Math 1342 Sample Test 3 Solutions
Question 1
25 points
A random sample of data was obtained
47
8
68
29
95
90
7
54
3
15
33
49
21
64
4
52
4
44
6
8
79
78
57
80
100
9
48
40
80
16
50
a. Use StatCrunch to generate Summary Statistics
Summary statistics:
Column n
var1
Mean
Variance
Std. dev. Median Range Min Max Q1 Q3 IQR
31 43.451613 977.58925 31.266424
b. Use StatCrunch to create a Box Plot
47
106
3 109
9 68
59
c. Based on your summary statistics, IQR and Box Plot does the Data Contain Outliers?
Min Max Q1 Q3 IQR
3 109
9 68
59
An outlier lives outside of
(q1 – 1.5 IRQ, q3+ 1.5IRQ) = ( 9 – 1.5*59, 68 + 1.5*50)
And none of our data lives there.
d. Use StatCrunch to generate a 90% confidence interval for the mean
90% confidence interval results:
μ : Mean of variable
Standard deviation not specified.
Variable n Sample Mean
var1
31
Std. Err.
L. Limit
U. Limit
43.451613 5.6156155 34.214747 52.688478
e. Use StatCrunch and Hypothesis Testing (P-Value Method) for the claim that the mean is
different from 40 at a 94% confidence level.
Hypothesis test results:
μ : Mean of variable
H0 : μ = 40
HA : μ ≠ 40
Standard deviation not specified.
Variable n Sample Mean
var1
31
Std. Err.
Z-Stat
P-value
43.451613 5.6156155 0.61464552 0.5388
Alpha = 1 – Confidence = 1 – 0.94 = 0.06
Page 418
P-Value > alpha since
0.5388 > 0.06 So do NOT reject the Null Hypothesis
Page 415
Our claim was H1 and we had do NOT REJECT so conclude There is not enough evidence to
support the claim that the mean is different from 40.
Question 2
25 points
A random sample of data was obtained
61
12
6
40
27
38
93
5
13
40
a. Use StatCrunch to generate Summary Statistics
Summary statistics:
Column n Mean Variance
var1
10
Std. dev. Median Range Min Max Q1 Q3 IQR
33.5 766.05556 27.677709
32.5
88
5
b. Use StatCrunch to generate a Stem and Leaf Plot
0 : 56
1 : 23
2:7
3:8
4 : 00
5:
6:1
7:
8:
9:3
c. Use StatCrunch to generate a 95% confidence interval for the mean
95% confidence interval results:
μ : Mean of variable
Variable Sample Mean Std. Err. DF
var1
33.5 8.75246
L. Limit
U. Limit
9 13.70056 53.29944
93 12 40
28
d. Use StatCrunch to generate a 97% confidence interval for the variance
97% confidence interval results:
σ2 : Variance of variable
Variable Sample Var. DF
var1
766.05556
L. Limit
U. Limit
9 336.11211 2952.8578
e. Use StatCrunch and Hypothesis Testing (P-Value Method) for the claim that the mean is less
than 30 at a 96% confidence level.
Hypothesis test results:
μ : Mean of variable
H0 : μ = 30
HA : μ < 30
Variable Sample Mean Std. Err. DF
var1
33.5 8.75246
T-Stat
P-value
9 0.39988758 0.6507
Alpha = 1 – Confidence = 1 – 0.96 = 0.04
Page 418
P-Value > alpha since
0.6507 > 0.04 So do NOT reject the Null Hypothesis
Page 415
Our claim was H1 and we had do NOT REJECT so conclude There is not enough evidence to
support the claim that the mean is less than 30.
Question 3
25 points
A random sample of data was obtained
10.8
9.8
9.8
9.4
8.8
9.8
9.6
9.9
10.0
8.4
9.6
10.0
a. Use StatCrunch to generate Summary Statistics
Summary statistics:
Column n
var1
Mean
Variance
Std. dev.
13 9.6384615 0.3425641 0.58528976
Min Max Q1 Q3 IQR
8.4 10.8 9.4 9.9
b. Based on your summary statistics, IQR does the Data Contain Outliers?
Min Max Q1 Q3 IQR
8.4 10.8 9.4 9.9
0.5
An outlier would live outside of (q1 – 1.5IRQ, q3 + 1.5IRQ)
Or ( 9.4 – 0.75, 9.9 + 0.75) = (8.65, 10.65)
So outliers are 8.4 and 10.8
c. Use StatCrunch to generate a 92% confidence interval for the mean
92% confidence interval results:
μ : Mean of variable
Variable Sample Mean
var1
Std. Err.
DF L. Limit
U. Limit
9.6384615 0.16233017 12 9.3280354 9.9488877
d. Use StatCrunch to generate a 98% confidence interval for the mean
98% confidence interval results:
μ : Mean of variable
Variable Sample Mean
var1
Std. Err.
DF L. Limit
U. Limit
9.6384615 0.16233017 12 9.2032547 10.073668
0.5
9.4
e. Use StatCrunch and Hypothesis Testing (P-Value Method) for the claim that the mean is less
than 10 at a 99% confidence level.
Hypothesis test results:
μ : Mean of variable
H0 : μ = 10
HA : μ < 10
Variable Sample Mean
var1
Std. Err.
DF
T-Stat
9.6384615 0.16233017 12 -2.2271797
P-value
0.0229
Alpha = 1 – Confidence = 1 – 0.99 = 0.01
Page 418
P-Value > alpha since
0.0229 > 0.01 So do NOT reject the Null Hypothesis
Page 415
Our claim was H1 and we had do NOT REJECT so conclude There is not enough evidence to
support the claim that the mean is less than 10.
Question 4
25 points
A survey was done concerning – Do you favor saving the Astrodome? 1 = Yes, 0 = No and the
results were as follows
1
1
1
0
1
0
0
1
1
0
1
1
0
1
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1
1
0
1
1
0
0
0
a. Use StatCrunch to the 93% confidence interval for the proportion of people that FAVOR
saving the Astrodome.
Number of responses = 40, Number of Yes = 22, Number of No = 18
93% confidence interval results:
p : Proportion of successes
Method: Standard-Wald
Proportion Count Total Sample Prop.
p
22
40
Std. Err.
L. Limit
U. Limit
0.55 0.078660664 0.4074739 0.6925261
0
1
0
1
b. Use StatCrunch to the 97% confidence interval for the proportion of people that DO NOT
FAVOR saving the Astrodome.
In this case “success” is the number of NO
97% confidence interval results:
p : Proportion of successes
Method: Standard-Wald
Proportion Count Total Sample Prop.
p
18
40
Std. Err.
L. Limit
U. Limit
0.45 0.078660664 0.27929925 0.62070075
c. Use StatCrunch and Hypothesis Testing (P-Value Method) for the claim that the proportion of
people that favor saving the Astrodome is greater than 50% at 90% confidence level.
Hypothesis test results:
p : Proportion of successes
H0 : p = 0.5
HA : p > 0.5
Proportion Count Total Sample Prop.
p
22
40
Std. Err.
Z-Stat
0.55 0.079056942 0.63245553
P-value
0.2635
Alpha = 1 – Confidence = 1 – 0.90 = 0.1
Page 418
P-Value > alpha since
0.2635 > 0.1 So do NOT reject the Null Hypothesis
Page 415
Our claim was H1 and we had do NOT REJECT so conclude There is not enough evidence to
support the claim that proportion is greater than 50%.
d. Use StatCrunch and Hypothesis Testing (P-Value Method) for the claim that the proportion of
people that favor saving the Astrodome is different from 50% at 90% confidence level.
Hypothesis test results:
p : Proportion of successes
H0 : p = 0.5
HA : p ≠ 0.5
Proportion Count Total Sample Prop.
p
22
40
Std. Err.
Z-Stat
0.55 0.079056942 0.63245553
P-value
0.5271
Alpha = 1 – Confidence = 1 – 0.90 = 0.1
Page 418
P-Value > alpha since
0.5271 > 0.1 So do NOT reject the Null Hypothesis
Page 415
Our claim was H1 and we had do NOT REJECT so conclude There is not enough evidence to
support the claim that proportion is different from 50%.