Mathematics 376 – Probability and Statistics II Solutions – Problem Set 7 April 5, 2004

10.29/From problem 10.9, we have a lower tail Z-test with z = \frac{\overline{Y} −130}{2.1/\sqrt{40}}, so the rejection

region is \{z : z < −z_{.05}\} = \{z < −1.645\}, or \{\overline{Y} < 130 − 1.644(2.1)/\sqrt{40} = 129.45\}. The desired Type II error probability is

β = P (\overline{Y} ∉ RR|μ = 128)

= P (\overline{Y} > 129.45|μ = 128)

= P (\frac{\overline{Y} − 128}{2.1/\sqrt{40}} > \frac{129.45 − 128}{2.1/\sqrt{40}})

= P (z > 4.37) < .0000317

(from the closest entry in the standard normal table). Hence, β is “approximately zero”.

10.30/From problem 10.23, we have an α = .05 upper tail Z-test with rejection region (described in terms of p̂_{1} − p̂_{2}):

\{p̂_{1} − p̂_{2} > 1.645\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}} \}

To get β ≤ .2, given that p_{1} − p_{2} = .1, we want to choose n so that

β = .2 = P (p̂_{1} − p̂_{2} ∉ RR|p_{1} − p_{2} = .1)

= P (\frac{p̂_{1} − p̂_{2} − .1}{\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}}} < \frac{1.645\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}} − .1}{\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}}})

= P (z < \frac{1.645\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}} − .1}{\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}}})

This implies

\frac{1.645\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}} − .1}{\sqrt{\frac{p_{1}q_{1}}{n} + \frac{p_{2}q_{2}}{n}}} = −z_{.2} ≐ −.84

So after some algebra isolating n:

\sqrt{n} = 10(2.485)\sqrt{p_{1}q_{1} + p_{2}q_{2}}

At the risk of possibly taking n larger than necessary in some cases, we can replace p_{i}q_{i} on the right by .25, which is the maximum value of pq = p(1 − p) as per the hint:

\sqrt{n} ≤ 24.58\sqrt{.5} ⇒ n ≤ 308.76

Hence n = 309 or so will work for any p_{1} and p_{2} with p_{1} − p_{2} = .1. (Note: depending on how you round in the computations here, the answer can vary a bit.)

10.33/a) This is an upper-tail Z-test for the difference of the means. The test statistic is:

z = \frac{32.19 − 31.68}{\sqrt{\frac{(4.34)^{2}}{37} + \frac{(4.56)^{2}}{37}}} = \frac{.51}{1.035} ≐ .493

The rejection region for the test with α = .05 is RR = \{z > 1.645\}, or \{\overline{Y}_{1} − \overline{Y}_{2} > (1.645)(1.035) = 1.702}. Since z = .493 ∉ RR, we cannot reject the null hypothesis from this information.

b) Now, assuming that μ_{1} − μ_{2} = 3,

β = P (z ∉ RR|μ_{1} − μ_{2} = 3)

= P (\overline{Y}_{1} − \overline{Y}_{2} < 1.702|μ_{1} − μ_{2} = 3)

= P (\frac{\overline{Y}_{1} − \overline{Y}_{2} − 3}{1.035} < \frac{1.702 − 3}{1.035})

= P (z < −1.254)

= .1056

10.34/Continuing from 33, we want to determine n so that α = β = .05 if μ_{1} −μ_{2} = 3 (and we use the variance estimates as before). The rejection region for a level α test should be:

\{z : z > z_{.05} = 1.645\} ⇔ \{\overline{Y}_{1} − \overline{Y}_{2} > (1.645)\sqrt{\frac{(4.34)^{2}}{n} + \frac{(4.56)^{2}}{n}} = \frac{10.356}{\sqrt{n}}\}.

So we want

.05 = β = P (\overline{Y}_{1} − \overline{Y}_{2} ∉ RR|μ_{1} − μ_{2} = 3)

= P (z < \frac{\frac{10.356}{\sqrt{n}} − 3}{\frac{6.295}{\sqrt{n}}})

= P (z < \frac{10.356 − 3\sqrt{n}}{6.295})

This says

\frac{10.356 − 3\sqrt{n}}{6.295} = −z_{.05} = −1.645

so

\sqrt{n} = (2)(10.356)/3 ⇒ n = 47.7

Hence n = 48 will suffice.

10.36/We have

\begin{aligned}\frac{θ̂− θ_{0}}{σ_{θ}} > z_{α} & ⇒ θ̂− θ_{0} > z_{α}σ_{θ} \\ & ⇒ θ̂ > θ_{0} + z_{α}σ_{θ} \end{aligned}

This is exactly the formula for the (1 − α) × 100% upper confidence bound (see page 386).

10.38/We have

\begin{aligned}\frac{θ̂− θ_{0}}{σ_{θ}} < −z_{α} & ⇒ θ̂− θ_{0} < −z_{α}σ_{θ} \\ & ⇒ θ̂ < θ_{0} − z_{α}σ_{θ} \end{aligned}

This is exactly the formula for the (1 − α) × 100% lower confidence bound (see page 386).

10.41/For the two-tailed Z-test, our test statistic is

z = \frac{74 − 71}{\sqrt{9^{2}/50 + 10^{2}/50}} = 1.5768 ≐ z_{.0571}.

To find the p-value of the test, we see that .0571 is α/2 from the 2-tail test rejection region formula. The smallest α for which z ∈ RR is α = 2(.0571) = .1142, so p = .1142.

10.43 a) For this lower-tail Z-test,

z = \frac{3.6 − 3.8}{1.1/\sqrt{30}} = −.9959

This is approximately −z_{.1587}, so the p-value for the test is .1587. b) There is not enough evidence to reject the null hypothesis at that level. c) From the FVC data for the women,

z = \frac{2.7 − 3.1}{1.2/\sqrt{30}} = −1.83

This is approximately −z_{.0336}, so the p-value this time is p = .0336. d) There is sufficient evidence to reject the null hypothesis for the women.

10.51/Since n = 5 (small sample case), we use a lower-tail T -test. From the data, S^{2} = 69.5, and \overline{Y} = 795 so

t = \frac{795 − 800}{S/\sqrt{5}} = −1.3406

From the t-table in the text (with n−1 = 4 degrees of freedom), −t_{.05}(4) = −2.132. There is not sufficient evidence to reject H_{0} (μ = 800). The p-value of the test is > .1, since −t_{.1}(4) = −1.533 is the nearest tabulated value.

10.53 a) From the data, \overline{Y} = 39.65 and S^{2} ≐ 51.51. Since n = 18, this also calls for a lower-tail T -test.

t = \frac{39.56 − 45}{7.18/\sqrt{18}} ≐ −3.2145

From the book’s t-table, t_{.005} = 2.898, so we can say that the p-value of the test is p < .005. This is very convincing evidence to reject the null hypothesis and conclude that the average tax per gallon was less that $0.45

b) The 95% confidence interval is

39.56 ± t_{.025}(17)\frac{7.18}{\sqrt{18}} ≐ 39.56 ± 3.55

This is consistent with the conclusion in part a: 45 is not contained in this interval.