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1 Confidence intervals
1. Basketball. Suppose that against a certain opponent the number of points the MIT basketaball team scores
is normally distributed with unknown mean θ and unknown vari ance, σ2. Suppose that over the course of the
last 10 games between the two teams MIT scored the following points:
59, 62, 59, 74, 70, 61, 62, 66, 62, 75
(a)Compute a 95% t–confidence interval for θ. Does 95% confidence mean that the probability θ is in the interval
you just found is 95%?
(b) Now suppose that you learn that σ2 = 25. Compute a 95% z–confidence interval for
θ. How does this compare to the interval in (a)?
(c)Let X be the number of points scored in a game. Suppose that your friend is a confirmed Bayesian with a
priori belief θ ∼ N (60, 16) and that X ∼ N (θ, 25). He computes a 95% probability interval for θ, given the data in part
(a). How does this interval compare to the intervals in (a) and (b)?
(d) Which of the three intervals constructed above do you prefer? Why?
2.The volume in a set of wine bottles is known to follow a N(µ, 25) distribution. You take a sample of the bottles
and measure their volumes. How many bottles do you have to sample to have a 95% confidence interval for µ
with width 1?
3. Suppose data x1, . . . , xn are i.i.d. and drawn from N(µ, σ2), where µ and σ are unknown.
Suppose a data set is taken and we have n = 49, sample mean x¯= 92 and sample standard deviation s = 0.75.
Find a 90% confidence interval for µ.
4.You do a poll to see what fraction p of the population supports candidate A over candidate B.
(a) How many people do you need to poll to know p to within 1% with 95% confidence.
(b)Let p be the fraction of the population who prefer candidate A. If you poll 400 people, how many have to
prefer candidate A to so that the 90% confidence interval is entirely above p = 0.5.
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Problems

5. Suppose you made 40 confidence intervals with confidence level 95%. About how many of them
would you expect to be “wrong’? That is, how many would not actually contain the parameter
being estimated? Should you be surprised if 10 of them are wrong?
2 χ2 confidence interval
6. Hotel. A statistician chooses 27 randomly selected dates, and when examining the occupancy
records of a particular motel for those dates, finds a standard deviation of
5.86 rooms rented. If the number of rooms rented is normally distributed, find the 95% confidence
interval for the population standard deviation of the number of rooms rented.
3 Bootstrapping
7. Parametric bootstrap
Suppose we have a sample of size 100 drawn from a geom(p) distribution with unknown
p. The MLE estimate for p is given by by pˆ = 1/x¯. Assume for our data x¯ = 3.30, so
pˆ= 1/x¯ = 0.30303.
(a) Outline the steps needed to generate a parametric bootstrap 90% confidence interval.
(b)Suppose the following sorted list consists of 200 bootstrap means computed from a sample of
size 100 drawn from a geometric(0.30303) distribution. Use the list to construct a 90% CI for p.
2.68 2.77 2.79 2.81 2.82 2.84 2.84 2.85 2.88 2.89
2.91 2.91 2.91 2.92 2.94 2.94 2.95 2.97 2.97 2.99
3.00 3.00 3.01 3.01 3.01 3.03 3.04 3.04 3.04 3.04
3.04 3.05 3.06 3.06 3.07 3.07 3.07 3.08 3.08 3.08
3.08 3.09 3.09 3.10 3.11 3.11 3.12 3.13 3.13 3.13
3.13 3.15 3.15 3.15 3.16 3.16 3.16 3.16 3.17 3.17
3.17 3.18 3.20 3.20 3.20 3.21 3.21 3.22 3.23 3.23
3.23 3.23 3.23 3.24 3.24 3.24 3.24 3.25 3.25 3.25
3.25 3.25 3.25 3.26 3.26 3.26 3.26 3.27 3.27 3.27
3.28 3.29 3.29 3.30 3.30 3.30 3.30 3.30 3.30 3.31
3.31 3.32 3.32 3.34 3.34 3.34 3.34 3.35 3.35 3.35
3.35 3.35 3.36 3.36 3.37 3.37 3.37 3.37 3.37 3.37
3.38 3.38 3.39 3.39 3.40 3.40 3.40 3.40 3.41 3.42
3.42 3.42 3.43 3.43 3.43 3.43 3.44 3.44 3.44 3.44
3.44 3.45 3.45 3.45 3.45 3.45 3.45 3.45 3.46 3.46
3.46 3.46 3.47 3.47 3.49 3.49 3.49 3.49 3.49 3.50
3.50 3.50 3.52 3.52 3.52 3.52 3.53 3.54 3.54 3.54
3.55 3.56 3.57 3.58 3.59 3.59 3.60 3.61 3.61 3.61
3.62 3.63 3.65 3.65 3.67 3.67 3.68 3.70 3.72 3.72
3.73 3.73 3.74 3.76 3.78 3.79 3.80 3.86 3.89 3.91
8. Empirical bootstrap
Suppose we had 100 data points x1, . . . x100 with sample median q0ˆ.5= 3.3.
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(a)Outline the steps needed to generate an empirical bootstrap 90% confidence interval for the
median q0.5.
(b)Suppose now that the sorted list in the previous problems consists of 200 empirical bootstrap
medians computed from resamples of size 100 drawn from the original data. Use the list to
construct a 90% CI for q0.5.
4 Linear regression/Least squares
9. Fitting a line to data using theMLE.
Suppose you have bivariate data (x1, y1), . . . , (xn, yn). A common model is that there is a linear
relationship between x and y, so in principle the data should lie exactly along a line. However
since data has random noise and our model is probably not exact this will not be the case. What
we can do is look for the line that best fits the data. To do this we will use a simple linear
regression model.
For bivariate data the simple linear regression model assumes that the xi are not random but that
for some values of the parameters a and bthe value yi is drawn from the random variable
Yi ∼axi + b+ εi
where εi is a normal random variable with mean 0 and variance σ2. We assume all of the random
variables εi are independent.
Notes. 1. The model assumes that σ is a known constant, the same for each εi.
2. We think of εi as the measurement error, so the model says that
yi = axi + b+ random measurement error.
3. Remember that (xi, yi) are not variables. They are values of data.
(a) The distribution of Yi depends on a, b, σ and xi. Of these only a and bare not known.
Give the formula for the likelihood function f (yi |a, b, xi, σ) corresponding to one random
value yi. (Hint: yi −axi −b∼N(0, σ2).)
(b) (i) Suppose we have data (1, 8), (3, 2), (5, 1). Based on our model write down the likelihood
and log likelihood as functions of a, b, and σ.
(ii) For general data (x1, y1), . . . , (xn, yn) give the likelihood and and log likelihood functions (again
as functions of a, b, and σ).
(c) Assume σ is a constant, known value. For the data in part b(i) find the maximum likelihood
estimates for a and b
10. What is the relationship between correlation and least squares fit line?
11. You have bivariate data (xi, yi). You have reason to suspect the data is related by
yi = a/xi + Ui where Ui is a random variable with mean 0 and variance σ2 (the same for
all i).
Find the least squares estimate of a.
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12. Least Squares and MLE. In this problem we will see that the least squares ﬁt of a line is just
the MLE assuming the error terms are normally distributed.
For bivariate data (x1, y1), . . . , (xn, yn), the simple linear regression model says that yi is a random
value generated by a random variable
Yi = axi + b+ εi
where a, b, xi are ﬁxed (not random) values, and εi is a random variable with mean 0 and variance
σ2.
(a) Suppose that each εi ∼N(0, σ2). Show that Yi ∼N(axi + b,σ2).
(b) Give the formula for the pdf fYi (yi) of Yi.
(c) Write down the likelihood of the data as a function of a, b, and σ.
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z Φ(z) z Φ(z) z Φ(z) z Φ(z)
-4.00 0.0000 -2.00 0.0228 0.00 0.5000 2.00 0.9772
-3.95 0.0000 -1.95 0.0256 0.05 0.5199 2.05 0.9798
-3.90 0.0000 -1.90 0.0287 0.10 0.5398 2.10 0.9821
-3.85 0.0001 -1.85 0.0322 0.15 0.5596 2.15 0.9842
-3.80 0.0001 -1.80 0.0359 0.20 0.5793 2.20 0.9861
-3.75 0.0001 -1.75 0.0401 0.25 0.5987 2.25 0.9878
-3.70 0.0001 -1.70 0.0446 0.30 0.6179 2.30 0.9893
-3.65 0.0001 -1.65 0.0495 0.35 0.6368 2.35 0.9906
-3.60 0.0002 -1.60 0.0548 0.40 0.6554 2.40 0.9918
-3.55 0.0002 -1.55 0.0606 0.45 0.6736 2.45 0.9929
-3.50 0.0002 -1.50 0.0668 0.50 0.6915 2.50 0.9938
-3.45 0.0003 -1.45 0.0735 0.55 0.7088 2.55 0.9946
-3.40 0.0003 -1.40 0.0808 0.60 0.7257 2.60 0.9953
-3.35 0.0004 -1.35 0.0885 0.65 0.7422 2.65 0.9960
-3.30 0.0005 -1.30 0.0968 0.70 0.7580 2.70 0.9965
-3.25 0.0006 -1.25 0.1056 0.75 0.7734 2.75 0.9970
-3.20 0.0007 -1.20 0.1151 0.80 0.7881 2.80 0.9974
-3.15 0.0008 -1.15 0.1251 0.85 0.8023 2.85 0.9978
-3.10 0.0010 -1.10 0.1357 0.90 0.8159 2.90 0.9981
-3.05 0.0011 -1.05 0.1469 0.95 0.8289 2.95 0.9984
-3.00 0.0013 -1.00 0.1587 1.00 0.8413 3.00 0.9987
-2.95 0.0016 -0.95 0.1711 1.05 0.8531 3.05 0.9989
-2.90 0.0019 -0.90 0.1841 1.10 0.8643 3.10 0.9990
-2.85 0.0022 -0.85 0.1977 1.15 0.8749 3.15 0.9992
-2.80 0.0026 -0.80 0.2119 1.20 0.8849 3.20 0.9993
-2.75 0.0030 -0.75 0.2266 1.25 0.8944 3.25 0.9994
-2.70 0.0035 -0.70 0.2420 1.30 0.9032 3.30 0.9995
-2.65 0.0040 -0.65 0.2578 1.35 0.9115 3.35 0.9996
-2.60 0.0047 -0.60 0.2743 1.40 0.9192 3.40 0.9997
-2.55 0.0054 -0.55 0.2912 1.45 0.9265 3.45 0.9997
-2.50 0.0062 -0.50 0.3085 1.50 0.9332 3.50 0.9998
-2.45 0.0071 -0.45 0.3264 1.55 0.9394 3.55 0.9998
-2.40 0.0082 -0.40 0.3446 1.60 0.9452 3.60 0.9998
-2.35 0.0094 -0.35 0.3632 1.65 0.9505 3.65 0.9999
-2.30 0.0107 -0.30 0.3821 1.70 0.9554 3.70 0.9999
-2.25 0.0122 -0.25 0.4013 1.75 0.9599 3.75 0.9999
-2.20 0.0139 -0.20 0.4207 1.80 0.9641 3.80 0.9999
-2.15 0.0158 -0.15 0.4404 1.85 0.9678 3.85 0.9999
-2.10 0.0179 -0.10 0.4602 1.90 0.9713 3.90 1.0000
-2.05 0.0202 -0.05 0.4801 1.95 0.9744 3.95 1.0000
Standard normal table of left tail probabilities.
Φ(z) = P (Z ≤z) for N(0, 1).
(Use interpolation to estimate z values
to a 3rd decimal place.)
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dfp 0.005 0.010 0.015 0.020 0.025 0.030 0.040 0.050 0.100 0.200 0.300 0.400 0.500
1 63.66 31.82 21.20 15.89 12.71 10.58 7.92 6.31 3.08 1.38 0.73 0.32 0.00
2 9.92 6.96 5.64 4.85 4.30 3.90 3.32 2.92 1.89 1.06 0.62 0.29 0.00
3 5.84 4.54 3.90 3.48 3.18 2.95 2.61 2.35 1.64 0.98 0.58 0.28 0.00
4 4.60 3.75 3.30 3.00 2.78 2.60 2.33 2.13 1.53 0.94 0.57 0.27 0.00
5 4.03 3.36 3.00 2.76 2.57 2.42 2.19 2.02 1.48 0.92 0.56 0.27 0.00
6 3.71 3.14 2.83 2.61 2.45 2.31 2.10 1.94 1.44 0.91 0.55 0.26 0.00
7 3.50 3.00 2.71 2.52 2.36 2.24 2.05 1.89 1.41 0.90 0.55 0.26 0.00
8 3.36 2.90 2.63 2.45 2.31 2.19 2.00 1.86 1.40 0.89 0.55 0.26 0.00
9 3.25 2.82 2.57 2.40 2.26 2.15 1.97 1.83 1.38 0.88 0.54 0.26 0.00
10 3.17 2.76 2.53 2.36 2.23 2.12 1.95 1.81 1.37 0.88 0.54 0.26 0.00
16 2.92 2.58 2.38 2.24 2.12 2.02 1.87 1.75 1.34 0.86 0.54 0.26 0.00
17 2.90 2.57 2.37 2.22 2.11 2.02 1.86 1.74 1.33 0.86 0.53 0.26 0.00
18 2.88 2.55 2.36 2.21 2.10 2.01 1.86 1.73 1.33 0.86 0.53 0.26 0.00
19 2.86 2.54 2.35 2.20 2.09 2.00 1.85 1.73 1.33 0.86 0.53 0.26 0.00
20 2.85 2.53 2.34 2.20 2.09 1.99 1.84 1.72 1.33 0.86 0.53 0.26 0.00
21 2.83 2.52 2.33 2.19 2.08 1.99 1.84 1.72 1.32 0.86 0.53 0.26 0.00
22 2.82 2.51 2.32 2.18 2.07 1.98 1.84 1.72 1.32 0.86 0.53 0.26 0.00
23 2.81 2.50 2.31 2.18 2.07 1.98 1.83 1.71 1.32 0.86 0.53 0.26 0.00
24 2.80 2.49 2.31 2.17 2.06 1.97 1.83 1.71 1.32 0.86 0.53 0.26 0.00
25 2.79 2.49 2.30 2.17 2.06 1.97 1.82 1.71 1.32 0.86 0.53 0.26 0.00
30 2.75 2.46 2.28 2.15 2.04 1.95 1.81 1.70 1.31 0.85 0.53 0.26 0.00
31 2.74 2.45 2.27 2.14 2.04 1.95 1.81 1.70 1.31 0.85 0.53 0.26 0.00
32 2.74 2.45 2.27 2.14 2.04 1.95 1.81 1.69 1.31 0.85 0.53 0.26 0.00
33 2.73 2.44 2.27 2.14 2.03 1.95 1.81 1.69 1.31 0.85 0.53 0.26 0.00
34 2.73 2.44 2.27 2.14 2.03 1.95 1.80 1.69 1.31 0.85 0.53 0.26 0.00
35 2.72 2.44 2.26 2.13 2.03 1.94 1.80 1.69 1.31 0.85 0.53 0.26 0.00
40 2.70 2.42 2.25 2.12 2.02 1.94 1.80 1.68 1.30 0.85 0.53 0.26 0.00
41 2.70 2.42 2.25 2.12 2.02 1.93 1.80 1.68 1.30 0.85 0.53 0.25 0.00
42 2.70 2.42 2.25 2.12 2.02 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
43 2.70 2.42 2.24 2.12 2.02 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
44 2.69 2.41 2.24 2.12 2.02 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
45 2.69 2.41 2.24 2.12 2.01 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
46 2.69 2.41 2.24 2.11 2.01 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
47 2.68 2.41 2.24 2.11 2.01 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
48 2.68 2.41 2.24 2.11 2.01 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
49 2.68 2.40 2.24 2.11 2.01 1.93 1.79 1.68 1.30 0.85 0.53 0.25 0.00
Table of Student t critical values (right-tail)
The table shows tdf, p = the 1 −p quantile of t(df ).
We only give values for p ≤0.5. Use symmetry to ﬁnd the values for p > 0.5, e.g.
t5, 0.975 = −t5, 0.025
In R notation tdf, p = qt(1-p,df).
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dfp 0.010 0.025 0.050 0.100 0.200 0.300 0.500 0.700 0.800 0.900 0.950 0.975 0.990
1 6.63 5.02 3.84 2.71 1.64 1.07 0.45 0.15 0.06 0.02 0.00 0.00 0.00
2 9.21 7.38 5.99 4.61 3.22 2.41 1.39 0.71 0.45 0.21 0.10 0.05 0.02
3 11.34 9.35 7.81 6.25 4.64 3.66 2.37 1.42 1.01 0.58 0.35 0.22 0.11
4 13.28 11.14 9.49 7.78 5.99 4.88 3.36 2.19 1.65 1.06 0.71 0.48 0.30
5 15.09 12.83 11.07 9.24 7.29 6.06 4.35 3.00 2.34 1.61 1.15 0.83 0.55
6 16.81 14.45 12.59 10.64 8.56 7.23 5.35 3.83 3.07 2.20 1.64 1.24 0.87
7 18.48 16.01 14.07 12.02 9.80 8.38 6.35 4.67 3.82 2.83 2.17 1.69 1.24
8 20.09 17.53 15.51 13.36 11.03 9.52 7.34 5.53 4.59 3.49 2.73 2.18 1.65
9 21.67 19.02 16.92 14.68 12.24 10.66 8.34 6.39 5.38 4.17 3.33 2.70 2.09
10 23.21 20.48 18.31 15.99 13.44 11.78 9.34 7.27 6.18 4.87 3.94 3.25 2.56
16 32.00 28.85 26.30 23.54 20.47 18.42 15.34 12.62 11.15 9.31 7.96 6.91 5.81
17 33.41 30.19 27.59 24.77 21.61 19.51 16.34 13.53 12.00 10.09 8.67 7.56 6.41
18 34.81 31.53 28.87 25.99 22.76 20.60 17.34 14.44 12.86 10.86 9.39 8.23 7.01
19 36.19 32.85 30.14 27.20 23.90 21.69 18.34 15.35 13.72 11.65 10.12 8.91 7.63
20 37.57 34.17 31.41 28.41 25.04 22.77 19.34 16.27 14.58 12.44 10.85 9.59 8.26
21 38.93 35.48 32.67 29.62 26.17 23.86 20.34 17.18 15.44 13.24 11.59 10.28 8.90
22 40.29 36.78 33.92 30.81 27.30 24.94 21.34 18.10 16.31 14.04 12.34 10.98 9.54
23 41.64 38.08 35.17 32.01 28.43 26.02 22.34 19.02 17.19 14.85 13.09 11.69 10.20
24 42.98 39.36 36.42 33.20 29.55 27.10 23.34 19.94 18.06 15.66 13.85 12.40 10.86
25 44.31 40.65 37.65 34.38 30.68 28.17 24.34 20.87 18.94 16.47 14.61 13.12 11.52
30 50.89 46.98 43.77 40.26 36.25 33.53 29.34 25.51 23.36 20.60 18.49 16.79 14.95
31 52.19 48.23 44.99 41.42 37.36 34.60 30.34 26.44 24.26 21.43 19.28 17.54 15.66
32 53.49 49.48 46.19 42.58 38.47 35.66 31.34 27.37 25.15 22.27 20.07 18.29 16.36
33 54.78 50.73 47.40 43.75 39.57 36.73 32.34 28.31 26.04 23.11 20.87 19.05 17.07
34 56.06 51.97 48.60 44.90 40.68 37.80 33.34 29.24 26.94 23.95 21.66 19.81 17.79
35 57.34 53.20 49.80 46.06 41.78 38.86 34.34 30.18 27.84 24.80 22.47 20.57 18.51
40 63.69 59.34 55.76 51.81 47.27 44.16 39.34 34.87 32.34 29.05 26.51 24.43 22.16
41 64.95 60.56 56.94 52.95 48.36 45.22 40.34 35.81 33.25 29.91 27.33 25.21 22.91
42 66.21 61.78 58.12 54.09 49.46 46.28 41.34 36.75 34.16 30.77 28.14 26.00 23.65
43 67.46 62.99 59.30 55.23 50.55 47.34 42.34 37.70 35.07 31.63 28.96 26.79 24.40
44 68.71 64.20 60.48 56.37 51.64 48.40 43.34 38.64 35.97 32.49 29.79 27.57 25.15
45 69.96 65.41 61.66 57.51 52.73 49.45 44.34 39.58 36.88 33.35 30.61 28.37 25.90
46 71.20 66.62 62.83 58.64 53.82 50.51 45.34 40.53 37.80 34.22 31.44 29.16 26.66
47 72.44 67.82 64.00 59.77 54.91 51.56 46.34 41.47 38.71 35.08 32.27 29.96 27.42
48 73.68 69.02 65.17 60.91 55.99 52.62 47.34 42.42 39.62 35.95 33.10 30.75 28.18
49 74.92 70.22 66.34 62.04 57.08 53.67 48.33 43.37 40.53 36.82 33.93 31.55 28.94
Table of χ2 critical values (right-tail)
The table shows cdf, p = the 1 −p quantile of χ2(df ). In R
notation cdf, p = qchisq(1-p, df).
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1 Confidence intervals
To practice for the exam use the t and z-tables supplied at the end of this file. Be sure to learn to use these tables. Note
the t and z-tables give left tail probabilities and the χ2-table gives right tail critical values.
1. (a) We compute the data mean and variance x¯ = 65, s2 = 35.778. The number of degrees of
freedom is 9. We look up the critical value t9,0.025 = 2.262 in the t-table The 95% confidence interval is
√ √
9,0.025
t s t 9,0.025 s
x¯ − √ , x¯ + √ = 65 −2.262 3.5778, 65 + 2.262 3.5778 = [60.721,69.279]
n n
On the exam you will be expected to be able to use the t-table. We won’t ask you to compute
by hand the mean and variance of 10 numbers.
95% confidence means that in 95% of experiments the random interval will contain the true θ. It is
not the probability that θ is in the given interval. That depends on the prior distribution for θ, which
we don’t know.
(b) We can look in the z-table or simply remember that z0.025 = 1.96. The 95% confidence interval is
0.025
z σ 0.025
z σ 1.96 ·5 1.96 ·5
x¯ − √ , x¯ + √ = 65 − , 65 + = [61.901,68.099]
n n
√
10
√
10
This is a narrower interval than in part (a). There are two reasons for this, first the true variance
25 is smaller than the sample variance 35.8 and second, the normal distribution has narrower tails
than the t distribution.
(c) We use the normal-normal update formulas to find the posterior pdf for θ.
σ2
post
1 10 a60 + b65 1
a = , b = , µ = = 64.3, post = = 2.16.
16 25 a + b a + b
The posterior pdf is f (θ|data) = N(64.3, 2.16). The posterior 95% probability interval for
θ is
64.3 −z0.025
√
2.16, 64.3 + z0.025
√
2.16 = [61.442,67.206]
(d) There’s no one correct answer; each method has its own advantages and disadvantages. In
this problem they all give similar answers.
2. Suppose we have taken data x1, . . . , xn with mean x¯.The 95% confidence interval for
σ σ
the mean is x ±z0.025 √
n
. This has width 2 z0.025 √
n
. Setting the width equal to 1 and substitituting
values z0.025 = 1.96 and σ = 5 we get
5 √
2 ·1.96 √
n
= 1 ⇒ n = 19.6.
1
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Solutions

So, n = (19.6)2 = 384. .
x¯−µ
If we use our rule of thumb that z0.025 = 2 we have
√
n/10 = 2 ⇒ n = 400.
3. We need to use the studentized mean t = √ ..
s/ n
We know t ∼ t(n−1) = t(48). So we use the m = 48 line of the t table and find t0.05 = 1.677. Thus,
x¯−µ
P (−1.677 <
s/
√
n
< 1.677 | µ) = 0.90.
Unwinding this, we get the 90% confidence interval for µ is
s s 0.75
7
x¯ −√
n
· 1.677, x¯ + √
n
·1.677 = 92 − 0.75
7
· 1.677, 92 + · 1.677 = [91.82, 92.18].
4. (a) The rule-of-thumb is that a 95% confidence interval is x¯±1/
√
n. To be within
1% we need
1
√
n
= 0.01 ⇒ n = 10000.
Using z0.025 = 1.96 instead the 95% confidence interval is
z0.025
x¯±
2
√
n
.
To be within 1% we need z0.025
2
√
n
= 0.01 ⇒ n = 9604.
Note, we are still using the standard Bernoulli approximation σ ≤1/2.
√
1
(b) The 90% confidence interval is x ±z0.05 · . Since z0.05 = 1.64 and n = 400 our
2 n
confidence interval is
1
x ±1.64 ·
40
= x ±0.041
If this is entirely above 0.5 we have x −0.041 > 0.5, so x > 0.541. Let T be the number
out of 400 who prefer A. We have x = T > 0.541, so T > 216 .
400
5. A 95% confidence means about 5% = 1/20 will be wrong. You’d expect about 2 to be wrong.
With a probability p = 0.05 of being wrong, the number wrong follows a Binomial(40, p)
g
distribution. This has expected value 2, and standard deviation 40(0.05)(0.95) = 1.38.
10 wrong is (10-2)/1.38 = 5.8 standard deviations from the mean. This would be surprising.
2 χ2 confidence interval
6. We have n = 27 and s2 = 5.862. If we fix a hypothesis for σ2 we know
(n −1)s2
2
∼χn −1
σ2
Σ Σ Σ Σ
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We can take square roots to find the 95% confidence interval for σ
[4.6148, 8.0307]
On the exam we will give you enough of a table to compute the critical values you need for
χ2 distributions.
3 Bootstrapping
7. (a) Step 1. We have the point estimate p ≈pˆ= 0.30303.
Step 2. Use the computer to generate many (say 10000) size 100 samples. (These are called the bootstrap
samples.)
Step 3. For each sample compute p∗ = 1/x¯∗and δ∗ = p∗ −pˆ.
Step 4. Sort the δ∗ and find the critical values δ0.95 and δ0.05. (Remember δ0.95 is the 5th percentile etc.)
Step 5. The 90% bootstrap confidence interval for p is
[pˆ −δ0.05, pˆ −δ0.95]
(b) It’s tricky to keep the sides straight here. We work slowly and carefully: The 5th and 95th
percentiles for x¯∗are the 10th and 190th entries
2.89, 3.72
(Here again there is some ambiguity on which entries to use. We will accept using the 11th or the 191st entries
or some interpolation between these entries.)
So the 5th and 95th percentiles for p∗ are
1/3.72 = 0.26882, 1/2.89 = 0.34602
So the 5th and 95th percentiles for δ∗ = p∗ −pâre
We used R to find the critical values. (Or use the χ2 table at the end of this file.)
c025 = qchisq(0.975,26) = 41.923 c975 = qchisq(0.025,26) = 13.844
The 95% confidence interval for σ2 is
(n −1) ·s (n −1) ·s
2 2 26 ·5.86 26 ·5.86
2 2
, = , = [21.2968,64.4926]
c0.025 c0.975 41.923 13.844
−0.034213,
These are also the 0.95 and 0.05 critical values. So
the 90% CI for p is
0.042990
[0.30303 −0.042990, 0.30303 + 0.034213] = [0.26004, 0.33724]
8. (a) The steps are the same as in the previous problem except the bootstrap samples are
generated in different ways.
Σ Σ Σ Σ
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Step 1. We have the point estimate q0.5 ≈qˆ0.5= 3.3.
Step 2. Use the computer to generate many (say 10000) size 100 resamples of the original data.
Step 3. For each sample compute the median q0
∗
.5and δ∗ = q0
∗
.5−qˆ0.5.
Step 4. Sort the δ∗ and ﬁnd the critical values δ0.95 and δ0.05. (Remember δ0.95 is the 5th percentile etc.)
Step 5. The 90% bootstrap conﬁdence interval for q0.5 is
[qˆ0.5−δ0.05, qˆ0.5 −δ0.95]
(b) This is very similar to the previous problem. We proceed slowly and carefully to get terms on the
correct side of the inequalities.
The 5th and 95th percentiles for q0
∗
.5are
2.89, 3.72
So the 5th and 95th percentiles for δ∗ = q0
∗
.5−qˆ0.5 are
[2.89 −3.3, 3.72 −3.3] = [−0.41, 0.42]
These are also the 0.95 and 0.05 critical values.
So the 90% CI for p is
[3.3 −0.42, 3.3 + 0.41] = [2.91, 3.71]
4 Linear regression/Least squares
9. (a) The density fε i for the normal distribution is known.
i i
(y − a x −b) 2
1 −
i
i i ε i i
f (y |a, b, x , σ) = f (y −ax −b) = √ e 2σ2
.
σ 2π
(b) (i) The y values are 8, 2, 1. The likelihood function is a product of the densities found in part
(a)
3 2 2 2
−((8−a−b) +(2−3a−b) +(1−5a−b) )/2σ 2
f (y-data |a, b,σ, x-data) = 1
σ
√
2π
e
3 (8 −a −b)2 + (2 −3a −b)2 + (1 −5a −b)2
ln(f (y-data |a, b,σ, x-data)) = −3 ln(σ) −
2
ln(2π) −
2σ2
(ii) We just copy our answer in part (i) replacing the explicit values of xi and yi by their symbols
n �
1 j =1 j j
n 2
− (y −a x −b) /2σ 2
f (y1, . . . , yn |a, b,σ, x1, . . . , xn ) = √ e
σ 2π
n
n n
j j
2
ln(f (8, 3, 2 |a, b,σ)) = −n ln(σ) − ln(2π) − (y −ax −b) /2σ 2
2 j =1
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5
Post-exam 2 practice solutions, Spring 2014
(c) We set partial derivatives to 0 to try and find the MLE. (Don’t forget that σ is a contstant.)
ln(f (8, 3, 2 |a, b,σ)) = −
∂ −2(8 −a −b) −6(2 −3a −b) −10(1 −5a −b)
∂a 2σ2
=
−70a −18b + 38
2σ2
= 0
⇒ 70a + 18b = 38
−2(8 −a −b) −2(2 −3a −b) −2(1 −5a −b)
ln(f (8, 3, 2 |a, b,σ)) = −
∂
∂b 2σ2
=
−18a −6b + 22
2σ2
= 0
⇒ 18a + 6b = 22
We have two simultaneous equations: 70a + 18b = 38, 18a + 6b = 22. These
are easy to solve, e.g. first eliminate band solve for a. We get
7 107
a = −
4
b=
12
You can use R to plot the data and the regression line you found in part (c). R
code I used to make the plot
Here’s the
x = c(1,3,5)
y = c(8,2,1)
a = -7/4
b = 107/12
plot(x,y,pch=19,col="blue") abline(a=b,b=a,
col="magenta") 1 2 3
x
4 5
y
1
2
3
4
5
6
7
8
10. The correlation between x and y is the same as the coefficient b1 of the best fit line
to the standardized data
i
u = √
sx x
i
, v =
i i
x −x¯ y −y¯
√syy
11. The total squared error is
n .
a 2
S(a) = yi −
x
.
i
Taking the derivative and setting it to 0 gives
/ n 2
S (a) = − y i −
a
xi
= 0
xi
Σ
. Σ
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� �
This implies
a
n 1
x2
i
=
n yi y / x
i i
⇒ a
ˆ= _
xi 1/x2
i
.
12.
(a) We’re given εi ∼N(0, σ2). Since axi + bis a constant, Yi is simply a shift of εi. Thus
E(Yi) = axi + b+ E(εi ) = axi + b
Var(Yi) = Var(εi) = σ2.
Since a shifted normal random variable is still normal (you should be able to show this by
transforming cdf’s) we have
Yi ∼N(axi + b,σ2).
(b) The density for a normal distribution is known
i i
(y − a x −b) 2
1 −
i
Y i
f (y ) = √ 2σ2
e .
2π σ
(c)
f (data |σ, a, b) = fY1 (y1)fY2 (y2) ···fY n (yn)
n
n
n
= (2π)− 2 σ−n exp
−
1
2σ2
2
(yi − axi −b) .
i=1
(d) The log likelihood is
n
n
n
ln(f (data |σ, a, b)) = − log 2π −n log σ − (yi − axi −b) 2
2
1
2σ2 i=1
If σ is constant then the only part of the log likelihood that varies is the sum in the last
term. So, the maximum likelihood is found by maximizing this sum:
n
n
i i
2
− (y − ax − b) .
i=1
Notice the minus sign out front. This is exactly the same as minimizing
n
n
i i
2
(y − ax − b) .
i=1
This last expression is the expression minimized by least squares. Therefore, under our
normality assumptions, the values of a and bare the same for MLE and least squares.
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