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Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 1 of 22
This exam paper must not be removed from the venue
School of Mathematics & Physics
EXAMINATION
Semester Two Final Examinations, 2019
STAT1201 Analysis of Scientific Data
This paper is for Gatton Campus and St Lucia Campus students.
Examination Duration: 120 minutes
Reading Time: 10 minutes
Exam Conditions:
This is a Central Examination
This is a Closed Book Examination - no materials permitted
During reading time - write only on the rough paper provided
This examination paper will be released to the Library
Materials Permitted In The Exam Venue:
(No electronic aids are permitted e.g. laptops, phones)
Calculators - Casio FX82 series or UQ approved (labelled)
Materials To Be Supplied To Students:
None
Instructions To Students:
There are 54 marks available on this exam from 5 questions.
Write your answers in the spaces provided in pages 2–16 of this examination paper. Show your
working and state conclusions, where appropriate.
Pages 17–22 give formulas and statistical tables. Those pages will not be marked.
Additional exam materials (eg. rough paper) will be provided upon request.
Venue ____________________
Seat Number ________
Student Number |__|__|__|__|__|__|__|__|
Family Name _____________________
First Name _____________________
For Examiner Use Only
Question Mark
1
2
3
4
5
Total ________
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Question 1
14 marks Measurements of the enzyme lactate dehydrogenase (LDH) in the blood were taken on seven subjects before fasting and after fasting. The results (μmol/L) are shown below:
Subject 1 2 3 4 5 6 7
Before 0.95 0.96 0.92 1.07 1.38 0.89 1.06
After 0.76 1.11 0.88 0.90 1.17 0.88 1.00
Decrease 0.19 -0.15 0.04 0.17 0.21 0.01 0.06 Researchers are interested in whether blood LDH lowers during fasting. (a) Calculate the five-number summary of the decreases in LDH and sketch a (horizontal) box plot for the distribution in the space below. [3 marks]
Decrease (µmol/L)
−0.2 −0.1 0.0 0.1 0.2
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(b) The decrease in LDH had a mean of 0.0767 μmol/L with a standard deviation of 0.1270 μmol/L. Carry out a test to determine if there is evidence that blood LDH lowers with fasting. [3 marks]
(c) Construct a 95% confidence interval for the mean decrease in LDH after fasting. [2 marks]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(d) Carry out a sign test to determine if there is evidence that blood LDH lowers with fasting. [2 marks] (e) Carry out a signed-rank test to determine if there is evidence that blood LDH lowers with fasting. [3 marks] (f) Which of the three tests above do you feel is most appropriate for this data? [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Question 2
11 marks A study on the Islands in Semester 1, 2019 explored the effects of increased blood alcohol concentration on reaction time. Participants were randomly selected from three towns and given 60 mL of vodka. The blood alcohol concentration (BAC; g/dL) of participants was measured after 15 minutes and each participant then did a ruler test to measure their reaction distance (cm). The distances were then converted into reaction times (ms). The figure below shows a scatterplot of the results, along with the least-squares line for the relationship:
The output below shows the (edited) results of a linear regression in R for the relationship between reaction time and BAC:
lm(formula = ReactionTime ~ BAC)
Residuals:
Min 1Q Median 3Q Max
-27.0534 -8.7784 -0.8572 5.9786 27.7224
Coefficients:
Estimate Std. Error
(Intercept) 168.405 4.781
BAC 1444.037 272.379
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 12.03 on 65 degrees of freedom
Multiple R-squared: 0.3019, Adjusted R-squared: 0.2911
Blood Alcohol Concentration (g/mL)
Re
ac
tio
n
Ti
m
e
(m
s)
170
180
190
200
210
220
0.005 0.010 0.015 0.020 0.025 0.030
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(a) How many participants were in the study? [1 mark] (b) What is the estimated reaction time for people who have a blood alcohol concentration of 0.01 g/dL? [1 mark] (c) Briefly interpret the value 27.7224 in the regression output. [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(d) Does the regression analysis provide evidence of an association between reaction time and blood alcohol concentration? [3 marks]
(e) Give a 95% confidence interval for the underlying slope of the linear relationship between reaction time and blood alcohol concentration. [2 marks]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(f) The following figures were generated by R to help check the assumptions underlying the linear regression:
Comment on the validity of the assumptions underlying linear regression for this data, with reference to these figures. [3 marks]
Fitted
Re
sid
ua
ls
−20
−10
0
10
20
30
180 190 200 210
Theoretical Quantiles
Re
sid
ua
ls
−20
−10
0
10
20
30
−2 −1 0 1 2
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Question 3
10 marks A 2016 study investigated a possible association between social media use and sleep disturbance among young adults. A total of 1,788 young adults were recruited via random digit dialing and address-based sampling. Participants were asked to estimate the number of total minutes that they used social media per day for personal, non-work related use. Sleep disturbance was quantified using a measure which assessed problems with sleep, difficulty falling asleep, whether sleep was refreshing, and sleep quality over the past 7 days, with values coded as ‘Low’, ‘Medium’ or ‘High’ sleep disturbance. (a) Was this an observational or experimental study? Briefly justify your answer. [1 mark]
(b) There were 590 participants who were coded as having ‘High’ levels of sleep disturbance. Construct and interpret a 95% confidence interval for the proportion of all young adults having high levels of sleep disturbance. [3 marks]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(c) The following table gives the summary of results related to the question of whether social media usage is associated with the level of sleep disturbance:
Sleep Disturbance
Social Media Usage Low Medium High
Low (0-60 mins per day) 419 230 244
High (61+ mins per day) 290 259 346
Total 709 489 590
Based on this table, is there evidence of an association between the levels of sleep disturbance and social media usage? [5 marks] (d) The researchers also assessed social media usage through the number of visits per week to popular platforms. They again split the results into ‘Low’ and ‘High’ values and found strong evidence of an association between sleep disturbance and the number of site visits (p < 0.001). Based on this conclusion, should they recommend that young adults reduce their social media usage to improve their sleep quality? [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Question 4
12 marks A 2017 study into oral health practices asked 42 participants to complete a questionnaire that asked for the frequency of consumption of a range of specific food categories. The researchers used the responses to the questionnaire to estimate the total intake of added sugar (grams/day) in the diet from foods, snacks and beverages. They also asked participants how likely they were to floss daily, with the options ‘Highly unlikely’, ‘Unlikely’, ‘Neutral’, ‘Likely’ and ‘Highly likely’. The figure below shows a boxplot of the amount of added sugar, split by sex.
(a) Briefly describe the shape of the added sugar distribution for females. [1 mark]
Sex
Ad
de
d
Su
ga
r (
gr
am
s/d
ay
)
0
500
1000
1500
Female Male
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(b) Based on the shape of the original distributions, the researchers decided to first log-transform the sugar values, giving a revised boxplot for comparing the sexes:
The summary statistics for the log-transformed values are as follows:
Sex $ $
Female 27 1.716 0.4261
Male 15 1.977 0.4105 Based on the log-transformed values, is there any evidence that males tend to have higher levels of added sugar in their diets? [4 marks]
Sex
log
10
(A
dd
ed
Su
ga
r)
1.0
1.5
2.0
2.5
3.0
Female Male
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(c) The boxplot below shows the same log-transformed sugar values but split now by the response to how likely they were to floss daily.
The researchers conducted a one-way analysis of variance to test for an association between mean added sugar and how likely the person was to floss each day. This gave a sum of squares for the group variable of 2.992 with a residual sum of squares of 4.744. Based on these values and the study design, complete the following ANOVA table. [3 marks]
Source DF SS MS F
Residuals
Total
(d) What is the ' value for this model? Briefly interpret the value. [1 mark]
Daily Flossing
log
10
(A
dd
ed
Su
ga
r)
1.0
1.5
2.0
2.5
3.0
Highly unlikely Unlikely Neutral Likely Highly likely
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(e) Is there evidence of a difference in the mean amount of added sugar across the five levels of flossing? [2 marks] (f) Initially the researchers had coded flossing on a numeric scale from 1 (for ‘highly unlikely’) to 5 (for ‘highly likely’) and were going to use linear regression to test for an association between added sugar and flossing. Briefly explain why this would not have been appropriate. [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Question 5
7 marks A 2011 study used a randomised response technique to estimate the prevalence of negative attitudes towards people with disability. A total of 383 participants were given the following instructions:
If your mother was born in February, March or April, then please reply ‘‘yes’’ to the
following question independently of its content. If your mother, however, was born
in another month, then please answer truthfully. The question then read:
Do you feel uneasy in the presence of people with physical disabilities? There were 127 participants who said ‘Yes’. (a) Draw a tree diagram to describe the process described above. Indicate the probabilities on each branch of the tree, including the unknown probability, , of interest. [2 marks] (b) Suppose is the proportion of all people who would say ‘Yes’ to the randomised response question. Give an equation for in terms of . [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(c) Based on the data, we have an estimate * = ,'-./.. Use this to give an estimate, ̂, of the true proportion of all people who would say that they feel uneasy in the presence of people with physical disabilities. [1 mark] (d) Since the count of people saying ‘Yes’ is Binomial, the standard error of * can be calculated using se(*) = 5*(1 − *) = 0.0241 What is the standard error of the corresponding ̂? [1 mark] (e) Construct a 95% confidence interval for , the true proportion of all people who would say that they feel uneasy in the presence of people with physical disabilities. [2 marks]
END OF EXAMINATION
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Formulas and Statistical Tables
BASICS
x =
P
x j
n
s =
vtP(x j x )2
n 1
SENSITIVE QUESTIONS EXAMPLE
T
No12
Yes121
2
H
No1p
Yesp
1
2
P (A|B ) = P (A \B )
P (B )
EXPECTED VALUES
E(X ) =
X
x
P (X = x )x var(X ) =
X
x
P (X = x )(x µX )2 sd(X ) =pvar(X )
E(aX + b ) = aE(X ) + b sd(aX + b ) = |a |sd(X )
E(X1+X2) = E(X1) +E(X2) var(X1+X2) = var(X1) + var(X2), if independent
STANDARDISING
If X ⇠Normal(µ,) then Z = Xµ ⇠Normal(0,1)
BINOMIAL RANDOM VARIABLES
P (X = x ) =
✓
n
x
◆
p x (1p )nx (but is usually available in tables)
E(X ) = np sd(X ) =
∆
np (1p )
Pˆ =
X
n
E(Pˆ ) = p sd(Pˆ ) =
vtp (1p )
n
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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P -VALUES AND ERRORS
0 10.10.050.01
InconclusiveWeakModerate
Strong
Decision
Retain Reject
H0 is true Correct Type I Error
(1↵) (↵)
H0 is false Type II Error Correct
( ) (1 )
TESTS AND CONFIDENCE INTERVALS BASED ON STANDARD ERRORS
t =
estimatehypothesised
se(estimate)
estimate± t ⇤se(estimate)
se(x ) =
sp
n
se(x 1 x 2) =
vut s 21
n1
+
s 22
n2
se(r ) =
vt1 r 2
n 2
se(pˆ ) =
vt pˆ (1 pˆ )
n
se(pˆ1 pˆ2) =
vt pˆ1(1 pˆ1)
n1
+
pˆ2(1 pˆ2)
n2
Use t for means, correlation and regression. Use z for proportions.
ODDS AND ODDS RATIOS
Odds=
p
1p OR=
Odds for group B
Odds for group A
se(ln(OR)) =
vt 1
a
+
1
b
+
1
c
+
1
d
REGRESSION
y = b0+ b1x y = b0+ b1x + b2x1 x1 =
®
1, if Group B
0, if Group A
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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POOLED VARIANCE
s 2p =
(n11)s 21 + (n21)s 22
n1+n22
ANOVA TABLES
DFT= n 1 DFG= k 1
MS=
SS
DF
R 2 =
SSG
SST
sp =
p
MSR F =
MSG
MSR
BONFERRONI CORRECTION FOR k COMPARISONS
↵=
0.05
k
CHI-SQUARED TESTS
expected=
(row total)⇥ (column total)
overall total
2 =
X (observedexpected)2
expected
df= (# rows1)⇥ (# columns1)
SIGN TEST
Count of positive values is X ⇠Binomial(n ,0.5)
SIGNED-RANK TEST
S = sum of ranks corresponding to positive differences
E(S ) =
n (n +1)
4
sd(S ) =
vtn (n +1)(2n +1)
24
RANK-SUM TEST
W = sum of ranks in Group 1
E(W ) = n1
(n1+n2+1)
2
sd(W ) =
vtn1n2(n1+n2+1)
12
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Binomial Distribution
This table gives P (X x ), where X ⇠Binomial(n ,p ).
p
n x 0.01 0.05 0.10 0.20 0.25 0.30 0.40 0.50
7 1 0.068 0.302 0.522 0.790 0.867 0.918 0.972 0.992
2 0.002 0.044 0.150 0.423 0.555 0.671 0.841 0.938
3 0.004 0.026 0.148 0.244 0.353 0.580 0.773
4 0.003 0.033 0.071 0.126 0.290 0.500
5 0.005 0.013 0.029 0.096 0.227
6 0.001 0.004 0.019 0.062
7 0.002 0.008
Critical values of the F distribution
This table gives f ⇤ such that P (Fn ,d f ⇤) = p .
n
d p 1 2 3 4 5 6 7 8 9
36 0.100 2.85 2.46 2.24 2.11 2.01 1.94 1.89 1.85 1.81
0.050 4.11 3.26 2.87 2.63 2.48 2.36 2.28 2.21 2.15
0.010 7.40 5.25 4.38 3.89 3.57 3.35 3.18 3.05 2.95
0.001 12.8 8.42 6.74 5.84 5.26 4.86 4.56 4.33 4.14
37 0.100 2.85 2.45 2.24 2.10 2.01 1.94 1.89 1.84 1.81
0.050 4.11 3.25 2.86 2.63 2.47 2.36 2.27 2.20 2.14
0.010 7.37 5.23 4.36 3.87 3.56 3.33 3.17 3.04 2.93
0.001 12.8 8.37 6.70 5.80 5.22 4.82 4.53 4.30 4.11
38 0.100 2.84 2.45 2.23 2.10 2.01 1.94 1.88 1.84 1.80
0.050 4.10 3.24 2.85 2.62 2.46 2.35 2.26 2.19 2.14
0.010 7.35 5.21 4.34 3.86 3.54 3.32 3.15 3.02 2.92
0.001 12.7 8.33 6.66 5.76 5.19 4.79 4.49 4.26 4.08
39 0.100 2.84 2.44 2.23 2.09 2.00 1.93 1.88 1.83 1.80
0.050 4.09 3.24 2.85 2.61 2.46 2.34 2.26 2.19 2.13
0.010 7.33 5.19 4.33 3.84 3.53 3.30 3.14 3.01 2.90
0.001 12.7 8.29 6.63 5.73 5.16 4.76 4.46 4.23 4.05
40 0.100 2.84 2.44 2.23 2.09 2.00 1.93 1.87 1.83 1.79
0.050 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 2.12
0.010 7.31 5.18 4.31 3.83 3.51 3.29 3.12 2.99 2.89
0.001 12.6 8.25 6.59 5.70 5.13 4.73 4.44 4.21 4.02
41 0.100 2.83 2.44 2.22 2.09 1.99 1.92 1.87 1.82 1.79
0.050 4.08 3.23 2.83 2.60 2.44 2.33 2.24 2.17 2.12
0.010 7.30 5.16 4.30 3.81 3.50 3.28 3.11 2.98 2.87
0.001 12.6 8.21 6.56 5.67 5.10 4.70 4.41 4.18 4.00
42 0.100 2.83 2.43 2.22 2.08 1.99 1.92 1.86 1.82 1.78
0.050 4.07 3.22 2.83 2.59 2.44 2.32 2.24 2.17 2.11
0.010 7.28 5.15 4.29 3.80 3.49 3.27 3.10 2.97 2.86
0.001 12.5 8.18 6.53 5.64 5.07 4.68 4.38 4.16 3.97
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Probabilities for the Standard Normal distribution
This table gives P (Z z ) for Z ⇠Normal(0,1).
Second decimal place of z
z 0 1 2 3 4 5 6 7 8 9
0.0 0.500 0.496 0.492 0.488 0.484 0.480 0.476 0.472 0.468 0.464
0.1 0.460 0.456 0.452 0.448 0.444 0.440 0.436 0.433 0.429 0.425
0.2 0.421 0.417 0.413 0.409 0.405 0.401 0.397 0.394 0.390 0.386
0.3 0.382 0.378 0.374 0.371 0.367 0.363 0.359 0.356 0.352 0.348
0.4 0.345 0.341 0.337 0.334 0.330 0.326 0.323 0.319 0.316 0.312
0.5 0.309 0.305 0.302 0.298 0.295 0.291 0.288 0.284 0.281 0.278
0.6 0.274 0.271 0.268 0.264 0.261 0.258 0.255 0.251 0.248 0.245
0.7 0.242 0.239 0.236 0.233 0.230 0.227 0.224 0.221 0.218 0.215
0.8 0.212 0.209 0.206 0.203 0.200 0.198 0.195 0.192 0.189 0.187
0.9 0.184 0.181 0.179 0.176 0.174 0.171 0.169 0.166 0.164 0.161
1.0 0.159 0.156 0.154 0.152 0.149 0.147 0.145 0.142 0.140 0.138
1.1 0.136 0.133 0.131 0.129 0.127 0.125 0.123 0.121 0.119 0.117
1.2 0.115 0.113 0.111 0.109 0.107 0.106 0.104 0.102 0.100 0.099
1.3 0.097 0.095 0.093 0.092 0.090 0.089 0.087 0.085 0.084 0.082
1.4 0.081 0.079 0.078 0.076 0.075 0.074 0.072 0.071 0.069 0.068
1.5 0.067 0.066 0.064 0.063 0.062 0.061 0.059 0.058 0.057 0.056
1.6 0.055 0.054 0.053 0.052 0.051 0.049 0.048 0.047 0.046 0.046
1.7 0.045 0.044 0.043 0.042 0.041 0.040 0.039 0.038 0.038 0.037
1.8 0.036 0.035 0.034 0.034 0.033 0.032 0.031 0.031 0.030 0.029
1.9 0.029 0.028 0.027 0.027 0.026 0.026 0.025 0.024 0.024 0.023
2.0 0.023 0.022 0.022 0.021 0.021 0.020 0.020 0.019 0.019 0.018
2.1 0.018 0.017 0.017 0.017 0.016 0.016 0.015 0.015 0.015 0.014
2.2 0.014 0.014 0.013 0.013 0.013 0.012 0.012 0.012 0.011 0.011
2.3 0.011 0.010 0.010 0.010 0.010 0.009 0.009 0.009 0.009 0.008
2.4 0.008 0.008 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006
2.5 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.005 0.005
2.6 0.005 0.005 0.004 0.004 0.004 0.004 0.004 0.004 0.004 0.004
2.7 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003
2.8 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002
2.9 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001
3.0 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
3.1 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
3.2 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
3.3
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Critical values of Student’s T distribution
This table gives t ⇤ such that P (T t ⇤) = p , where T ⇠ Student(df).
Probability p
df 0.25 0.10 0.05 0.025 0.01 0.005 0.001 0.0005 0.0001
1 1.000 3.078 6.314 12.71 31.82 63.66 318.3 636.6 3183
2 0.816 1.886 2.920 4.303 6.965 9.925 22.33 31.60 70.70
3 0.765 1.638 2.353 3.182 4.541 5.841 10.21 12.92 22.20
4 0.741 1.533 2.132 2.776 3.747 4.604 7.173 8.610 13.03
5 0.727 1.476 2.015 2.571 3.365 4.032 5.893 6.869 9.678
6 0.718 1.440 1.943 2.447 3.143 3.707 5.208 5.959 8.025
7 0.711 1.415 1.895 2.365 2.998 3.499 4.785 5.408 7.063
8 0.706 1.397 1.860 2.306 2.896 3.355 4.501 5.041 6.442
9 0.703 1.383 1.833 2.262 2.821 3.250 4.297 4.781 6.010
10 0.700 1.372 1.812 2.228 2.764 3.169 4.144 4.587 5.694
11 0.697 1.363 1.796 2.201 2.718 3.106 4.025 4.437 5.453
12 0.695 1.356 1.782 2.179 2.681 3.055 3.930 4.318 5.263
13 0.694 1.350 1.771 2.160 2.650 3.012 3.852 4.221 5.111
14 0.692 1.345 1.761 2.145 2.624 2.977 3.787 4.140 4.985
15 0.691 1.341 1.753 2.131 2.602 2.947 3.733 4.073 4.880
16 0.690 1.337 1.746 2.120 2.583 2.921 3.686 4.015 4.791
17 0.689 1.333 1.740 2.110 2.567 2.898 3.646 3.965 4.714
18 0.688 1.330 1.734 2.101 2.552 2.878 3.610 3.922 4.648
19 0.688 1.328 1.729 2.093 2.539 2.861 3.579 3.883 4.590
20 0.687 1.325 1.725 2.086 2.528 2.845 3.552 3.850 4.539
30 0.683 1.310 1.697 2.042 2.457 2.750 3.385 3.646 4.234
40 0.681 1.303 1.684 2.021 2.423 2.704 3.307 3.551 4.094
50 0.679 1.299 1.676 2.009 2.403 2.678 3.261 3.496 4.014
60 0.679 1.296 1.671 2.000 2.390 2.660 3.232 3.460 3.962
65 0.678 1.295 1.669 1.997 2.385 2.654 3.220 3.447 3.942
1 0.674 1.282 1.645 1.960 2.326 2.576 3.090 3.291 3.719
Critical values of the 2 distribution
This table gives x ⇤ such that P (X 2 x ⇤) = p , where X 2 ⇠2(df).
Probability p
df 0.975 0.95 0.25 0.10 0.05 0.025 0.01 0.005 0.001
1 0.001 0.004 1.323 2.706 3.841 5.024 6.635 7.879 10.83
2 0.051 0.103 2.773 4.605 5.991 7.378 9.210 10.60 13.82
3 0.216 0.352 4.108 6.251 7.815 9.348 11.34 12.84 16.27
4 0.484 0.711 5.385 7.779 9.488 11.14 13.28 14.86 18.47
5 0.831 1.145 6.626 9.236 11.07 12.83 15.09 16.75 20.52
学霸联盟
学霸联盟
Page 1 of 22
This exam paper must not be removed from the venue
School of Mathematics & Physics
EXAMINATION
Semester Two Final Examinations, 2019
STAT1201 Analysis of Scientific Data
This paper is for Gatton Campus and St Lucia Campus students.
Examination Duration: 120 minutes
Reading Time: 10 minutes
Exam Conditions:
This is a Central Examination
This is a Closed Book Examination - no materials permitted
During reading time - write only on the rough paper provided
This examination paper will be released to the Library
Materials Permitted In The Exam Venue:
(No electronic aids are permitted e.g. laptops, phones)
Calculators - Casio FX82 series or UQ approved (labelled)
Materials To Be Supplied To Students:
None
Instructions To Students:
There are 54 marks available on this exam from 5 questions.
Write your answers in the spaces provided in pages 2–16 of this examination paper. Show your
working and state conclusions, where appropriate.
Pages 17–22 give formulas and statistical tables. Those pages will not be marked.
Additional exam materials (eg. rough paper) will be provided upon request.
Venue ____________________
Seat Number ________
Student Number |__|__|__|__|__|__|__|__|
Family Name _____________________
First Name _____________________
For Examiner Use Only
Question Mark
1
2
3
4
5
Total ________
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 2 of 22
Question 1
14 marks Measurements of the enzyme lactate dehydrogenase (LDH) in the blood were taken on seven subjects before fasting and after fasting. The results (μmol/L) are shown below:
Subject 1 2 3 4 5 6 7
Before 0.95 0.96 0.92 1.07 1.38 0.89 1.06
After 0.76 1.11 0.88 0.90 1.17 0.88 1.00
Decrease 0.19 -0.15 0.04 0.17 0.21 0.01 0.06 Researchers are interested in whether blood LDH lowers during fasting. (a) Calculate the five-number summary of the decreases in LDH and sketch a (horizontal) box plot for the distribution in the space below. [3 marks]
Decrease (µmol/L)
−0.2 −0.1 0.0 0.1 0.2
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 3 of 22
(b) The decrease in LDH had a mean of 0.0767 μmol/L with a standard deviation of 0.1270 μmol/L. Carry out a test to determine if there is evidence that blood LDH lowers with fasting. [3 marks]
(c) Construct a 95% confidence interval for the mean decrease in LDH after fasting. [2 marks]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(d) Carry out a sign test to determine if there is evidence that blood LDH lowers with fasting. [2 marks] (e) Carry out a signed-rank test to determine if there is evidence that blood LDH lowers with fasting. [3 marks] (f) Which of the three tests above do you feel is most appropriate for this data? [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Question 2
11 marks A study on the Islands in Semester 1, 2019 explored the effects of increased blood alcohol concentration on reaction time. Participants were randomly selected from three towns and given 60 mL of vodka. The blood alcohol concentration (BAC; g/dL) of participants was measured after 15 minutes and each participant then did a ruler test to measure their reaction distance (cm). The distances were then converted into reaction times (ms). The figure below shows a scatterplot of the results, along with the least-squares line for the relationship:
The output below shows the (edited) results of a linear regression in R for the relationship between reaction time and BAC:
lm(formula = ReactionTime ~ BAC)
Residuals:
Min 1Q Median 3Q Max
-27.0534 -8.7784 -0.8572 5.9786 27.7224
Coefficients:
Estimate Std. Error
(Intercept) 168.405 4.781
BAC 1444.037 272.379
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 12.03 on 65 degrees of freedom
Multiple R-squared: 0.3019, Adjusted R-squared: 0.2911
Blood Alcohol Concentration (g/mL)
Re
ac
tio
n
Ti
m
e
(m
s)
170
180
190
200
210
220
0.005 0.010 0.015 0.020 0.025 0.030
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(a) How many participants were in the study? [1 mark] (b) What is the estimated reaction time for people who have a blood alcohol concentration of 0.01 g/dL? [1 mark] (c) Briefly interpret the value 27.7224 in the regression output. [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 7 of 22
(d) Does the regression analysis provide evidence of an association between reaction time and blood alcohol concentration? [3 marks]
(e) Give a 95% confidence interval for the underlying slope of the linear relationship between reaction time and blood alcohol concentration. [2 marks]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 8 of 22
(f) The following figures were generated by R to help check the assumptions underlying the linear regression:
Comment on the validity of the assumptions underlying linear regression for this data, with reference to these figures. [3 marks]
Fitted
Re
sid
ua
ls
−20
−10
0
10
20
30
180 190 200 210
Theoretical Quantiles
Re
sid
ua
ls
−20
−10
0
10
20
30
−2 −1 0 1 2
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 9 of 22
Question 3
10 marks A 2016 study investigated a possible association between social media use and sleep disturbance among young adults. A total of 1,788 young adults were recruited via random digit dialing and address-based sampling. Participants were asked to estimate the number of total minutes that they used social media per day for personal, non-work related use. Sleep disturbance was quantified using a measure which assessed problems with sleep, difficulty falling asleep, whether sleep was refreshing, and sleep quality over the past 7 days, with values coded as ‘Low’, ‘Medium’ or ‘High’ sleep disturbance. (a) Was this an observational or experimental study? Briefly justify your answer. [1 mark]
(b) There were 590 participants who were coded as having ‘High’ levels of sleep disturbance. Construct and interpret a 95% confidence interval for the proportion of all young adults having high levels of sleep disturbance. [3 marks]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 10 of 22
(c) The following table gives the summary of results related to the question of whether social media usage is associated with the level of sleep disturbance:
Sleep Disturbance
Social Media Usage Low Medium High
Low (0-60 mins per day) 419 230 244
High (61+ mins per day) 290 259 346
Total 709 489 590
Based on this table, is there evidence of an association between the levels of sleep disturbance and social media usage? [5 marks] (d) The researchers also assessed social media usage through the number of visits per week to popular platforms. They again split the results into ‘Low’ and ‘High’ values and found strong evidence of an association between sleep disturbance and the number of site visits (p < 0.001). Based on this conclusion, should they recommend that young adults reduce their social media usage to improve their sleep quality? [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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Question 4
12 marks A 2017 study into oral health practices asked 42 participants to complete a questionnaire that asked for the frequency of consumption of a range of specific food categories. The researchers used the responses to the questionnaire to estimate the total intake of added sugar (grams/day) in the diet from foods, snacks and beverages. They also asked participants how likely they were to floss daily, with the options ‘Highly unlikely’, ‘Unlikely’, ‘Neutral’, ‘Likely’ and ‘Highly likely’. The figure below shows a boxplot of the amount of added sugar, split by sex.
(a) Briefly describe the shape of the added sugar distribution for females. [1 mark]
Sex
Ad
de
d
Su
ga
r (
gr
am
s/d
ay
)
0
500
1000
1500
Female Male
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 12 of 22
(b) Based on the shape of the original distributions, the researchers decided to first log-transform the sugar values, giving a revised boxplot for comparing the sexes:
The summary statistics for the log-transformed values are as follows:
Sex $ $
Female 27 1.716 0.4261
Male 15 1.977 0.4105 Based on the log-transformed values, is there any evidence that males tend to have higher levels of added sugar in their diets? [4 marks]
Sex
log
10
(A
dd
ed
Su
ga
r)
1.0
1.5
2.0
2.5
3.0
Female Male
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 13 of 22
(c) The boxplot below shows the same log-transformed sugar values but split now by the response to how likely they were to floss daily.
The researchers conducted a one-way analysis of variance to test for an association between mean added sugar and how likely the person was to floss each day. This gave a sum of squares for the group variable of 2.992 with a residual sum of squares of 4.744. Based on these values and the study design, complete the following ANOVA table. [3 marks]
Source DF SS MS F
Residuals
Total
(d) What is the ' value for this model? Briefly interpret the value. [1 mark]
Daily Flossing
log
10
(A
dd
ed
Su
ga
r)
1.0
1.5
2.0
2.5
3.0
Highly unlikely Unlikely Neutral Likely Highly likely
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
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(e) Is there evidence of a difference in the mean amount of added sugar across the five levels of flossing? [2 marks] (f) Initially the researchers had coded flossing on a numeric scale from 1 (for ‘highly unlikely’) to 5 (for ‘highly likely’) and were going to use linear regression to test for an association between added sugar and flossing. Briefly explain why this would not have been appropriate. [1 mark]
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Question 5
7 marks A 2011 study used a randomised response technique to estimate the prevalence of negative attitudes towards people with disability. A total of 383 participants were given the following instructions:
If your mother was born in February, March or April, then please reply ‘‘yes’’ to the
following question independently of its content. If your mother, however, was born
in another month, then please answer truthfully. The question then read:
Do you feel uneasy in the presence of people with physical disabilities? There were 127 participants who said ‘Yes’. (a) Draw a tree diagram to describe the process described above. Indicate the probabilities on each branch of the tree, including the unknown probability, , of interest. [2 marks] (b) Suppose is the proportion of all people who would say ‘Yes’ to the randomised response question. Give an equation for in terms of . [1 mark]
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 16 of 22
(c) Based on the data, we have an estimate * = ,'-./.. Use this to give an estimate, ̂, of the true proportion of all people who would say that they feel uneasy in the presence of people with physical disabilities. [1 mark] (d) Since the count of people saying ‘Yes’ is Binomial, the standard error of * can be calculated using se(*) = 5*(1 − *) = 0.0241 What is the standard error of the corresponding ̂? [1 mark] (e) Construct a 95% confidence interval for , the true proportion of all people who would say that they feel uneasy in the presence of people with physical disabilities. [2 marks]
END OF EXAMINATION
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 17 of 22
Formulas and Statistical Tables
BASICS
x =
P
x j
n
s =
vtP(x j x )2
n 1
SENSITIVE QUESTIONS EXAMPLE
T
No12
Yes121
2
H
No1p
Yesp
1
2
P (A|B ) = P (A \B )
P (B )
EXPECTED VALUES
E(X ) =
X
x
P (X = x )x var(X ) =
X
x
P (X = x )(x µX )2 sd(X ) =pvar(X )
E(aX + b ) = aE(X ) + b sd(aX + b ) = |a |sd(X )
E(X1+X2) = E(X1) +E(X2) var(X1+X2) = var(X1) + var(X2), if independent
STANDARDISING
If X ⇠Normal(µ,) then Z = Xµ ⇠Normal(0,1)
BINOMIAL RANDOM VARIABLES
P (X = x ) =
✓
n
x
◆
p x (1p )nx (but is usually available in tables)
E(X ) = np sd(X ) =
∆
np (1p )
Pˆ =
X
n
E(Pˆ ) = p sd(Pˆ ) =
vtp (1p )
n
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 18 of 22
P -VALUES AND ERRORS
0 10.10.050.01
InconclusiveWeakModerate
Strong
Decision
Retain Reject
H0 is true Correct Type I Error
(1↵) (↵)
H0 is false Type II Error Correct
( ) (1 )
TESTS AND CONFIDENCE INTERVALS BASED ON STANDARD ERRORS
t =
estimatehypothesised
se(estimate)
estimate± t ⇤se(estimate)
se(x ) =
sp
n
se(x 1 x 2) =
vut s 21
n1
+
s 22
n2
se(r ) =
vt1 r 2
n 2
se(pˆ ) =
vt pˆ (1 pˆ )
n
se(pˆ1 pˆ2) =
vt pˆ1(1 pˆ1)
n1
+
pˆ2(1 pˆ2)
n2
Use t for means, correlation and regression. Use z for proportions.
ODDS AND ODDS RATIOS
Odds=
p
1p OR=
Odds for group B
Odds for group A
se(ln(OR)) =
vt 1
a
+
1
b
+
1
c
+
1
d
REGRESSION
y = b0+ b1x y = b0+ b1x + b2x1 x1 =
®
1, if Group B
0, if Group A
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 19 of 22
POOLED VARIANCE
s 2p =
(n11)s 21 + (n21)s 22
n1+n22
ANOVA TABLES
DFT= n 1 DFG= k 1
MS=
SS
DF
R 2 =
SSG
SST
sp =
p
MSR F =
MSG
MSR
BONFERRONI CORRECTION FOR k COMPARISONS
↵=
0.05
k
CHI-SQUARED TESTS
expected=
(row total)⇥ (column total)
overall total
2 =
X (observedexpected)2
expected
df= (# rows1)⇥ (# columns1)
SIGN TEST
Count of positive values is X ⇠Binomial(n ,0.5)
SIGNED-RANK TEST
S = sum of ranks corresponding to positive differences
E(S ) =
n (n +1)
4
sd(S ) =
vtn (n +1)(2n +1)
24
RANK-SUM TEST
W = sum of ranks in Group 1
E(W ) = n1
(n1+n2+1)
2
sd(W ) =
vtn1n2(n1+n2+1)
12
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 20 of 22
Binomial Distribution
This table gives P (X x ), where X ⇠Binomial(n ,p ).
p
n x 0.01 0.05 0.10 0.20 0.25 0.30 0.40 0.50
7 1 0.068 0.302 0.522 0.790 0.867 0.918 0.972 0.992
2 0.002 0.044 0.150 0.423 0.555 0.671 0.841 0.938
3 0.004 0.026 0.148 0.244 0.353 0.580 0.773
4 0.003 0.033 0.071 0.126 0.290 0.500
5 0.005 0.013 0.029 0.096 0.227
6 0.001 0.004 0.019 0.062
7 0.002 0.008
Critical values of the F distribution
This table gives f ⇤ such that P (Fn ,d f ⇤) = p .
n
d p 1 2 3 4 5 6 7 8 9
36 0.100 2.85 2.46 2.24 2.11 2.01 1.94 1.89 1.85 1.81
0.050 4.11 3.26 2.87 2.63 2.48 2.36 2.28 2.21 2.15
0.010 7.40 5.25 4.38 3.89 3.57 3.35 3.18 3.05 2.95
0.001 12.8 8.42 6.74 5.84 5.26 4.86 4.56 4.33 4.14
37 0.100 2.85 2.45 2.24 2.10 2.01 1.94 1.89 1.84 1.81
0.050 4.11 3.25 2.86 2.63 2.47 2.36 2.27 2.20 2.14
0.010 7.37 5.23 4.36 3.87 3.56 3.33 3.17 3.04 2.93
0.001 12.8 8.37 6.70 5.80 5.22 4.82 4.53 4.30 4.11
38 0.100 2.84 2.45 2.23 2.10 2.01 1.94 1.88 1.84 1.80
0.050 4.10 3.24 2.85 2.62 2.46 2.35 2.26 2.19 2.14
0.010 7.35 5.21 4.34 3.86 3.54 3.32 3.15 3.02 2.92
0.001 12.7 8.33 6.66 5.76 5.19 4.79 4.49 4.26 4.08
39 0.100 2.84 2.44 2.23 2.09 2.00 1.93 1.88 1.83 1.80
0.050 4.09 3.24 2.85 2.61 2.46 2.34 2.26 2.19 2.13
0.010 7.33 5.19 4.33 3.84 3.53 3.30 3.14 3.01 2.90
0.001 12.7 8.29 6.63 5.73 5.16 4.76 4.46 4.23 4.05
40 0.100 2.84 2.44 2.23 2.09 2.00 1.93 1.87 1.83 1.79
0.050 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 2.12
0.010 7.31 5.18 4.31 3.83 3.51 3.29 3.12 2.99 2.89
0.001 12.6 8.25 6.59 5.70 5.13 4.73 4.44 4.21 4.02
41 0.100 2.83 2.44 2.22 2.09 1.99 1.92 1.87 1.82 1.79
0.050 4.08 3.23 2.83 2.60 2.44 2.33 2.24 2.17 2.12
0.010 7.30 5.16 4.30 3.81 3.50 3.28 3.11 2.98 2.87
0.001 12.6 8.21 6.56 5.67 5.10 4.70 4.41 4.18 4.00
42 0.100 2.83 2.43 2.22 2.08 1.99 1.92 1.86 1.82 1.78
0.050 4.07 3.22 2.83 2.59 2.44 2.32 2.24 2.17 2.11
0.010 7.28 5.15 4.29 3.80 3.49 3.27 3.10 2.97 2.86
0.001 12.5 8.18 6.53 5.64 5.07 4.68 4.38 4.16 3.97
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 21 of 22
Probabilities for the Standard Normal distribution
This table gives P (Z z ) for Z ⇠Normal(0,1).
Second decimal place of z
z 0 1 2 3 4 5 6 7 8 9
0.0 0.500 0.496 0.492 0.488 0.484 0.480 0.476 0.472 0.468 0.464
0.1 0.460 0.456 0.452 0.448 0.444 0.440 0.436 0.433 0.429 0.425
0.2 0.421 0.417 0.413 0.409 0.405 0.401 0.397 0.394 0.390 0.386
0.3 0.382 0.378 0.374 0.371 0.367 0.363 0.359 0.356 0.352 0.348
0.4 0.345 0.341 0.337 0.334 0.330 0.326 0.323 0.319 0.316 0.312
0.5 0.309 0.305 0.302 0.298 0.295 0.291 0.288 0.284 0.281 0.278
0.6 0.274 0.271 0.268 0.264 0.261 0.258 0.255 0.251 0.248 0.245
0.7 0.242 0.239 0.236 0.233 0.230 0.227 0.224 0.221 0.218 0.215
0.8 0.212 0.209 0.206 0.203 0.200 0.198 0.195 0.192 0.189 0.187
0.9 0.184 0.181 0.179 0.176 0.174 0.171 0.169 0.166 0.164 0.161
1.0 0.159 0.156 0.154 0.152 0.149 0.147 0.145 0.142 0.140 0.138
1.1 0.136 0.133 0.131 0.129 0.127 0.125 0.123 0.121 0.119 0.117
1.2 0.115 0.113 0.111 0.109 0.107 0.106 0.104 0.102 0.100 0.099
1.3 0.097 0.095 0.093 0.092 0.090 0.089 0.087 0.085 0.084 0.082
1.4 0.081 0.079 0.078 0.076 0.075 0.074 0.072 0.071 0.069 0.068
1.5 0.067 0.066 0.064 0.063 0.062 0.061 0.059 0.058 0.057 0.056
1.6 0.055 0.054 0.053 0.052 0.051 0.049 0.048 0.047 0.046 0.046
1.7 0.045 0.044 0.043 0.042 0.041 0.040 0.039 0.038 0.038 0.037
1.8 0.036 0.035 0.034 0.034 0.033 0.032 0.031 0.031 0.030 0.029
1.9 0.029 0.028 0.027 0.027 0.026 0.026 0.025 0.024 0.024 0.023
2.0 0.023 0.022 0.022 0.021 0.021 0.020 0.020 0.019 0.019 0.018
2.1 0.018 0.017 0.017 0.017 0.016 0.016 0.015 0.015 0.015 0.014
2.2 0.014 0.014 0.013 0.013 0.013 0.012 0.012 0.012 0.011 0.011
2.3 0.011 0.010 0.010 0.010 0.010 0.009 0.009 0.009 0.009 0.008
2.4 0.008 0.008 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006
2.5 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.005 0.005
2.6 0.005 0.005 0.004 0.004 0.004 0.004 0.004 0.004 0.004 0.004
2.7 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003
2.8 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002
2.9 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001
3.0 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
3.1 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
3.2 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
3.3
Semester Two Final Examinations, 2019 STAT1201 Analysis of Scientific Data
Page 22 of 22
Critical values of Student’s T distribution
This table gives t ⇤ such that P (T t ⇤) = p , where T ⇠ Student(df).
Probability p
df 0.25 0.10 0.05 0.025 0.01 0.005 0.001 0.0005 0.0001
1 1.000 3.078 6.314 12.71 31.82 63.66 318.3 636.6 3183
2 0.816 1.886 2.920 4.303 6.965 9.925 22.33 31.60 70.70
3 0.765 1.638 2.353 3.182 4.541 5.841 10.21 12.92 22.20
4 0.741 1.533 2.132 2.776 3.747 4.604 7.173 8.610 13.03
5 0.727 1.476 2.015 2.571 3.365 4.032 5.893 6.869 9.678
6 0.718 1.440 1.943 2.447 3.143 3.707 5.208 5.959 8.025
7 0.711 1.415 1.895 2.365 2.998 3.499 4.785 5.408 7.063
8 0.706 1.397 1.860 2.306 2.896 3.355 4.501 5.041 6.442
9 0.703 1.383 1.833 2.262 2.821 3.250 4.297 4.781 6.010
10 0.700 1.372 1.812 2.228 2.764 3.169 4.144 4.587 5.694
11 0.697 1.363 1.796 2.201 2.718 3.106 4.025 4.437 5.453
12 0.695 1.356 1.782 2.179 2.681 3.055 3.930 4.318 5.263
13 0.694 1.350 1.771 2.160 2.650 3.012 3.852 4.221 5.111
14 0.692 1.345 1.761 2.145 2.624 2.977 3.787 4.140 4.985
15 0.691 1.341 1.753 2.131 2.602 2.947 3.733 4.073 4.880
16 0.690 1.337 1.746 2.120 2.583 2.921 3.686 4.015 4.791
17 0.689 1.333 1.740 2.110 2.567 2.898 3.646 3.965 4.714
18 0.688 1.330 1.734 2.101 2.552 2.878 3.610 3.922 4.648
19 0.688 1.328 1.729 2.093 2.539 2.861 3.579 3.883 4.590
20 0.687 1.325 1.725 2.086 2.528 2.845 3.552 3.850 4.539
30 0.683 1.310 1.697 2.042 2.457 2.750 3.385 3.646 4.234
40 0.681 1.303 1.684 2.021 2.423 2.704 3.307 3.551 4.094
50 0.679 1.299 1.676 2.009 2.403 2.678 3.261 3.496 4.014
60 0.679 1.296 1.671 2.000 2.390 2.660 3.232 3.460 3.962
65 0.678 1.295 1.669 1.997 2.385 2.654 3.220 3.447 3.942
1 0.674 1.282 1.645 1.960 2.326 2.576 3.090 3.291 3.719
Critical values of the 2 distribution
This table gives x ⇤ such that P (X 2 x ⇤) = p , where X 2 ⇠2(df).
Probability p
df 0.975 0.95 0.25 0.10 0.05 0.025 0.01 0.005 0.001
1 0.001 0.004 1.323 2.706 3.841 5.024 6.635 7.879 10.83
2 0.051 0.103 2.773 4.605 5.991 7.378 9.210 10.60 13.82
3 0.216 0.352 4.108 6.251 7.815 9.348 11.34 12.84 16.27
4 0.484 0.711 5.385 7.779 9.488 11.14 13.28 14.86 18.47
5 0.831 1.145 6.626 9.236 11.07 12.83 15.09 16.75 20.52
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学霸联盟
