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(ISMT111)final_fall05sol.pdf
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ISMT 111 Fall 05 Final Exam Suggested solution
Question 1:
(a)
The Hypothesis statement is H0: =23 vs H1 >23. Naturally we need to assume that that either the data are normally distributed in which case we assume that the CLT applies. The sample size n=56 is large enough and we can use the Z-test and the test statistic is

Z=
56 (23.5 C 23)/10.2=.367. We can use the normal table to find the critical value of 1.645 at alpha =.05. In this case since .367<1.645 we cannot reject the null hypothesis and must conclude that the evidence indicates that their claim is false.

(b)
The 95% confidence interval is approximately 23.5 2.67

(c) To find A, solve the equation
56 (23.5 C A)/10.2 1.645, the solution is A 21.258. Hence set A=21. Note: Question (c) can be adjusted for different values of the standard deviation.


Question 2:
(a)
Using the formula for the differences where the population variances are assumed equal but unknown and the populations are normally distributed. 2K=16.14+6.14 or K=11.14, but K=T.025 SE with 10 degrees of freedom T.025=2.228. Hence the Standard Error is 5.

(b)
T.05 is 1.812, x1 . x2 = (16.14 . 6.14) / 2 = 5 . Thus the 90% CI is given by
55 1.812 =[-4.06, 14.06]


(c)
The observed test statistic is -1, which is larger than -1.812 hence we cannot reject the null hypothesis.

(d)
Since SE is 5 and the hypothesized difference is 10 it follows that the difference of the sample means must be 5 (based on the result of T = - 1). But this can also be calculated as in (b)


Question 3:
(a)
E(Y ) = 1+ 0.6 5 + 0.04 20 = 4.8; s.d. = 1

(b)
P(Y < 4) = P(Z <.0.8) = 0.2119 .

(c)
E(Y ) = 4.8 s.e. = 1/

(d)
P(4.5 < Y < 5) = P(.1.47 < Z < 0.98) = 0.8365 . 0.0708 = 0.7657

(e)
P(Y < 4.5) = P(Z <.0.3) = 0.3821 s.d. =



24 =0.204
0.3821(1. 0.3821) / 24 = 0.09918

(f) P(0.05 < p. < 0.1) = P(.3.35 < Z <.2.84) = 0.4996 . 0.4977 = 0.0019
5

Question 4:
(a) Let be the time reduced by the new system H0: = 6 vs H1: < 6
Type I error: The actual time reduced is not less than 6 minutes, but we regard the time
reduced is less than 6.
Type II error: We think that the time reduced is not less than 6, but in fact it is actually
less than 6.

. x . 6 5.7 . 6 .
(b) = P(reject H0| H0 is true) = P(x < 5.7) = P.
<
.
=
..
1.5 100 1.5 100
.
. = P(z<.2) = 0.0228 x . 6 5.9 . 6
(c) p-value = P(x < 5.9) = P(
<
) = P(z <.0.667) = 0.2514

1.5/ 100 1.5/ 100

(d)
Let n be the sample size required . 2 2.576 1.5 .2


n =. .= 1493.0496 1494
. 0.2 .
Question 5:
.

Question 6:
a) b1 = [8340 . 8(134 / 8)(410 / 8)]/[3020 . 8(134 / 8)2 ] = 1.8988 ,
b0 = (410 / 8) .1.8988 (134 /8) = 19.4451, cost = 19.4451+1.8988 year b) 1. (24000 .19.4451 410 .1.89888340) /[24000 . 8(410 / 8)2 ] = 0.94 c) observed t: 9.3583,for 6 degrees of freedom t = 2.447
0.025
d) the interval 38.43