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(MATH144)[2010](s)final~1487^_10050.pdf
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Math 144: Solutions to the Final examination (Spring 2010)
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This exam will be 2 hours long.

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Please show all your work so that you do not lose any marks.

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Unless otherwise speci.ed, numerical answers should be either exact or correct to 4
decimal places.



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1. (a) [5 marks] Consider the following hypothesis testing
.H0 : = 120 . . H1 : = 100

Suppose that the test procedure is to reject H0 if X< 70.65, where X N(, 30). Then draw a picture and explain why we cannot control Type I and II errors simultaneously.
(b) [5 marks] The following are the weights, in decagrams, of 10 packages of grass seed distributed by a certain company:
46.4, 46.1, 45.8, 47.0, 46.1, 45.9, 45.8, 46.9, 45.2, and 46.0.
Assume that the package of grass seed distributed by this company has a normal distribution. Find a 95% con.dence interval for 1/.
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(b) Based on the above data, we have s=0.2862222 or s=0.2862. From the chi-squared table with the degree of freedom n . 1 = 9, we have 2 = 19.023 and 2 =2.7. Hence, the
0.025 0.975
.19.023
95% C.I. for 1/ is .. 2.7 , . = [1.0238, 2.7175].
9s2 9s2
2. A recent article in the British journal Lancet reports that babies who were fed mothers milk tended to have a higher IQ than formula-fed babies. Suppose that two groups of babies are compared, a group fed mothers milk and a group fed formula. The resulting IQ scores are as follows:
Mother: 121, 105, 111, 119, 108, 101, 90, 131, 106, 112 Formula: 101, 110, 107, 98, 89, 103, 86, 117, 113, 87
Assume that the IQ scores are normally distributed with population mean 1 and population variance 2 for mother-fed babies; population mean 2 and population variance 2 for formula-fed babies.
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(a)
[7 marks] Assume that the two population variances are NOT equal. Construct a 95% con.dence interval for the di.erence 1 . 2 between the mean IQ scores. What can be concluded from this con.dence interval?

(b)
[3 marks] At a 0.05 level of signi.cance, test H0 : 1 .


= 2 + 25 against H1 : 1 = 2 + 25, when the two population variances are assumed to be unequal.
Solution: (a) Based on the above data, we have
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x = 110.4,s = 130.2667,n = 10; y = 101.1,s = 121.6556,m = 10.
XY
Since the two population variances are NOT equal, we consider the degree of freedom
2 X
+ s
2 Y
( s
n
( 130.2667 + 121.6556 )2
10 10
)2 = 17.97899.
m
v =
=
2 X 2 Y
1 ( 130.2667 )2 + 1 ( 121.6556 )2
910 910
( s
n
( s
m
1 n.1
1
)2 +
)2
m.1
2 Y
2 X
Thus, we take v = 17, and from the table of the t-distribution with v = 17, we have t17,0.025 =2.110. Hence, the 95% con.dence interval for the di.erence 1 . 2 between the mean IQ scores is
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.ss.130.2667 121.6556
XY
(x . y) tv,