Mathematics 376 – Probability and Statistics II Problem Set 1 Solutions January 30, 2004

7.3/We have n = 9 samples from an N(μ, 4) population, so by Theorem 7.1, the sample mean \overline{X} is N(μ, 4/\sqrt{9}) = N(μ, 4/3). Then

P (−2 < \overline{X} − μ < 2) = P (−1.5 < 3 (\frac{\overline{X} − μ}{4}) < 1.5)
= 2(.5 − .0668)
= .8664

(by the standard normal table). Note: 3 (\frac{\overline{X}−μ}{4}) is N(0, 1).

7.4/Continuing from the previous problem, with an unknown number of samples n, we have \sqrt{n} (\frac{\overline{X}−μ}{4}) is N(0, 1), so from the standard normal table again,

.90 = P (−1 < \overline{X} − μ < 1) = P (\frac{−\sqrt{n}}{4} < \sqrt{n} (\frac{\overline{X} − μ}{4}) < \frac{\sqrt{n}}{4})

when \sqrt{n}/4 ≥ 1.645. Rearranging and squaring, we get n ≥ 43.3, so n ≥ 44.

7.7/a) By linearity of expected values and Theorem 7.1, E(\overline{X} − \overline{Y}) = μ_{1} − μ_{2}. b) Because the two samples are independent, Cov(\overline{X}, \overline{Y}) = 0, and

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

c) Using parts a, b and Theorem 7.1, we get that \overline{X} − \overline{Y} is N(μ_{1} − μ_{2}, \sqrt{\frac{σ_{1}^{2}}{m} + \frac{σ_{2}^{2}}{n}}). Hence, we proceed as in 7.4. With σ_{1}^{2} = 2, σ_{2}^{2} = 2.5, and m = n, we want to choose n so that

P (\frac{−\sqrt{n}}{\sqrt{4.5}} < Z < \begin{matrix}√ & \overline{n} \\ √ & \overline{4.5} \end{matrix}) = .95

This means we want the two “tails” of the standard normal to have probability .025 each, so

\begin{matrix}√ & \overline{n} \\ √ & \overline{4.5} \end{matrix} = 1.96 ⇒ n = 17.28

or n = 18 after going to the next whole number.

7.10/ a) By definition the χ^{2}(ν) distribution is the same as the gamma distribution with α = ν/2 and β = 2. Hence E(U) = α · β = ν, and V (U) = α · β^{2} = 2ν using our formulas from last semester. b) By Theorem 7.3, \frac{(n−1)S^{2}}{σ^{2}} is χ^{2}(n − 1). So \frac{n−1}{σ^{2}}E(S^{2}) = n − 1, or E(S^{2}) = σ^{2}. Similarly,

\frac{(n − 1)^{2}}{σ^{4}} V (S^{2}) = 2(n − 1) ⇒ V (S^{2}) = \frac{2σ^{4}}{n − 1}

7.11/For this problem, you want to use Table 6 in the Appendix in the text book (“Percentage points of the χ^{2} distributions”). The entry marked χ_{α}^{2} in the table is the probability in a right tail of the distribution. a) Since n = 20, σ^{2} = 1.4, from Theorem 7.3, we know \frac{19S^{2}}{1.4} is χ^{2}(19). To get P (S^{2} ≤ b) = .975, we want the probability of the right tail to be .025, so we take the table entry χ_{.025}^{2} for ν = 19 degrees of freedom and

\frac{19b}{1.4} = 32.8523 ⇒ b = 2.241

b) Similarly to get P (a ≤ S^{2}) = .975, we want χ_{.975}^{2} for ν = 19 d.f. We get

\frac{19a}{1.4} = 8.90655 ⇒ a = .6563

c) Using the sum rule, P (A ∩ B) = P (A) + P (B) − P (A ∪ B). Letting A be the event a ≤ S^{2} and B be the event S^{2} ≤ b, we get P (a ≤ S^{2} ≤ b) = P (A ∩ B) = 2(.975) − 1 = .95.

7.14/a) As always, Z is N(0, 1), so E(Z) = 0 and E(Z^{2}) = V (Z) + (E(Z))^{2} = 1 + 0 = 1.

b) We want to use Theorem 5.9 here (independence ⇒ expected values multiply). Since

T = Z · \begin{matrix}√ & \overline{ν} \\ √ & \overline{Y} \end{matrix} , we have

E(T) = E(Z) · E (\begin{matrix}√ & \overline{ν} \\ √ & \overline{Y} \end{matrix} ) = 0

(unless ν = 1 – in that case, \sqrt{Y} is also N(0, 1) and E(T) is undefined.) For the variance, we proceed as follows:

V (T) = E(T^{2}) − (E(T))^{2}

= E(T^{2})

= νE(Z^{2} · \frac{1}{Y} )

= νE(Z^{2})E (\frac{1}{Y} ) using Th. 5.9

We know E(Z^{2}) = 1 from above. To get E(Y^{−1}) use the formula given with a = −1 (note: ν > 2 ⇒ ν > −2a), and the functional equation Γ(x + 1) = xΓ(x):

V (T) = νE (\frac{1}{Y} )

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

= ν · \frac{1}{ν/2 − 1} · \frac{1}{2}

= \frac{ν}{ν − 2}

7.19/a) W is χ^{2}(5) using Theorem 7.2. b) U is χ^{2}(4) by Theorem 7.3. c) The sum of independent χ^{2}(ν_{1}) and χ^{2}(ν_{2}) random variables is χ^{2}(ν_{1} +ν_{2}), so this is χ^{2}(4+1) = χ^{2}(5).

7/20/a) This is t(5) by Definition 7.2. b) t(4) using 7.19 b. c) By Theorem 7.1, \overline{Y} is N(0, 1/\sqrt{5}), so \sqrt{5} · \overline{Y} is N(0, 1), so 5\overline{Y}^{2} is χ^{2}(1). Then 5\overline{Y}^{2} + Y_{6}^{2} is χ^{2}(2). Finally the ratio is F (2, 4) by Definition 7.3.