Midterm 1 Practice Questions
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IBEHS 4C03: Winter 2024 MIDTERM #1 PRACTICE QUESTIONS| IBEHS 4C03: STATISTICAL METHODS IN BIOMEDICAL ENGINEERING Question 1 [4 marks] The following data are the stress values for a particular strain values on the elastic deformation range. Label the boxplot using Tukey Hinges with the 5-number summary elements of the figure. Also label the whisker values, the IQR, and the outliers (if any). 85 78 99 80 93 82 81 93 73 Question 2 [4 marks] The corrosion resistance of a new coating for Sports Utility Vehicles (SUVs) is modelled using a normal distribution with a mean of 0.04 [mm/year] and a standard deviation of 0.005 [mm/year].
a)
What is the probability that corrosion resistance is greater than 0.047 [mm/year]?
b)
What corrosion resistance is worse for 85% of coatings?
Question 3 [4 marks] 1000 Iron bars are tested, obtaining the number of fractures per unit of volume - The average number of fractures was 60.2. 25 random bars are sampled into a group. The average number of fractures for those in the selected group was 63.6 and the corresponding standard deviation was 8.3. Did the bars in the group get a different number of fractures?
IBEHS 4C03: Winter 2024 Question 4 [6 marks] For a specialized electronic microchip application, a determined number of samples of GaAs were tested, where there were 3 possible crystalline phases (a-GaAs, b-GaAs, g-GaAs) present; resulting in 4 performance types: useless, fair, good, and excellent. The results (number of observations) are displayed in:
a
b
g
Useless
60
56
65
Fair
79
100
77
Good
113
105
132
Excellent
41
45
27
Determine the joint and marginal probabilities:
a
b
g
Marginal Probabilities for rows
Useless
Fair
Good
Excellent
Marginal Probabilities for columns
IBEHS 4C03: Winter 2024 Question 5 [6 marks] Four random samples from the density [
?𝑔/?
3
] of the metal employed for the compressor of a jet engine turbine are presented. Each individual sample mean, and deviation is calculated. Samples Individual 1 2 3 4 1 99 182 99 103 2 113 203 125 117 3 135 102 63 76 4 84 97 118 118 5 102 100 136 113 6 142 117 131 125 7 106 152 115 89 8 136 130 142 112 9 126 132 123 124 ?
̅
115.89 135 116.89 108.55 s 19.86 37.48 23.79 16.48 a)
Compute the 95% Confidence Intervals of the sample means. b)
Interpret your results, include commentary on what this means in terms of the population.
Question 6 [7 marks]
A flying drone manufacturer wants to find if a new alloy increases the ability of the wings to withstand mechanical vibration. This property is tested by mounting the wings on a jig that can be vibrated at a fixed frequency but with various amplitudes. The amplitude at which the wing fails is the criterion used to evaluate the strength of the wing. The distribution of amplitudes of the currently used alloy is normally distributed with a mean of 23 cm and a variance of 36 cm2. A sample of n=16 wings are fabricated with the new alloy and subjected to testing, with the following failure amplitudes:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
25.7
22.8
21.5
28.3
24.2
25.3
26.1
20.9
22.4
22.6
23.5
26.8
26.5
23.1
28.1
20.5
a)
State the null and alternative hypotheses.
b)
State which test you would use.
c)
State the alpha level.
d)
Calculate the test statistic
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IBEHS 4C03: Winter 2024 e)
Support your recommendation based on a 95% confidence interval (2-sided)
Question 7 [9 marks]
Electrical engineers are considering the speed-up of cellular neural networks (CNN) for a parallel general-purpose computing architecture. The data follow (of speed-up times): 3.78
3.35
4.22
4.03
4.64
4.14
4.4
4.82
4.27
4.58
4.93
4.32
4.6
a)
Is there sufficient evidence to reject the claim that the mean speed-up exceeds 4.0? Assume that alpha = 0.05. The sample mean is 4.3132 and the sample standard deviation is 0.4328. i.
State the null and alternative hypotheses.
ii.
Calculate the test statistic
iii.
Using the critical value method identify if there is statistical significance or not.
b)
Given the following plot and statistic, would you assume that this data is normally distributed? Explain why or why not.
Shapiro-Wilk Test for mean difference, p =
0.7301892
c)
Calculate the 95% two-sided CI on mean speed-up time.
IBEHS 4C03: Winter 2024 Question 8 [3 marks]
Batches that consist of 50 coil springs from a production process are checked for conformance to customer requirements. The mean number of nonconforming coil springs in a batch is five. Assume that the number of nonconforming springs in a batch, denoted as X, is a binomial random variable. a) What are n and p? (b) What is P(X ≤ 2)? (c) What is P(X ≥ 49)?
Question 9 [8 marks]
Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Because catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield (does not show a difference). A test is run in the pilot plant and results in the data shown in the summary table below; additional statistical information is provided in the other table. Is there any difference between the mean yields at the 95% Confidence level? Sample Mean
Sample Standard Deviation
Catalyst 1
91.5
94.1
8
92.1
8
95.3
9
91.7
9
89.0
7
94.7
2
89.
21
92.255
2.385
Catalyst 2
89.1
9
80.9
5
80.4
6
93.2
1
87.1
9
87.0
4
81.0
7
92.
75
86.483
5.194
Shapiro-Wilks (Catalyst 1)
p=0.444
Shapiro-Wilks (Catalyst 1)
p=0.180
Levene’s Test
p = 0.065
a)
What statistical test would you use?
b)
What is the null and alternative hypothesis?
c)
What assumptions can be made about the data?
d)
Calculate the test statistic.
e)
Using the critical value method, do you have a statistically significant result?
Question 10 [4 marks]
IBEHS 4C03: Winter 2024 Water is used constantly in a pharmaceutical production facility and it is important that it’s pH is monitored. The water was sampled over 30 days resulting in a sample mean value of 7.1 and a sample standard deviation of 0.8. a)
Calculate the 99% confidence interval for the mean pH.
b)
Hypothetically, your company has a massive budget so you consider sampling more. What will happen to the confidence interval as you do this?
Question 11 [10 marks]
The following data show the part performance satisfaction of the testing team's experience of a new sensor dispositive, as rated on the overall likelihood of them using the sensor after a test of the device (with the rating being 1 out of 10 with 10 = "definitely will use", and 1 "definitely will not use")
Senior worker
8
10
10
10
9
Junior worker
2
2
2
3
4
The purpose of the worker experience test is to determine if there is a difference in ratings between senior and junior workers at the 0.05 alpha level. a)
What is the appropriate statistical test? b)
State an assumption that you are making by using this test. c)
What is the null and alternative hypothesis?
d)
Calculate the test-statistic What is your conclusion based on the critical value method? Question 12 [7 marks]
Students were sampled on overall opinion of an app changed before and after viewing a short video. Students were asked before and after the video to rate their opinion on a scale of 0 to 100. Student
Before
After
1
0
15
2
50
30
3
70
100
4
50
35
5
20
50
6
10
0
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IBEHS 4C03: Winter 2024 7
50
75
The goal is to determine whether watching the video changes how students perceive the app using a 𝛼 = 0.05.
a)
What is the appropriate statistical test? b)
State an assumption that you are making by using this test. c)
What is the null and alternative hypothesis?
d)
Calculate the test-statistic What is your conclusion based on the critical value method? Question 13 [4 marks]
Students at a university were surveyed on what their favourite discipline to study was. The following chart is displayed below. What problems do you note and what would you do to improve the visualization? Question 14 [2 marks]
Given the following Q-Q plot, describe what your conclusions about this dataset would be. Favourite Discipline
Arts
Science
Engineering
Medicine
Social Sciences
Humanities
IBEHS 4C03: Winter 2024 Question 15 [4 marks]
Diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples are taken from each machine with n
1
= 15 and n
2
= 17. The corresponding samples means and sample variances of 8.73 and 0.35 cm for machine 1 and 8.68 and 0.4 cm for machine 2. Assuming that the data is normally distributed and that variances are equal, is there evidence to suggest that the machines produce rods with different mean diameters?
IBEHS 4C03: Winter 2024 Answers Question 1 0.5 marks for each (4 marks max) Minimum sample value: 73 25
th
percentile (1
st
quartile): 80 Median –
50
th
percentile (2
nd
quartile): 82 75
th
percentile (3
rd
quartile): 93 Max sample value: 99 Interquartile Range (IQR) –
Distance from 1
st
to 3
rd
quartile: 13 Whisker values: Lower (Q1- 1.5*IQR = 60.5
) Upper (Q3 + 1.5*IQR = 112.5
) Outliers: None Question 2:
a)
M= 0.04; SD = 0.005 ; Z = (x – M)/SD = 1.4 →
p(z < 2.0) = 0.9192 →
complement rule required: 1-0.9192 = 0.0808
1 mark for correct formula, 1 mark for correct answer
b)
x exceeded by 85% of coatings →
phi = 0.1492 →
z = -1.04 z = (x – 0.04)/0.005 →
x = 0.0348 [mm/year]
1 mark for correct formula, 1 mark for correct answer
Question 3:
Answer:
1 mark for hypothesis statement
H
o
: µ= µ
o
=60.2 H
1
: µ≠ µ
o
≠ 60.2
X (mean of selected group) = 63.6
mo = 63.6 = mean of all bars = hypothesized population mean
S = 8.3
1 mark for correct t statistic (0.5 if right formula, wrong answer)
t = (X – mo) / [ s/(sqrt(n))] = (63.6 – 60.2) / [8.3/5] = 2.05
df = n -1 = 24, alpha = 0.05
1 mark for critical value
C
t
(0.975,24) = 2.064
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IBEHS 4C03: Winter 2024 1 mark for correct interpretation
The test statistic is not less extreme than the critical value, we fail to reject Ho and deny H1.
Question 4:
Answer:
0.5 marks for each cell (except for the marginal probabilities), 6 marks max
(a)
Count the number of samples: 900.
(b)
Divide everything by the number of samples to find the probability distribution.
(c)
Sum the total by columns and rows to find the marginal probability.
a
b
g
Marginal Probabilities for rows
Useless
0.0666
0.0622
0.0722
0.2011
Fair
0.0877
0.1111
0.0855
0.2844
Good
0.1255
0.1166
0.1466
0.3888
Excellent
0.0455
0.05
0.03
0.1255
Marginal Probabilities for columns
0.3255
0.34
0.3344
Question 5:
Answers:
a)
(1 mark for each confidence interval, 0.5 marks if right formula, wrong answer, 4 marks max)
df = 9 -1 = 8 ; alpha = 0.05 --à Ct (8, 0.975) = 2.306 Ct (8, 0.025) = -2.306 s/sqrt(n) 6.62 12.493 7.93 5.4933 ct*s/sqrt(n) 15.266 28.81 18.287 12.668 Lower bound 100.62 106.19 98.603 95.882 Upper bound 131.16 163.81 135.18 121.22 1: (100.62, 131.16) 2: (106.19, 163.81). 3: (98.603, 135.18) 4: (95.882, 121.22)
IBEHS 4C03: Winter 2024 b)
(2 marks)
-
If we sampled many times, 95% of the intervals would cover µ
-
Before we draw the sample, we have a 95% chance of constructing an interval that will cover µ Interval suggests the range of plausible values for µ
Question 6
a)
H
o
: µ= µ
o
, H
1
: µ> µ
o
(1 mark)
b)
One-sided z test (1 mark)
c)
Alpha = 0.05 (1 mark)
d)
?
=
388.3
16
= 24.3
(1 mark)
? =
𝑥
−𝜇
𝜎
√𝑛
=
24.3−23
√36
√16
= 0.865
(1 mark)
?
for 0.95 →
1.645
0.865 ??. 1.645 → 𝑎𝑐𝑐𝑒?? ???? ℎ????ℎ𝑒???
(1 mark)
e)
(
𝑿
̅
- 1.96 𝝈
√?
⁄
, 𝑿
̅
+ 1.96 𝝈
√?
⁄
) (21.36,27.24) (1 mark, 0.5 if right formula but wrong answer)
Question 7 a)
i.
H
o
: µ= µ
o
= 4, H
1
: µ > µ
o
(1 mark)
ii.
? =
𝑥
−𝜇
𝑠
√𝑛
=
4.3132−4
0.4328
√13
= 2.609
(1 mark, -0.25 if formula states 𝝈 𝒊????𝒂? ?? ?
)
iii.
?
for 0.95 with n-1 = 12 DF →
1.782 (1 mark)
2.609 ??. 1.782 → ?𝑒?𝑒𝑐? ???? ℎ????ℎ𝑒???
(1 mark)
b) Assume that it is normally distributed (1 mark)
Justification: q-q plot shows points following 45
o
line (1 mark),
Shapiro-wilk test result shows p-value > 0, the null hypothesis is accepted (1 mark)
c) (
𝑿
̅
- C
T
?
√?
⁄
, 𝑿
̅
+ C
T
?
√?
⁄
) C
T(12,1-0.05/2) = 2.179 (1 mark)
(4.2916,4.575)
(1 mark)
IBEHS 4C03: Winter 2024 Question 8
a) n = 50, p = 5/50 = 0.1 (1 mark)
b) 𝑃(𝑋 ≤ 2) = (
50
0
) 0.1
0
(0.9)
50
+ (
50
1
) 0.1
1
(0.9)
49
+ (
50
2
) 0.1
2
(0.9)
48
= 0.1117
(1 mark)
c) 𝑃(𝑋 ≥ 49) = (
50
49
) 0.1
49
(0.9)
1
+ (
50
50
) 0.1
50
(0.9)
1
= 0
(1 mark)
Question 9
a)
What statistical test would you use?
Difference of means two-sample t-test (1 mark)
b)
What is the null and alternative hypothesis?
𝐻
0
: 𝜇
1
− 𝜇
2
= 0, 𝐻
1
: 𝜇
1
− 𝜇
2
> 0
(1 mark)
c)
What assumptions can be made about the data?
Data for both catalysts is normal (Shapiro-wilk’s p-values > 0.05) (1 mark)
Can assume equal variance (levene’s test) (1 mark)
d)
Calculate the test statistic.
?
𝑃
2
=
(?
?
−1)?
?
2
+(?
?
−1)?
?
2
?
?
−1+?
?
−1
= 16.3
(1 mark)
? =
(𝑥̅
?
−𝑥̅
?
)−(𝜇
?
−𝜇
?
)
√?
𝑝
2
(
1
𝑛
?
+
1
𝑛
?
)
=
(92.255−86.483)−0
√16.3(
1
8
+
1
8
)
= 2.85
(1 mark)
e)
Determine the critical value. Do you have a statistically significant result?
2.145, for df = 8 -1 + 8 – 1 = 14 (1 mark)
2.146 vs. 2.85, result is statistically significant (1 mark)
Question 10 a)
Calculate the 99% confidence interval for the mean pH. What assumptions are you making to calculate this interval?
Assumptions: normality, independence (1 mark each)
(
𝑿
̅
- C
T
?
√?
⁄
, 𝑿
̅
+ C
T
?
√?
⁄
) C
T(29,1-0.01/2) = 2.756 (1 mark)
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IBEHS 4C03: Winter 2024 (6.69,7.50)
(1 mark)
b)
Hypothetically, your company has a massive budget so you consider sampling more. What will happen to the confidence interval as you do this?
-increasing sample size will make the bounds tighter (1 mark)
Question 11 The following data show the part performance satisfaction of the testing team's experience of a new sensor dispositive, as rated on the overall likelihood of them using the sensor after a test of the device (with the rating being 1 out of 10 with 10 = "definitely will use", and 1 "definitely will not use")
Senior worker
8
10
10
10
9
Junior worker
2
2
2
3
4
The purpose of the worker experience test is to determine if there is a difference in ratings between senior and junior workers at the 0.05 alpha level. a)
What is the appropriate statistical test? 1 mark
-Non-parametric independent 2 sample test (Wilcoxon Rank Sum Test/Mann-Whitney U Test)
b)
State an assumption that you are making by using this test. 1 mark
-non-normal data
-data is independent
c)
What is the null and alternative hypothesis?
0.5 marks for each, 1 mark max
-H
o
= median senior = median junior -H
1
= median senior ≠ median junior
d)
Calculate the test-statistic What is your conclusion based on the critical value method? (For your information: The critical value of. Also note; that you do not have to calculate a supporting p-value or confidence interval) 2 mark for correct ranking
IBEHS 4C03: Winter 2024 Worker
Score
Rank (Junior)
Rank (Senior)
Junior
2
2
Junior
2
2
Junior
2
2
Junior
3
4
Junior
4
5
Senior
8
6
Senior
9
7
Senior
10
9
Senior
10
9
Senior
10
9
Rank Sum
15
40
1 mark each for correct calculation for Usenior and Ujunior (0.5 if right formula, wrong answer)
𝑈
𝐽𝑢?𝑖??
= ?
1
?
2
+ 0.5(?
1
+ 1) − 𝑅
𝐽𝑢?𝑖??
= (5)(5) + (0.5)(5 + 1) − 15 = 25
𝑈
𝑆𝑒?𝑖??
= ?
1
?
2
+ 0.5(?
2
+ 1) − 𝑅
𝑆𝑒?𝑖??
= (5)(5) + (0.5)(5 + 1) − 40 = 0
1 mark for correct UStat U
stat = 0
1 mark for correct critical value
U statistic at n1=5,n2=5 at 0.05=2
1 mark for interpretation
Since U
stat < Critical U, there is a significant difference, reject null hypothesis
Question 12 a)
What is the appropriate statistical test? 1 mark
-Non-parametric paired test (Wilcoxon Rank Signed Test)
b)
What is the null and alternative hypothesis?
1 mark
-H
o
= paired median differences = 0 -H
1
= paired median differences ≠ 0
IBEHS 4C03: Winter 2024 c)
Calculate the test-statistic What is your conclusion based on the critical value method? 3 mark for correct ranking (for differences, one for signed rank, 1 for correct statistics)
Student
Before
After
Difference
Rank
Signed Rank
1
0
15
-15
2.5
-2.5
2
50
30
20
4
4
3
70
100
-30
6.5
-6.5
4
50
35
15
2.5
2.5
5
20
50
-30
6.5
-6.5
6
10
0
10
1
1
7
50
75
-25
5
-5
Positive Sum
Negative Sum
19.5
7.5
Select 7.5 as W (test statistic)
1 mark for correct critical value
W statistic at n = 7 at 0.05 = 2
1 mark for interpretation
Since W
stat > Critical W, there is no significant difference, accept null hypothesis
Question 13 Up to 4 points for any of the following Issues
Potential Improvements
-legend font is small
-make legend font larger
-no labels
-add labels
-pie chart is not great visualization tool
-use table/bar plot
-colour for pie slices is poor (lots of similar colour)
-better contrast between slices
-title is not descriptive
Question 14 2 marks total
-The plot is not normal (1 mark), data is off the 45-angle line x=y. (1 mark)
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IBEHS 4C03: Winter 2024 Question 15 ?
𝑃
2
=
(?
1
−1)?
1
2
+(?
2
−1)?
2
2
?
1
−1+?
2
−1
= 0.377
(1 mark)
? =
(𝑥̅
2
−𝑥̅
2
)−(𝜇
2
−𝜇
1
)
√?
𝑝
2
(
1
𝑛
1
+
1
𝑛
2
)
=
(8.68−8.73)−0
√0.377(
1
15
+
1
17
)
= 0.230
(1 mark)
2.042, for df = 15 -1 + 17 – 1 = 30 (1 mark)
2.042 vs. 0.230, accept null hypothesis. There is no statistically significant difference (1 mark)
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