Mathematics 376 – Probability and Statistics II Solutions for Problem Set 1 February 4, 2010

7.10/ Since each Y_{i} ∼ N(μ, σ^{2}), the mean of Y_{1}, …, Y_{n} is \overline{Y} ∼ N(μ, σ^{2}/n), and Z = \sqrt{n}(\overline{Y} − μ)/σ has an N(0, 1) distribution (standard normal).

a) With n = 9,

P (μ − .3 < \overline{Y} < μ + .3) = P (−.45 < Z < .45) = 1 − 2(.3264) = .3472

b)

\begin{matrix}n = 25 & P (−.75 < Z < .75) = 1 − 2(.2266) = .5468 \\ n = 36 & P (−.9 < Z < .9) = 1 − 2(.1841) = .6318 \\ n = 49 & P (−1.05 < Z < 1.05) = 1 − 2(.1469) = .7062 \\ n = 64 & P (−1.2 < Z < 1.2) = 1 − 2(.1151) = .7698 \end{matrix}

c) They increase as n increases.

d) The probabilities in this question are smaller and increase more slowly than those in 7.1.

7.13/ We have P (|\overline{Y} − μ| < .5) =

P (−.5\sqrt{10/.4} < \frac{\overline{Y} − μ}{\sqrt{.4/10}} < .5\sqrt{10/.4}) = P (−2.5 < Z < 2.5) = 1 − 2(.0062) = .9876.

7.14/ To get P (−.5\sqrt{n/.4} < Z < .5\sqrt{n/.4}) = .95, from the standard normal table, take .5\sqrt{n/.4} = 1.96, so n = 4(.4)(1.96)^{2} ≐ 6.15. Since n must be an integer, any n ≥ 7 will do.

7.15/ a) E(\overline{X} − \overline{Y}) = E(\overline{X}) − E(\overline{Y}) = μ_{1} − μ_{2} by results from last semester. b) Similarly, because the samples are independent,

V (\overline{X} − \overline{Y}) = V (\overline{X}) + (−1)^{2}V (\overline{Y}) = \frac{σ_{1}^{2}}{m} + \frac{σ_{2}^{2}}{n} .

c) With m = n, we want

1.96 = 1/\sqrt{\frac{σ_{1}^{2}}{n} + \frac{σ_{2}^{2}}{n}} = \sqrt{n}/\sqrt{4.5},

so n = 17.29, or n ≥ 18.

7.16/ With m = n = 10 and σ_{1}^{2} = .4, σ_{2}^{2} = .8,

P (\overline{X} − \overline{Y} ≥ 1) = P (\frac{(\overline{X} − \overline{Y}) − (μ_{1} − μ_{2})}{\sqrt{σ_{1}^{2}/n + σ_{2}^{2}/n}} ≥ \sqrt{n}/\sqrt{σ_{1}^{2} + σ_{2}^{2}}) = P (Z > 2.886) = .0019.

7.21/ a) If S^{2} is the sample variance, then by the results in §7.2, \frac{19S^{2}}{1.4} has a χ^{2}(19) distribution, and

P (S^{2} ≤ b) = P (\frac{19S^{2}}{1.4} ≤ \frac{19b}{1.4}) = .975

From the χ^{2} table, this means that \frac{19b}{1.4} = χ_{.025}^{2}(10) = 32.8523, and b = 2.420. b) Similarly if

P (S^{2} ≥ a) = P (\frac{19S^{2}}{1.4} ≥ \frac{19a}{1.4}) = .975

From the χ^{2} table, this means that \frac{19a}{1.4} = χ_{.975}^{2}(10) = 8.90655, and a = .65627. c) By basic rules for probability, P (a ≤ S ≤ b) = 1 − (.025) − (.025) = .95.

7.30/ Since Z is standard normal, E(Z) = 0 and E(Z^{2}) = 1. (i) By independence of Z and Y, Theorem 5.9 applies, and E(T) = E(Z)E(1/\sqrt{Y/ν}) = 0 · E(1/\sqrt{Y/ν}) = 0. (ii) Similarly, E(T^{2}) = E(Z^{2})E(1/(Y/ν)) = 1 · ν · E(1/Y). Since ν > 2, using the formula for the expected values of powers of Y given in the problem,

E(1/Y) = E(Y^{−1}) = \frac{Γ(ν/2 − 1)}{Γ(ν/2)} · \frac{1}{2}

By the functional equation for the gamma function (i.e. Γ(1 + x) = xΓ(x) from last semester), if x = ν/2 − 1, then Γ(ν/2 − 1)/Γ(ν/2) = \frac{1}{ν/2−1}. This gives

E(1/Y) = \frac{ν}{2(ν/2 − 1)} = \frac{ν}{ν − 2}.

Then

V (T) = E(T^{2}) − E(T)^{2} = \frac{ν}{ν − 2},

since E(T) = 0.

7.37/ a) χ^{2}(5) distribution by Theorem 7.2. b) χ^{2}(4) distribution by Theorem 7.3. c) χ^{2}(5) by part b and Theorem 7.2

7.38/ a) t(5) distribution by definition. b)

2Y_{6}/\sqrt{U} = Y_{6}/\sqrt{U/4}

a standard normal, divided by the square root of a χ^{2}(4) over its number of degrees of freedom. This is t(4) by definition. c) \overline{Y} is N(0, 1/\sqrt{5}), so \sqrt{5} · \overline{Y} is N(0, 1). Hence

2(5\overline{Y}^{2} + Y_{6}^{2})/U = \frac{((\sqrt{5} · \overline{Y})^{2} + Y_{6}^{2})/2)}{(U/4)}

is the quotient of two χ^{2}’s, each divided by its number of degrees of freedom. By definition, this has an F (2, 4) distribution.

8.6/ a) Since θ̂_{1}, θ̂_{2} are unbiased,

E(θ̂_{3}) = aE(θ̂_{1}) + (1 − a)θ̂_{2} = aθ + (1 − a)θ = θ

This shows θ̂_{3} is also unbiased.

b) Since we are given θ̂_{1}, θ̂_{2} are independent,

V (θ̂_{3}) = a^{2}V (θ̂_{1}) + (1 − a)^{2}V (θ̂_{2})

To find the minimum, we apply optimization techniques from one-variable calculus – differentiate with respect to a and set equal zero to find the critical points, etc.:

0 = \frac{d}{da}(a^{2}σ_{1}^{2} + (1 − a)^{2}σ_{2}^{2}) = 2aσ_{1}^{2} − 2(1 − a)σ_{2}^{2}

when a = \frac{σ_{1}^{2}}{σ_{1}^{2}+σ_{2}^{2}} . This is a minimum by the second derivative test since

\frac{d^{2}}{da^{2}} (a^{2}σ_{1}^{2} + (1 − a)^{2}σ_{2}^{2}) = 2σ_{1}^{2} + 2σ_{2}^{2} > 0.

8.8/ a) θ̂_{1} is unbiased since E(θ̂_{1}) = θ. θ̂_{2} is unbiased since E(θ̂_{2}) = θ/2 + θ/2 = θ. θ̂_{3} is unbiased since E(θ̂_{3}) = θ/3 + 2θ/3 = θ. θ̂_{5} is unbiased since E(θ̂_{5}) = θ/3 + θ/3 + θ/3 = θ.

For θ̂_{4}, we need to use the formulas for the pdfs of the order statistics that we saw in class. We have

f(y) = 3(1 − (1 − e^{y/θ}))^{2}e^{y/θ}/θ = \frac{3}{θ}e^{−3y/θ}.

Hence we can recognize that θ̂_{4} also has an exponential distribution, but with parameter θ/3. Hence E(θ̂_{4}) = θ/3 so θ̂_{4} is a biased estimator for θ. b) By standard facts for variances (note samples are independent!)

\begin{aligned}V (θ̂_{1}) & = θ^{2} \\ V (θ̂_{2}) & = (\frac{1}{2})^{2} θ^{2} + (\frac{1}{2})^{2} θ^{2} = \frac{θ^{2}}{2} \\ V (θ̂_{3}) & = (\frac{1}{3})^{2} θ^{2} + (\frac{2}{3})^{2} θ^{2} = \frac{5θ^{2}}{9} \\ V (θ̂_{5}) & = 3 · (\frac{1}{3})^{2} θ^{2} = \frac{θ^{2}}{3} \end{aligned}

(General patterns: For a fixed n, the usual mean gives a lower variance than any other weighted average; the variance decreases with n for the usual mean.)

8.10/a) The mean \overline{Y} is one such estimator since

E(\overline{Y}) = \frac{1}{n}(E(Y_{1}) + ⋯ E(Y_{n})) = \frac{1}{n} · nλ = λ.

b) C = Y^{2} + 3Y, so

E(C) = E(Y^{2}) + 3E(Y) = V (Y) + E(Y)^{2} + 3E(Y) = (λ + λ^{2}) + 3λ = λ^{2} + 4λ.

c) We have E(\overline{Y}^{2}) = λ^{2} + λ/n. So one unbiased estimator for C is

\overline{C} = \overline{Y}^{2} + (4 − \frac{1}{n}) \overline{Y}

because

E(\overline{C}) = λ^{2} + λ/n + (4 − \frac{1}{n}) λ = λ^{2} + 4λ.

8.12/a) E(\overline{Y}) = \frac{θ+1 +θ}{2} = θ + \frac{1}{2}, (so \overline{Y} is a biased estimator for θ). b) The bias is B(\overline{Y}) = 1/2. \overline{Y} − 1/2 gives an unbiased estimator for θ. c) Using the formula on page 367:

MSE(\overline{Y}) = V (\overline{Y}) + B(\overline{Y})^{2} = \frac{1}{12n} + \frac{1}{4}

Note that with the interval [θ, θ + 1], the formula for V (Y) for a uniform Y gives V (Y) = 1/12. But we want the variance of the sample mean,

V (\overline{Y}) = \frac{1}{n^{2}} · n · \frac{1}{12} = \frac{1}{12n}.

8.13 a) Using

v̂ = n · \frac{Y}{n} · (1 − \frac{Y}{n} ) = Y − \frac{1}{n}Y^{2}

as estimator for the variance npq of a binomial random variable,

E(v̂) = E(Y) − \frac{1}{n}E(Y^{2}) = np − \frac{1}{n}(npq + n^{2}p^{2}) = npq − pq = (n − 1)pq

This is biased, since V (Y) = npq. But b) – if we use

v̂_{1} = \frac{n^{2}}{n − 1} \frac{Y}{n} (1 − \frac{Y}{n} )

then we get an unbiased estimator.

8.22/ The estimator value is μ̂ = \overline{x} = 7.2. The 2σ bound is 2\frac{s}{\sqrt{n}} = 2\frac{5.6}{200} ≐ .79. (Note that even though the data are given as percents – a form of proportion – the target parameter is a mean, and the \frac{s}{\sqrt{n}} formula for the standard error is used.)

8.23/a) (Mean case) Estimator: μ̂ = 11.3. Error bound: We use the sample variance to estimate the population variance σ: 2\frac{s}{\sqrt{n}} = 2\frac{16.6}{\sqrt{467}} ≐ 1.54.

b) (Difference of means case) Estimator: μ̂_{C} − μ_{R} = 46.4 − 45.1 = 1.3. Error

bound: We use the sample variances to estimate the population variances: 2\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}} ≐

2\sqrt{\frac{(10.2)^{2}}{467} + \frac{(9.8)^{2}}{191}} ≐ 1.7. (Note the interval [1.3 − 1.7, 1.3 + 1.7] contains both positive and negative values. This means the evidence does not support the conclusion μ_{C} − μ_{R} > 0.)

c) (Difference of proportions case) Estimator: p̂_{C} − p_{R} = .78−.61 = .17. Error bound:

We use the sample proportions to estimate p_{i}: 2\sqrt{\frac{p_{1}q_{1}}{n_{1}} + \frac{p_{2}q_{2}}{n_{2}}} ≐ \sqrt{\frac{(.78)(.22)}{467} + \frac{(.61)(.39)}{191}} ≐ .08. (This time, the interval [.17 − .08, .17 + .08] contains only positive values, so evidence indicates p_{C} > p_{R}.)

8.24/ (Proportion case) Estimator: p̂ = .69. The 2σ̂ error bound in this case is 2\sqrt{\frac{(.69)(.31)}{1001}} ≐ .029, which explains the 3% sampling error reported. Since .69 − .3 = .66 is still (much) larger than .5, yes, we can conclude that a majority of the individuals in the 18+ age group thought that the cost of gasoline was a crisis or major problem.

8.25/ (Difference of means case) (a) Estimator: μ̂_{2} − μ_{1} = 3.1 − 2.4 = .7. (b) Error bound:

2\sqrt{\frac{1.44}{100} + \frac{2.64}{100}} ≐ .404.

8.32/ Estimate for the total accounts receivable T is T̂ = 500\overline{Y} = 500(197.1) = 98, 550. Using the sample variance s^{2} = \frac{1}{19} ∑_{i=1}^{20}(Y_{i} − \overline{Y})^{2} ≐ 8255.04 to estimate the population variance, and s = \sqrt{8255.04} ≐ 90.86, the 2 standard error bound for the total is

2 · 500 · \frac{90.86}{\sqrt{20}} ≐ 20316.30

For the mean account value (not the total), we use the 2 standard error bound

2 · \frac{90.86}{\sqrt{20}} ≐ 40.63

Since 197.1 + 40.63 = 237.7 < 250, it unlikely that the average account value is as large as 250.

8.33/ Estimator p̂ = \frac{6}{20} = .3. Error bound: 2\sqrt{\frac{(.3)(.7)}{20}} ≐ .205. The interval [.3 − .205, .3 + .205] does include .2 = 1 − .8, so 20% not in compliance, or 80% in compliance is a possible value based on this evidence. (You could also look at the interval .7 ± .205 for the proportion of accounts that do comply.)

8.34/ For the Poisson random variables, E(Y) = V (Y) = λ. This says that if we have samples Y_{1}, …, Y_{n} from a Poisson distribution, then a reasonable estimator for λ is λ̂ = \overline{Y}. (It’s unbiased if nothing else!) Then V (\overline{Y}) = \frac{λ}{n}, so the standard error is σ\overline{Y} ≐ \sqrt{\frac{λ}{n}}. But of course, the exact value of λ is what we’re trying to estimate by sampling – we don’t know it. So we estimate the standard error by replacing the λ with \overline{Y}:

σ\overline{Y} ≐ \sqrt{\frac{\overline{Y}}{n}}

(This is not an exact equality; it’s an approximation that is used in this situation.)

8.36/ For the exponential case, μ = θ, and σ^{2} = θ^{2}. Estimator θ̂ = \overline{Y} = 1020. Error

bound: 2σ_{θ} ≐ 2\frac{\overline{Y}}{\sqrt{n}}.

8.40/ a) For a normal Y with σ^{2} = 1, Y − μ has a known (N(0, 1) distribution) so it can be used as a pivotal quantity: P (−a < Y − μ < a) = P (Y − a < μ < Y + a) = .95 when a = z_{0.25} = 1.96 (from the standard normal table). The interval is [Y − 1.96, Y + 1.96].

b) P (μ < a) = .95 when P (Y −μ > Y −a) = .95, so Y −a = −1.645, and a = Y +1.645. The interval is (−∞, Y + 1.645].

c) P (μ > a) = .95 when P (Y −μ < Y −a) = .95, so Y −a = 1.645, and a = Y −1.645. Interval is [Y − 1.645, +∞).

8.41/ a) \frac{\overline{Y}^{2}}{σ^{2}} has a χ^{2} distribution with ν = 1 degree of freedom. So from the χ^{2} table,

.95 = P (χ_{.975}^{2}(1) < \frac{Y^{2}}{σ^{2}} < χ_{.025}^{2}(1)) = P (.0009821 < \frac{Y^{2}}{σ^{2}} < 5.02389)

Now invert and multiply through by Y^{2}:

= P (.1990Y^{2} < σ^{2} < 1018.2Y^{2}).

b) Similarly,

.95 = P (χ_{.95}^{2}(1) < \frac{Y^{2}}{σ^{2}} ) = P (σ^{2} < \frac{Y^{2}}{χ_{.95}^{2}(1)}) = P (σ^{2} < 254.32Y^{2})

c) Similarly again,

.95 = P (\frac{Y^{2}}{σ^{2}} < χ_{.05}^{2}(1)) = P (σ^{2} > \frac{Y^{2}}{χ_{.05}^{2}(1)}) = P (σ^{2} > .2603Y^{2})

8.58/From the standard normal table, z_{.025} = 1.96. So the interval is

5.4 ± (1.96) · \frac{3.1}{\sqrt{500}} = 5.4 ± .27.

8.59/(a) From the standard normal table, z_{.05} ≐ 1.645. So the interval is .78±1.645\sqrt{\frac{(.78)(.22)}{1030}} = .78 ± .021.

(b) Since the whole confidence interval is in the range p > .75, yes, it is very likely that more than 75% of the “tweens” enjoy ads that are silly in nature.

8.61/ (two-sided 95% confidence interval for a difference of means): μ̂_{1} − μ_{2} = 167.1 − 140.9 = 26.2. Then interval is

[26.2 − (1.96)\sqrt{\frac{(24.3)^{2}}{30} + \frac{(17.6)^{2}}{30}} , 26.2 + (1.96)\sqrt{\frac{(24.3)^{2}}{30} + \frac{(17.6)^{2}}{30}} ] = [15.46, 36.94].

(Note that the interval contains only positive numbers. This indicates that the evidence supports the conclusion that μ_{1} > μ_{2} pretty strongly.)

8.62/(a) From the standard normal table, z_{.005} ≐ 2.575. So the interval is

24.8 − 21.3 ± 2.575\sqrt{\frac{(7.1)^{2}}{33} + \frac{(8.1)^{2}}{41}} = 3.5 ± 4.52.

(b) Since the interval contains both positive and negative numbers, there is not sufficient evidence to conclude that the normal males have longer time to molt than the “split” males.

8.63/(a) From the standard normal table, z_{.025} = 1.96. So the interval is

.78 ± (1.96) · \sqrt{\frac{(.78)(.22)}{1000}} = .78 ± 0.026.

(b) Although .031 is rather close to .026, there is a problem with the error reported in the study. There are no real solutions of the equation

1.96\sqrt{\frac{p(1 − p)}{1000}} = .031(!)

But p = .5 gives something quite close. Since p = .5 maximizes the standard error in this case, the authors of the study were probably just being conservative in reporting the estimation error.

8.64/(a) From the standard normal table, z_{.01} ≐ 2.33. So the interval is

.77 − .35 ± 2.33\sqrt{\frac{(.77)(.23)}{150} + \frac{(.35)(.65)}{340}} = .42 ± .100.

(b) Since the confidence interval is (.32, .52), it is not likely that the difference is as big as .6.

8.65/(a) The confidence interval for the difference of proportions of defectives is

.18 − .12 ± 2.33\sqrt{\frac{(.18)(.82)}{100} + \frac{(.12)(.88)}{100}} = .06 ± .117.

(b) No – the confidence interval contains both positive and negative values, so there is not enough evidence to say that either line produces more defectives.

8.66/(a) The 90% CI for the mean posttest score for the BACC students is

18.50 ± 1.645 · \frac{8.03}{\sqrt{365}} = 18.50 ± .691.

(b) The 95% CI for the difference between the mean posttest scores for BACC and traditionally-taught students is:

18.50 − 16.50 ± 1.96 · \sqrt{\frac{(8.03)^{2}}{365} + \frac{(6.96)^{2}}{298}} = 2.0 ± 1.14.

(c) Yes. The CI contains only positive values, so there is evidence that the BACC students do better on the posttest than the traditionally-taught students.

8.71/Using the 2θ̂ error bound, we want to choose n so that

2 · \frac{10}{\sqrt{n}} = 2,

so n = 100.

8.81/ Start by computing the sample mean and standard deviation: \overline{Y} = 60.8 and s = 7.969. Then since n = 10, we must use the small sample confidence interval:

60.8 ± t_{.025}(9)\frac{7.969}{\sqrt{10}} = 60.8 ± (2.262)(2.52) = 60.8 ± 5.7 = [55.1, 66.5]

8.82/ With n = 20, again the small-sample formulas are indicated: a) confidence interval for the SAT verbal scores:

419 ± t_{.05}(19)\frac{57}{\sqrt{20}} = 419 ± (1.729)(12.1746) = [396.96, 441.04]

b) This interval does include the national average 422. So that number would be consistent with this data. In other words, this data would not support the conclusion that the national mean is different from 422. (But it is also true that any other number in the interval would also be consistent at the 90% confidence level.) c) for the SAT math scores:

455 ± (1.729)\frac{69}{\sqrt{20}} = [428.32, 481.68]

d) Again the value μ = 474 is in the interval, so it would be consistent with the data. In other words, this data would not support the conclusion that the national mean is different from 474.

Extra problem: 1) By definition, if Y has an F (m, n) distribution, then 1/Y has an F (n, m) distribution – the number of degrees of freedom in the numerator and denominator get switched. So

P (Y ≥ f_{α}(m, n)) = α
⇔ P (1/Y ≤ \frac{1}{f_{α}(m, n)}) = α
⇔ P (1/Y ≥ \frac{1}{f_{α}(m, n)}) = 1 − α

So by the definition,

f_{1−α}(n, m) = \frac{1}{f_{α}(m, n)}.

(This says that the other percentage points for F distributions can be obtained from the ones in the table by taking reciprocals, and switching m, n, so that’s why they are not given.)

2) Using this observation,

f_{.99}(7, 10) = \frac{1}{f_{.01}(10, 7)} = \frac{1}{6.62} ≐ .151.

Text problems:

8.88/Compute \overline{Y} = 9 and s = 6.4244 first. With n = 12, we need the small sample formula:

\overline{Y} ± t_{.05}(11)\frac{s}{\sqrt{n}} = 9 ± (1.796)\frac{6.4244}{\sqrt{12}} = 9 ± 3.3308

8.89/ a) We have \overline{Y} = 3.57, s = 3.6692 (the large value of s comes because of the difference between the first data point and the other two(!)). The 90% confidence interval is

\overline{Y} ± t_{.05}(2)\frac{s}{\sqrt{3}} = 3.57 ± (2.920)\frac{3.6692}{\sqrt{3}} = 3.57 ± 6.1858.

Since the quantity measured is a concentration, it cannot be negative. So you could also report this confidence interval as [0, 9.75].

b) (See comments at end). For small sample confidence intervals for differences of means, to get a pivotal quantity with a t-distribution, we have to assume that the subpopulation variances are the same. This means that the best estimator for the variance is a “pooled” estimator which lumps the two subpopulations together:

s_{p}^{2} = \frac{(n_{1} − 1)s_{1}^{2} + (n_{2} − 1)s_{2}^{2}}{(n_{1} + n_{2} − 2)} = \frac{(11)(6.4244)^{2} + (2)(3.6692)^{2}}{13} ≐ 36.9945

So s_{p} = 6.0823. Then the confidence interval for the difference of the means is (9−3.57)±t_{0.5}(13)s_{p}\sqrt{1/n_{1} + 1/n_{2}} = 5.43±(1.771)(6.0823)\sqrt{1/12 + 1/3} = [−1.52, 12.38] We are assuming the subpopulation variances are equal and the LC50 levels are normally distributed, independent, and the samples are random. 8.96/ We have s^{2} ≐ 63.505. To derive the confidence interval for σ^{2}, we use the idea from 8.41 on Problem Set 2: \frac{(n−1)S^{2}}{σ^{2}} has a χ^{2}(n − 1) distribution, so

P (\frac{(n − 1)S^{2}}{χ_{α/2}^{2}} ≤ σ^{2} ≤ \frac{(n − 1)S^{2}}{χ_{1−α/2}^{2}} ) = 1 − α

Here with α = .1 and n = 10,

\frac{(9)(63.505)}{16.919} ≤ σ^{2} ≤ \frac{(9)(63.505)}{3.32511}

or 33.78 ≤ σ^{2} ≤ 171.89. 8.100/ We compute the sample standard deviation s = 186.69. For a 99% upper confidence bound on σ, we proceed as in 8.37 from Problem Set 2 with n = 20:

P (χ_{.99}^{2}(19) ≤ \frac{(19)S^{2}}{σ^{2}} ) = .99

so

P (σ^{2} ≤ \frac{(19)S^{2}}{χ_{.99}^{2}(19)}) = .99 ⇔ P (σ ≤ \frac{\sqrt{19} · S}{\sqrt{χ_{.99}^{2}(19)}}) = .99

We have

\frac{\sqrt{19} · S}{\sqrt{χ_{.99}^{2}(19)}} = \frac{\sqrt{19} · (186.69)}{\sqrt{7.63273}} ≐ 294.549.

Mathematics 376 – Probability and Statistics II Solutions for Problem Set 5 March 18, 2010

9.1/Using the results from Exercise 8.8 we have

\begin{aligned}eff(θ̂_{1}, θ̂_{5}) & = \frac{θ^{2}/3}{θ^{2}} = \frac{1}{3} \\ eff(θ̂_{2}, θ̂_{5}) & = \frac{θ^{2}/3}{θ^{2}/2} = \frac{2}{3} \\ eff(θ̂_{3}, θ̂_{5}) & = \frac{θ^{2}/3}{5θ^{2}/9} = \frac{3}{5} \end{aligned}

(In intuitive terms, this means that θ̂_{5} is the best of all – three times as efficient as θ̂_{1}, and so forth.)

9.2/ a) All of these are weighted averages of the Y_{i}: linear combinations a_{1}Y_{1} + ⋯ + a_{n}Y_{n} with ∑ a_{i} = 1. For any such weighted average

E(a_{1}Y_{1} + ⋯ + a_{n}Y_{n}) = a_{1}E(Y_{1}) + ⋯ + a_{n}E(Y_{n}) = (a_{1} + ⋯ + a_{n})μ = μ,

so this is an unbiased estimator for μ. b) Efficiency of μ̂_{3} relative to μ̂_{1}:

\frac{V (μ̂_{3})}{V (μ̂_{1})} = \frac{\frac{1}{2}σ^{2}}{\frac{1}{n}σ^{2}} = \frac{n}{2}

(That is, the larger n is, the more efficient μ̂_{3} is relative to μ̂_{1}.) Efficiency of μ̂_{3} relative to μ̂_{2}:

eff(μ̂_{1}, μ̂_{3}) = \frac{V (μ̂_{3})}{V (μ̂_{1})} = \frac{(2 · \frac{1}{16} + (n − 2) · \frac{1}{4(n−2)^{2}} ) σ^{2}}{\frac{1}{n}σ^{2}} = \frac{n}{8} + \frac{n}{4(n − 2)}

9.7/ By the given information E(θ̂_{1}) = θ (unbiased), and θ^{2} = MSE(θ̂_{1}) = V (θ̂_{1}) + B(θ̂_{1})^{2} = V (θ̂_{1}). By our general formulas, V (θ̂_{2}) = n · \frac{1}{n^{2}} V (Y_{i}) = θ^{2}/n. So

eff(θ̂_{1}, θ̂_{2}) = \frac{\frac{1}{n}θ^{2}}{θ^{2}} = \frac{1}{n}

9.17/ First, the given estimator is unbiased, since

E(\overline{X} − \overline{Y}) = E(\overline{X}) − E(\overline{Y}) = μ_{1} − μ_{2}.

Then

V (\overline{X} − \overline{Y}) = V (\overline{X}) + V (\overline{Y}) = \frac{σ_{1}^{2}}{n} + \frac{σ_{2}^{2}}{n} .

This clearly → 0 as n → ∞. Hence Theorem 9.1 applies and the family of estimators is consistent.

9.18/ We will use the fact that if X_{1}, …, X_{n} are a random sample from a normal N(μ, σ) population, then

\frac{(n − 1)S^{2}}{σ^{2}} ∼ χ^{2}(n − 1).

Note that the given estimator for σ^{2} is the pooled estimator as in our small sample confidence interval formulas:

S = \frac{(n − 1)S_{X}^{2} + (n − 1)S_{Y}^{2}}{2(n − 1)} .

Hence with the given information

E(S) = \frac{1}{2(n − 1)}E (σ^{2} \frac{(n − 1)S_{X}^{2}}{σ^{2}} + \frac{(n − 1)S_{Y}^{2}}{σ^{2}} )

= \frac{1}{2(n − 1)} (E (σ^{2} \frac{(n − 1)S_{X}^{2}}{σ^{2}} ) + E (σ^{2} \frac{(n − 1)S_{Y}^{2}}{σ^{2}} ))

= \frac{σ^{2}}{2(n − 1)} (E (\frac{(n − 1)S_{X}^{2}}{σ^{2}} ) + E (\frac{(n − 1)S_{Y}^{2}}{σ^{2}} ))

= \frac{σ^{2}}{2(n − 1)}((n − 1) + (n − 1))

= σ^{2},

using the formula for expected value of a χ^{2} random variable (see table inside back cover of text!) This shows that S is an unbiased estimator for σ. Then, to apply Theorem 9.1, we need to know V (S). This is computed by the same kind of

V (S) = \frac{1}{4(n − 1)^{2}} V (σ^{2} \frac{(n − 1)S_{X}^{2}}{σ^{2}} + \frac{(n − 1)S_{Y}^{2}}{σ^{2}} )

= \frac{1}{4(n − 1)^{2}} (V (σ^{2} \frac{(n − 1)S_{X}^{2}}{σ^{2}} ) + V (σ^{2} \frac{(n − 1)S_{Y}^{2}}{σ^{2}} ))

= \frac{σ^{4}}{4(n − 1)^{2}} (V (\frac{(n − 1)S_{X}^{2}}{σ^{2}} ) + V (\frac{(n − 1)S_{Y}^{2}}{σ^{2}} ))

= \frac{σ^{4}}{4(n − 1)^{2}} (2(n − 1) + 2(n − 1))

= \frac{σ^{4}}{n − 1}

by the formula for the variance of a χ^{2} random variable. This tends to 0 as n → ∞. So the family of estimators given by S (for increasing n) is consistent.

9.19/ We use the statement of Theorem 9.1 (or Example 9.2).

E(\overline{Y}) = E(Y) = ∫_{0}^{1} y · θy^{θ−1} dy = \frac{θ}{θ + 1}

So \overline{Y} is an unbiased estimator for \frac{θ}{θ+1}. Then V (\overline{Y}) = \frac{1}{n}V (Y_{i}) and

V (Y) = ∫_{0}^{1} y^{2} · θy^{θ−1} dy − (\frac{θ}{θ + 1})^{2} = \frac{θ}{θ + 2} − (\frac{θ}{θ + 1})^{2}

is finite. Hence lim_{n→∞} V (\overline{Y}) = lim_{n→∞} \frac{1}{n}V (Y_{i}) = 0. This shows that \overline{Y} is a consistent estimator for \frac{θ}{θ+1}. Note: The same kind of argument will show that \overline{Y} is consistent, whenever it is unbiased and the variance of the underlying distribution (i.e. the variance of one of the samples) is finite.

9.43/ We want to show that the factorization criterion from Theorem 9.4 is satisfied. Writing π = ∏_{i=1}^{n} y_{i}, the likelihood function factors as follows (note that the problem says θ is known, so this is just a constant in the formula):

\begin{aligned}L(y_{1}, …, y_{n}|α) & = (\frac{αy_{1}^{α−1}}{θ^{α}} ) ⋯ (\frac{αy_{n}^{α−1}}{θ^{α}} ) \\ & = \frac{α^{n}}{θ^{nα}} π^{α−1} \end{aligned}

This has the form g(π, α)h(y_{1}, …, y_{n}), where h is the constant 1 (this is OK, since the theorem just requires this factor to be independent of α). Hence π is a sufficient statistic for estimation of α.

9.44/ We want to show that the factorization criterion from Theorem 9.4 is satisfied. Writing π = ∏_{i=1}^{n} y_{i}, the likelihood function factors as follows (note that the problem says θ is known, so this is just a constant in the formula):

\begin{aligned}L(y_{1}, …, y_{n}|α) & = (αβ^{α}y_{1}^{−α−1}) ⋯ (αβ^{α}y_{n}^{−α−1}) \\ & = α^{n}β^{nα}π^{−α−1} \end{aligned}

This has the form g(π, α)h(y_{1}, …, y_{n}), where h is the constant 1 (this is OK, since the theorem just requires this factor to be independent of α). Hence π is a sufficient statistic for estimation of α.

9.49/The factorization criterion from Theorem 9.4 is less convenient here, so we can go back to the definition of sufficient statistics. The conditional density for the samples, given the value of the sample maximum is the ratio

\frac{(\frac{1}{θ})n}{n(y/θ)^{n−1}(1/θ)} = \frac{1}{ny^{n−1}} .

Since this is independent of θ, it follows that the sample maximum Y_{(n)} is a sufficient statistic for θ.

9.70/ Using the method of moments, use λ̂ = \overline{Y} (the sample mean).

9.71/ The second moments about 0 are

μ′_{2} = E(Y^{2}) = σ^{2} + μ^{2} = σ^{2}

(since it is given that μ = 0), and

m′_{2} = \frac{1}{n} ∑_{i=1}^{n} Y_{i}^{2}

So, by the method of moments, use

σ̂^{2} = \frac{1}{n} ∑_{i=1}^{n} Y_{i}^{2}.

9.80/ a) The likelihood function is

L(y_{1}, …, y_{n}|λ) = \frac{e^{−λ}λ^{y_{1}}}{y_{1}!} · · · · · \frac{e^{−λ}λ^{y_{n}}}{y_{n}!} = \frac{e^{−nλ}λ^{y_{1}+···+y_{n}}}{y_{1}! ⋯ y_{n}!}

To find the maximum likelihood estimator, we maximize ln(L):

ln(L) = − ln(y_{1}! ⋯ y_{n}!) − nλ + (y_{1} + ⋯ + y_{n}) ln(λ)

as a function of λ. The first term is constant with respect to λ, so the derivative with respect to λ is:

\frac{d}{dλ}(ln(L)) = −n + \frac{(y_{1} + ⋯ + y_{n})}{λ}

This is = 0 when λ = λ̂ = \overline{Y} = (y_{1} + \overline{+}y_{n})/n (sample mean). (We can also see that the second derivative with respect to λ:

\frac{d^{2}}{dλ^{2}} (ln(L)) = \frac{−(y_{1} + ⋯ + y_{n})}{λ^{2}}

is negative (since all y_{i} ≥ 0 from a Poisson sample), so this is a maximum by the second derivative test

b) E(λ̂) = λ, so this is an unbiased estimator. V (λ̂) = \frac{1}{n}λ.

c) Since λ̂ is unbiased and lim_{n→∞} V (λ̂) = lim_{n→∞} \frac{λ}{n} = 0, the estimator is consistent (Theorem 9.1). (See comment in solution for 9.13 above.)

d) By the invariance property of MLE’s (see discussion on page 452), the MLE for P (Y = 0) = e^{−λ} is e^{−\overline{Y}}

9.81/ The pdf is f(y) = \frac{e^{−y/θ}}{θ} for y > 0 and zero otherwise. Hence the likelihood function is

L = \frac{e^{−y_{1}/θ}}{θ} · · · · · \frac{e^{−y_{n}/θ}}{θ} = \frac{e^{−(y_{1}+···+y_{n})/θ}}{θ^{n}}

and

ln(L) = \frac{−(y_{1} + ⋯ + y_{n})}{θ} − n ln(θ)

So

\frac{d}{dθ} (ln(L)) = \frac{(y_{1} + ⋯ + y_{n})}{θ^{2}} − \frac{n}{θ}

This is zero when θ = θ̂_{ML} = (y_{1} + ⋯ + y_{n})/n = \overline{Y}, and when θ = \overline{Y},

\frac{d^{2}}{dθ^{2}} (ln(L)) = \frac{−2(y_{1} + ⋯ + y_{n})}{θ^{3}} + \frac{n}{θ^{2}} = \frac{−2n^{4} + n^{3}}{(\overline{Y})^{2}} < 0

for all n ≥ 1, so we have a maximum of the likelihood function by the second derivative test. By the invariance property, since σ^{2} = θ^{2}, which is a one-to-one function of θ ≥ 0, the maximum likelihood estimator for σ^{2} is

σ̂^{2}_{ML} = (\overline{Y})^{2}

9.83/a) We have f(y) = \frac{1}{2θ+1} for 0 ≤ y ≤ 2θ + 1. Note this says 2θ + 1 > 0, so θ > −1/2. This means the likelihood function:

L(y_{1}, …, y_{n}) = \frac{1}{(2θ + 1)^{n}}

is strictly decreasing (there’s a vertical asymptote of L at θ = −1/2, and the graph of L tends to the horizontal axis as θ increases. In this case there are no critical points of L and to maximize L given the sample data, we have to ask, what is the biggest value of L that is consistent with the information in a sample y_{1}, …, y_{n}? In other words, what is the smallest value of θ (recall L is decreasing!) that is consistent with the samples? The smallest value θ could have if we observe y_{1}, …, y_{n} uniform on the interval [0, 2θ + 1] is if 2θ + 1 = y_{(n)} (the order statistic giving the maximum of the sample values). Hence

θ̂_{ML} = \frac{Y_{(n)} − 1}{2}.

b) By the invariance property, the MLE for σ^{2} = (2θ + 1)^{2}/12 is

σ̂^{2}_{ML} = \frac{(2θ̂_{ML} + 1)^{2}}{12} = \frac{(Y_{(n)})^{2}}{12}

10.2/a) A Type I error occurs if p = .8 (that is, H_{0} is true), but we observe 12 or fewer insomniacs for which the drug induces sleep, so we reject H_{0}. b)

α = P (Type I error) = ∑_{y=0}^{12} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .032

c) A Type II error occurs if p ≠ .8, but we observe 13 or more insomniacs for which the drug induces sleep, so we do not reject H_{0}.

d) If p = .6,

β = P (Type II error) = ∑_{y=13}^{20} \binom{20}{y}(.6)^{y}(.4)^{20−y} ≐ .416

e) If p = .4,

β = P (Type II error) = ∑_{y=13}^{20} \binom{20}{y}(.4)^{y}(.6)^{20−y} ≐ .021

(Note – as we might intuitively expect, here the probability of a Type II error decreases as the probability gets farther away from the H_{0} value of .8.)

10.5/ Y_{1}, Y_{2} are independent and uniformly distributed on the interval (θ, θ + 1). In Test 1, we have the Type I error probability

α = P (reject|H_{0}) = P (Y_{1} > .95|θ = 0) = .05.

Then in Test 2, we must pick c to make α = .05 in that test as well. Under the null hypothesis, the joint density of Y_{1}, Y_{2} equals 1 on the unit square (0, 1) × (0, 1) and zero outside. The lines y_{1} + y_{2} = c have slope −1 and intersect the square as long as 0 < c < 2. Since c = 1 gives P (Y_{1} + Y_{2} > c|θ = 0) = 1/2, we clearly want c > 1. Then the region within the unit square defined by y_{1} + y_{2} > 2 is an isosceles right triangle with side s < 1. The area equals .05 when s^{2}/2 = .05, so s ≐ .316. This means that c can be computed from y_{1} + y_{2} at the point where the line cuts through the right edge of the unit square: c = 1 + (1 − .316) = 1.684.

10.18/Test has H_{0}: μ = 13.20 and H_{a} : μ < 13.20 (note the wording “substandard wages” indicates a one-tailed Z-test). We have RR: z < −z_{.01} = −2.33. The test statistic is

z = \sqrt{40} · \frac{12.20 − 13.20}{2.50} = −2.53

This is in the rejection region, so we reject the null hypothesis; there is evidence to support the statement that the wages are substandard (at the α = .01 level).

10.19/We have H_{0}: θ = 130 and H_{a}: θ < 130 (“... against the alternative that it is less than 130”). The rejection region is z < −z_{.05} = −1.645. The test statistic is:

Z = \frac{128.6 − 130}{2.2/\sqrt{40}} ≐ −4.22.

This is well into the rejection region so we reject H_{0} and conclude there is strong evidence that θ < 130.

10.21/Test has H_{0}: μ_{1} − μ_{2} = 0 and H_{a}: μ_{1} − μ_{2} ≠ 0 (“appears to differ”). The rejection region is |z| > z_{.005} = 2.575. The test statistic is

z = \frac{1.65 − 1.43 − 0}{\sqrt{\frac{(.26)^{2}}{30} + \frac{(.22)^{2}}{35}}} ≐ 3.65.

This is in the rejection region, so the data indicates that there is a difference in the average shear strengths. (Also note the zero in the numerator of the test statistic – that’s the value of μ_{1} − μ_{2} under the assumption that H_{0} is true.)

10.26/The test has H_{0}: p = .45 and H_{a}: p ≠ .45 (two-tailed). With α = .01, the rejection region is |z| > z_{.005} = 2.575. The test statistic is

z = \frac{\frac{32}{80} − .45}{\sqrt{\frac{(.45)(.55)}{80}}} ≐ −.899

Note the way the standard error for the estimator (that is, the \sqrt{pq/n}) is calculated using p = .45 from H_{0}. The z-value is not in the rejection region, so there is not sufficient evidence to conclude that the proportion of brown-eyed people in this region is different from the p = .45 reported in the Washington Post.

10.34/The test has H_{0}: μ = 5 and H_{a}: μ > 5 (one-tailed). The rejection region for α = .05 is z > z_{.05} = 1.645. The test statistic is

z = \frac{5.4 − 5}{3.1/\sqrt{500}} ≐ 2.88

Since this is in the rejection region, there is sufficient evidence to reject H_{0}.

Mathematics 376 – Probability and Statistics II Solutions for Problem Set 7 April 8, 2010

10.51/Using a 2-tailed Z-test (note n_{i} = 50 is in the large sample range), the test statistic is

Z = \frac{74 − 71}{\sqrt{\frac{9^{2}}{50} + \frac{10^{2}}{50}}} ≐ 1.577.

The smallest value p of α for which this would be in the rejection region would have z_{p/2} = 1.577, so p = 2(.0571) = .1142. This is the attained significance level. If we desired an α = .05, then there would not be sufficient evidence to reject the null hypothesis.

10.55/Let π be the proportion of bad checks (after the new verification system was introduced). We want to test H_{0}: π = .05 against the alternative H_{a}: π < .05 (we want to know whether the new system has reduced the proportion). Since we have a sample of size n = 1124, we use a lower-tail z-test. With α = 0.01, the rejection region is z < −z_{.01} = −2.33. The estimate for π is 45/1124 ≐ .04. Then the test statistic is

z = \frac{.04 − .05}{\sqrt{\frac{(.05)(.95)}{1124}}} ≐ −1.538.

(Note that the values of π, 1 − π in the formula for the standard error in this case come from the π = .05 in the null hypothesis.) This is not in the rejection region, so we cannot reject the null hypothesis. Using the approximation z ≐ −1.54, the attained significance level is p = .0618 (the smallest α-value for which the test statistic would be in the rejection region).

10.66/ The test has H_{0}: μ = 100 and H_{a}: μ < 100. Since n is only 20, we use the smallsample (t) test. Rejection region is t < −t_{.01}(19) = −2.539 (from the t-table). The test statistic is

t = \frac{\overline{Y} − μ_{0}}{s/\sqrt{n}} = \frac{89.85 − 100}{14.90/\sqrt{20}} ≐ −3.046

This is in the rejection region, so we reject H_{0}. The attained significance p is p < .005 using the t-table. Using Maple:

                             TCDF(19,-3.046);

returns a much closer approximation to the exact p-value: .0033238808.

10.67/ The point of this problem is to show that the confidence intervals we studied earlier and the machinery of hypothesis testing that we have studied more recently are essentially equivalent – two different ways of saying the same thing. a) Using an upper-tail t-test (i.e H_{a} is μ > 280), we have

t = \frac{358 − 280}{54/\sqrt{10}} ≐ 4.568

With 9 degrees of freedom, t_{.01} = 2.821. So the t-statistic is in the rejection region and we reject H_{0}: The potato wedges apparently have a higher average calorie content than claimed.

b) The 99% lower confidence bound is

μ = 358 − t_{.01}(9)\frac{54}{\sqrt{10}} = 309.8.

c) Since 280 < 309.8, the claimed average calorie content is less than the 99% lower confidence bound. So the value 280 is not a believable value for the average calorie content. Note that this is the same conclusion as in part (a).

10.73/ a) For the small-sample test for equality of two subpopulation means, in order to get a test statistic with an exact t-distribution, the two subpopulations must be normal, with equal variances. This means that the best estimate for the sample variance is the pooled estimator

s_{p}^{2} = \frac{(n_{1} − 1)s_{1}^{2} + (n_{2} − 1)s_{2}^{2}}{(n_{1} + n_{2} − 2)}

given on page 492 of the text. (The number of degrees of freedom is the n_{1} + n_{2} − 2 = 10 + 10 − 2 = 18 in this problem. We have

s_{p}^{2} = \frac{(9)(3.92)^{2} + (9)(3.98)^{2}}{18} ≐ 15.6034

The test statistic is

t = \frac{14.5 − 11.1 − 0}{\sqrt{15.6034(1/10 + 1/10)}} ≐ 1.92

The rejection region for the α = .05 two-tailed test is |t| > t_{.025}(18) = 2.101. The test statistic is not in the rejection region, so there is not sufficient evidence to reject H_{0}. The attained significance (using Maple) is about p = 2(.034) = .068 (slightly larger than .05).

b) Similar:

t = \frac{12.2 − 11.5}{\sqrt[\sqrt{\frac{(9)(3.49)^{2}+(9)(4.95)^{2}}{18}}]{1/10 + 1/10}} ≐ .365

The test statistic is not in the rejection region, so there is not sufficient evidence to reject H_{0}. The attained significance (using Maple) is quite large: about p = 2(.36) = .72. The evidence is even weaker to reject H_{0} in this part.

Mathematics 376 – Probability and Statistics II Problem Set 8 Solutions April 12, 2006

A)

1) From the linear model, ϵ_{i} = Y_{i} − β_{0} − β_{1}x_{i} is normally distributed (N(0, σ^{2})). Hence the pdf for ϵ_{i} is

f(ϵ) = \frac{1}{\sqrt{2πσ^{2}}} e^{−ϵ^{2}/2σ^{2}}

From observations (x_{1}, y_{1}), …, (x_{n}, y_{n}) of Y, the likelihood of these observations (as a function of β_{0} and β_{1}) is given by

\begin{aligned}L(β_{0}, β_{1}) & = (\frac{1}{\sqrt{2πσ^{2}}} )^{n} exp ((− ∑_{i=1}^{n} ϵ_{i}^{2}) \frac{1}{2σ^{2}} ) \\ & = (\frac{1}{\sqrt{2πσ^{2}}} )^{n} exp ((− ∑_{i=1}^{n}(y_{i} − β_{0} − β_{1}x_{i})^{2}) \frac{1}{2σ^{2}} ) \end{aligned}

2) Continuing from the last line, we have

ln(L) = \frac{−n}{2} ln(2πσ^{2}) − \frac{1}{2σ^{2}} ∑_{i=1}^{n}(y_{i} − β_{0} − β_{1}x_{i})^{2}

Hence

\frac{∂ ln(L)}{∂β_{0}} = \frac{−1}{2σ^{2}} ∑_{i=1}^{n} 2(y_{i} − β_{0} − β_{1}x)(−1)

= \frac{1}{σ^{2}} ((Σy_{i}) − nβ_{0} − (Σx_{i})β_{1})

\frac{∂ ln(L)}{∂β_{1}} = \frac{−1}{2σ^{2}} ∑_{i=1}^{n} 2(y_{i} − β_{0} − β_{1}x)(−x_{i})

= \frac{1}{σ^{2}} ((Σx_{i}y_{i}) − (Σx_{i})β_{0} − (Σx_{i}^{2})β_{1})

Setting \frac{∂ ln(L)}{∂β_{0}} = \frac{∂ ln(L)}{∂β_{1}} = 0 and multiplying by σ^{2}, we get exactly the same normal equations as from the least-squares approach:

\begin{aligned}nβ_{0} + (Σx_{i})β_{1} & = Σy_{i} \\ (Σx_{i})β_{0} + (Σx_{i}^{2})β_{1} & = Σx_{i}y_{i} \end{aligned}

Thus, the maximum likelihood estimators for β_{0}, β_{1} are the same as the least-squares estimators.

3) To check that (β̂_{0}, β̂_{1}) gives a maximum of the likelihood, we continue on to the second derivative test:

\frac{∂^{2} ln(L)}{∂β_{0}^{2}} = \frac{−n}{σ^{2}}
\frac{∂^{2} ln(L)}{∂β_{1}^{2}} = \frac{−(Σx_{i}^{2})}{σ^{2}}
\frac{∂^{2} ln(L)}{∂β_{1}∂β_{0}} = \frac{−(Σx_{i})}{σ^{2}}

The Hessian determinant is

H = \frac{∂^{2} ln(L)}{∂β_{0}^{2}} \frac{∂^{2} ln(L)}{∂β_{1}^{2}} − (\frac{∂^{2} ln(L)}{∂β_{1}∂β_{0}})^{2} = \frac{n(Σx_{i}^{2}) − (Σx_{i})^{2}}{σ^{4}}

The numerator here is nS_{xx} > 0 (see B below). Hence H > 0 at all points (including the critical point found in part 2). Furthermore, \frac{∂^{2} ln(L)}{∂β_{0}^{2}} = \frac{−n}{σ^{2}} < 0, so (β̂_{0}, β̂_{1}) is a local (and global) maximum.

Comment: If you missed some points on this problem, be sure to review the second derivative test for functions of two variables again and be ready to use it correctly on the final exam(!)

B)

1) We have

S_{xx} = Σ(x_{i} − \overline{x})^{2}
= (Σx_{i}^{2}) − 2\overline{x}(Σx_{i}) + n\overline{x}^{2}
= (Σx_{i}^{2}) − 2n\overline{x}^{2} + n\overline{x}^{2}
= (Σx_{i}^{2}) − n\overline{x}^{2}

Similarly,

S_{yy} = Σ(x_{i} − \overline{x})(y_{i} − \overline{y})
= (Σx_{i}y_{i}) − \overline{x}(Σy_{i}) − \overline{y}(Σx_{i}) + n\overline{x}\overline{y}
= (Σx_{i}y_{i}) − 2n\overline{x}\overline{y} + n\overline{x}\overline{y}
= (Σx_{i}y_{i}) − n\overline{x}\overline{y}

Hence

\frac{S_{xy}}{S_{xx}} = \frac{(Σx_{i}y_{i}) − n\overline{x}\overline{y}}{(Σx_{i}^{2}) − n\overline{x}^{2}}

= \frac{n(Σx_{i}y_{i}) − n^{2}\overline{x}\overline{y}}{n(Σx_{i}^{2}) − n^{2}\overline{x}^{2}}

= \frac{n(Σx_{i}y_{i}) − (Σx_{i})(Σy_{i})}{n(Σx_{i}^{2}) − (Σx_{i})^{2}}

= β̂_{1} 2) Then we have

\overline{y} − \overline{x}β̂_{1} = \overline{y} − \overline{x}\frac{n(Σx_{i}y_{i}) − (Σx_{i})(Σy_{i})}{n(Σx_{i}^{2}) − (Σx_{i})^{2}}

= \frac{\overline{y}(n(Σx_{i}^{2}) − (Σx_{i})^{2}) − \overline{x}(n(Σx_{i}y_{i}) − (Σx_{i})(Σy_{i}))}{n(Σx_{i}^{2}) − (Σx_{i})^{2}}

= \frac{\frac{1}{n}(Σy_{i})(n(Σx_{i}^{2}) − (Σx_{i})^{2}) − \frac{1}{n}(Σx_{i})(n(Σx_{i}y_{i}) − (Σx_{i})(Σy_{i}))}{n(Σx_{i}^{2}) − (Σx_{i})^{2}}

= \frac{(Σy_{i})(Σx_{i}^{2}) − (Σx_{i})(Σx_{i}y_{i})}{n(Σx_{i}^{2}) − (Σx_{i})^{2}}

= β̂_{0} 11.27/a) By various portions of our “Big Theorem” on the distribution of the least squares estimators β̂_{i}, we know that, under the assumption ε_{i} ∼ N(0, σ^{2}) for all i:

• β̂_{i} are normally distributed with mean β_{i0}, and variance c_{ii}σ^{2}.

• S^{2} = SSE/(n − 2) is an unbiased estimator for σ^{2},

• \frac{(n−2)S^{2}}{σ^{2}} ∼ χ^{2}(n − 2), and

• S^{2} and β̂_{i} are independent. Thus,

\frac{β̂_{i} − β_{i0}}{σ\sqrt{c_{ii}}} ∼ N(0, 1).

If we replace σ by the estimator S:

\begin{aligned}\frac{β̂_{i} − β_{i0}}{S\sqrt{c_{ii}}} & = \frac{β̂_{i} − β_{i0}}{σ\sqrt{c_{ii}}} /\frac{S}{σ} \\ & = \frac{β̂_{i} − β_{i0}}{σ\sqrt{c_{ii}}} /\sqrt{\frac{(n − 2)S^{2}}{σ^{2}} /(n − 2)} \end{aligned}

This last has the form \frac{Z}{\sqrt{W/ν}}, where Z is standard normal, W has a χ^{2}(ν) distribution, and

Z, W are independent. By definition 7.2 in the book, this says \frac{_{β_{i}}−β_{i0}}{S\sqrt{c_{ii}}} has a t-distribution with ν = n − 2 degrees of freedom.

b) The quantity \frac{_{β_{i}}−β_{i}}{S\sqrt{c_{ii}}} is a pivotal quantity for β_{i} so we can use it to derive confidence intervals as usual:

1 − α = P (−t_{α/2} < \frac{β̂_{i} − β_{i}}{S\sqrt{c_{ii}}} < t_{α/2})
= P (−t_{α/2}S\sqrt{c_{ii}} < β̂_{i} − β_{i} < t_{α/2}S\sqrt{c_{ii}})
= P (−β̂_{i} − t_{α/2}S\sqrt{c_{ii}} < −β_{i} < −β̂_{i} + t_{α/2}S\sqrt{c_{ii}})
= P (β̂_{i} − t_{α/2}S\sqrt{c_{ii}} < β_{i} < β̂_{i} + t_{α/2}S\sqrt{c_{ii}})

(after multiplying through by −1 and flipping inequalities back so the terms increase left to right). This gives the formula for the confidence interval:

β_{i} = β̂_{i} ± t_{α/2}S\sqrt{c_{ii}}.

See Maple worksheet for other solutions.