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CHEE 209 Midterm Solutions
Fall 2007
This version: November 8, 2007
1. (14 marks total) First, we reorder the data (in s−1 ) from highest to lowest:
396, 382, 375, 372, 358, 348, 329, 314, 312. (1 mark)
The median of the data is the 5th entry, 358 s−1 (2 marks). The first quartile Q1 and third quartile
Q3 are found as follows:
9+1 382 + 375
Q1 = = {2.5th entry} = = 378.5s−1 , (2 marks)
4 2
3 329 + 314
Q3 = (9 + 1) = {7.5th entry} = = 321.5s−1 . (2 marks)
4 2
The interquartile range is then IQR = Q1 − Q3 = 378.5 − 321.5 = 57 (1 mark). Therefore the whiskers
can extend to 378.5 + 1.5 · 57 = 464 (1 mark) and 321.5 − 1.5 · 57 = 236 (1 mark). No data points are
greater than 464s−1 or are less than 236s−1 , so there are no outliers (1 mark). Hence the whiskers
extend to 396s−1 and 312s−1 . The box-and-whisker plot is shown in Figure 1.
2
2. (a) (9 marks total) We would like to test if the variance σnew associated with the new manufacturing
−2
technique is significantly different from 4100s . From the data, we have that σnew = 919.2s−2 (1
2
2
mark). The general form for a 95% confidence interval around the variance σ is
(n − 1)s2 (n − 1)s2
< σ2 < 2 .
χ2
n−1,0.025 χn−1,0.975
Here n = 9, so we need to determine the values of χ2 2
8,0.025 (2 marks) and χ8,0.975 (1 mark) to construct
2
our confidence interval. Consulting the appropriate tables, we see that χ8,0.025 = 17.53 (1 mark) and
χ2 2
8,0.975 = 2.18 (1 mark). Our 95% confidence interval for σnew is therefore given (approximately) by
8 · 919.2 2 8 · 919.2
< σnew <
17.53 2.18
2
419 < σnew < 3373. (2 marks)
We conclude that σnew is significantly different from 4100s−2 .
2
(b) (9 marks total) This question involves a hypothesis test concerning the mean. In particular, we
would like to test the null hypothesis H0 : µ = 327.3s−1 (1 mark) against the alternative hypothesis
HA : µ > 327.3s−1 (1 mark). To do so, we compute the test statistic
x−µ 354 − 327.3
√ = √ ≈ 2.642. (2 marks)
s n 30.319/ 9
For this test at the 1% significance level, the fence value t8,0.01 (2 marks) is
t8,0.01 = 2.896. (1 mark)
1
Figure 1: Box-and-whisker plot for NMR transverse relaxation rate R2 (in units of s−1 ). (3 marks)
Since 2.642 < 2.896 (1 mark), we see that the mean is not significantly larger at the 1% significance
level (1 mark), i.e., the 99% confidence level.
(c) (4 marks total) Our assumptions were that:
• The mean and variance associated with the old manufacturing technique are known. (1 mark)
• Valeria’s data comprise random, independent samples (1 mark) from a normal distribution (1
mark). (The data seems to fit these assumptions, since there do not appear to be trends in
the data table, and the boxplot shows no outliers. If anything, the distribution appears slightly
asymmetric.) (1 mark)
3. (a) (7 marks total) The probability of success of each letter, call it p, is p = 0.1 (1 mark). Let X denote
the random variable which describes the number of “successful” letters out of the 35 letters written.
The random variable X has a binomial distribution (1 mark). Since raising exactly $10000 dollars
corresponds to obtaining exactly two successes from the 35 trials (1 mark), we compute the binomial
probability
35 2
P (X = 2) = p (1 − p)33 (2 marks)
2
35 · 34
= (0.1)2 (0.9)33
2
≈ 0.1839. (1 mark)
Therefore, there is approximately an 18% chance of Professor McKay raising exactly $10000 (1 mark).
(b) (3 marks total) Since the mean of a binomial(n, p) distribution is equal to np, we see that the
expected number of “successful” letters is µX = np = 35 · 0.1 = 3.5 (2 marks). Therefore, the expected
amount of money raised is 3.5 · 10000 = $17500 (1 mark).
(c) (8 marks total) The probability of raising $15000 or more is the probability of obtaining greater
than or equal to 3 successes (1 mark). We compute this as follows. First,
P (X ≥ 3) = 1 − P (X = 0) − P (X = 1) − P (X = 2). (1 mark)
We must evaluate the three probabilities appearing on the right-hand side of the above equation. We
2
have
35
P (X = 0) = (0.1)0 (0.9)35 ≈ 0.0250; (2 marks)
0
35
P (X = 1) = (0.1)1 (0.9)34 ≈ 0.0973; (2 marks)
1
35
P (X = 2) = (0.1)2 (0.9)33 ≈ 0.1839.
2
Then we have
P (X ≥ 3) ≈ 1 − 0.0260 − 0.0973 − 0.1839 = 0.694. (1 mark)
Therefore, there is approximately a 69.4% chance of Professor McKay raising $15000 or more.
(d) (4 marks total) With p = 0.12 (1 mark), the probability of Professor Mckay raising $15000 or more
becomes
35 35 35
P (X ≥ 3) = 1 − (0.12)0 (0.88)35 − (0.12)1 (0.88)34 − (0.12)2 (0.88)33
0 1 2
≈ 1 − 0.0114 − 0.0544 − 0.1261
= 0.808. (2 marks)
Therefore, Professor McKay cannot “be 95% sure” of raising $15000 or more (1 mark), only approxi-
mately “80.8% sure.”
(e) (1 mark total) Our assumption was that the outcomes of distinct letters were independent. In
practical terms, this might correspond to the recipients of distinct letters not communicating with one
another (1 mark).
4. (12 marks total) This question requires the use of Bayes’ theorem (1 mark). We want to evaluate the
probability P (Hot Dog | Poisoning). We know the following probabilities:
P (Poisoning | Hot Dog) = 0.35, (1 mark)
P (Poisoning | Pretzel) = 0.05, (1 mark)
P (Poisoning | Restaurant) = 0.001, (1 mark)
P (Hot Dog) = 10/58, (1 mark)
P (Pretzel) = 2/58, (1 mark)
P (Restaurant) = 46/58. (1 mark)
Then, using Bayes’ theorem, we have
P (Poisoning | Hot Dog) · P (Hot Dog)
P (Hot Dog | Poisoning) = . (1 mark)
P (Poisoning)
Both probabilities appearing in the numerator are known. As for the probability appearing in the
denominator, we have
P (Poisoning) = P (Poisoning ∩ Hot Dog) + P (Poisoning ∩ Pretzel) + P (Poisoning ∩ Restaurant)
= P (Poisoning | Hot Dog)P (Hot Dog) + P (Poisoning | Pretzel)P (Pretzel)
+ P (Poisoning | Restaurant)P (Restaurant)
10 2 46
= 0.35 · + 0.05 · + 0.001 ·
58 55 58
≈ 0.0622.
3
Therefore, we have
0.35 · 10/58
P (Hot Dog | Poisoning) ≈ ≈ 97%. (4 marks)
0.0622
4
