS:APTS:1
22.ANS:APTS:1
23.ANS:APTS:1
24.ANS:DPTS:1
25.ANS:BPTS:1
26.ANS:DPTS:1
27.ANS:CPTS:1
28.ANS:DPTS:1
29.ANS:BPTS:1
Short Answer Problem Solutions
1. So we want to see if our sample value of 11.8 is significantly far from our hypothesized value of 12. We also note that
since the data is normally distributed we can use our Z-stat if we wanted, but since we have only a sample standard
deviation and have n < 30 we will use the t-stat instead.
a) Hypotheses:
H0: μ≥ 12
Ha: μ < 12
b) Critical Values:
dof = n -1 = 25-1 = 24 with α = 0.05
c) TestStat =t-test stat = = = -2; where
d) Since our test stat does lies in the tail, we reject Ho and conclude with 95% confidence that the
mean coffee can size is significantly different from 12 (i.e. so since we got 11.8 it is far enough away from 12 to
say that it is not 12 ounces). We have evidence to suggest it is smaller.
2.
a. = 40
b.s2= Σ [ (38-40)2 + (40-40)2+ …+ (46-40)2 + (42-40)2] / 5 = 22.4
So s ≈4.73
c. Since we have as sample size of 6 then the df = n-1 = 5. We use a t-distribution since we have an unknown
variance and n < 30. So we reject H0 if t > 2.015
d.
e. Test Stat: t= = ≈1.04
So we do not reject H0 since the value does not lie in the critical region. It is less than the critical value in c (i.e. it is not in
the tail.
3.
a. Table
Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F-StatTreatments / 126 / 8-1= 7
(for the number of groups) / 126/7 = 18 / 18/4 = 4.5
Error / 240 / 67-7 = 60 / 240/60 =4
Total / 126+240=366 / 67
b.
(1) Ho: μ1= μ2=…=μ8 (i.e. there is no difference in the mean for the 8 groups)
Ha: At least one of the means is different
(2) Critical Value: Taken from table A.8 since α = 0.01 with p-1 = 7 numerator degrees of freedom and n-p =68-8 = 60
Denominator degrees of freedom – 2.95
(3) Test Stat – F = MST/MSE = 18/4=4.5
(4) Conclusion – Since 4.5>2.95 (i.e. it is in the tail) we reject Ho and conclude there is enough evidence
to suggest that there is some difference with at least one of the treatment means from the others.
4.
Day\Treatment / Store 1 / Store 2 / Store 31 / 9 / 10 / 6
2 / 8 / 11 / 7
3 / 7 / 10 / 8
4 / 8 / 13 / 11
Avg / 8 / 11 / 8
The first thing we want to do is get the overall mean of all the values
= 9
So we know that SST = = 4[(8 – 9)2 + (11 – 9)2 + (8 – 9-)2] = 24
SSE = = [(9-8)2 + (8-8)2 + (7-8)2 + (8-8)2 ] + [(10-11)2 + (11-11)2
+ (10-11)2 + (13-11)2 ] + [(6-8)2 + (7-8)2 + (8-8)2 + (11-8)2 ] = 22
MST= SST /( p-1) = 24/2=12
MSE = SSE / n-p = 22/9= 2.44
(1) Ho: μA = μB =μC (ie there is no difference in the mean test score for each study method)
Ha: At least one of the means is different
(2) Critical Value: Taken from table A.6 with p-1 = 2 numerator degrees of freedom and n-p =12-4 = 8 denominator
degrees offreedom – 4.26
(3) Test Stat – F = MST/MSE = 12/2.44 ≈4.92
(4) Conclusion – Since 4.92> 4.26 (i.e. it is in the tail) we fail to reject Ho and conclude there is enough evidence
to suggest that there is a difference in sales between the 3 store location on a daily basis with 95% confidence.
5.
a.Reject H0 if z < -1.96 or if z > 1.96
b.= = 0.01767
c.Z = ) / 0.1767 = 4.53;
Since 4.53 > > 1.96 reject H0, there is sufficient evidence at = .05 to conclude that the population proportion is significantly different from 0.5
d.It is about zero. The P ( Z > 4.53) is not even on our z-table since it is so small. It is roughly < 0.004 (note we multiply our lowest p-value of -3.49 by 2 since it a two tailed test to get this)
6.
a.H0: p 0.22
Ha: p < 0.22
= = 0.0207
Z = ) / 0.0207 = -0.97; therefore, do not reject H0, there is not sufficient evidence at = 5% to conclude that fewer than 22% of the population like the new soft drink
b.P ( Z < -1 ) = 0.1660
-since it is only one tailed we don’t have to multiply the value by 2.
7.
a.Since we have larger samples we can use the standard normal if we want (even with the unknown standard deviations. But lets be conservative and use the t-stat. Since the standard deviations are roughly the same (ie the larger is less than 3X the smaller) we can use the pooling estimate as before and note that the df = 50+45-2 = 93…and since we always round down it will be t = 1.98
Since the standard deviations are roughly the same (ie the larger is less than 3X the smaller) we can use the pooling estimate.
= ≈8.24
So
Recall: () ± tα/2 2 ± 1.98= 2 ± 1.17 = 0.83 to 3.17
So we are 95% confident the true population mean lies within 0.83 and 3.17
b. To see if there is a difference in the mean we run the following procedure:
1) Hypotheses:
Ho: u1 - u2 = 0
Ha: u1 - u2≠ 0
2) Critical Values. These are the values used in the CI. So we compare our test stat to ± 1.98. If they are larger or smaller (ie they lie in the tail) then we would reject Ho.
3) Test Stat: t =
4) Conclusion: Since 3.39 > 1.98 (ie it lies in the tail) then we are 95% confident that the mean health fees charged at the two clinics is in fact different from one another.
NOTE: If we would have used the Z-stat instead since we have large samples the only difference would be that instead of t = ± 1.98…we would have used a z = ± 1.96. This is a good example of why when you have very larger samples you default to simply use the standard normal since it is much easier to use and generally gives a very similar result.
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