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Question

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(c) What justification is needed for part (b) that isn’t needed for part (a)? (Give some indication or argument about whether our confidence interval in (b) is reliable.)

(e) Is the justification given in (c) needed for (d)? (Again, you can now assume the data are normally distributed.) (f) The CI in (d) does not contain the value 21.1. So can we use this to conclude that the means are different?

Explanation / Answer

Back-up Theory

Under assumption of Normality,

(Y bar - µY) ~ N(0. 2Y/n1) and (X bar - µX) ~ N(0. 2X/n2) …………………………. (1),

where n1 and n2 are the respective sample sizes.

From (1): {(Y bar – X bar) – (µY - µX)} ~ N[0. {(2Y/n1) + (2X/n2)}] ………………(2)

and hence {(Y bar – X bar) – (µY - µX)}/sq,rt{(2Y/n1) + (2X/n2)} ~ N(0, 1) ………..(3)

Based on (3), 100(1 - )% Confidence Interval for (µY - µX) is given by:

(Y bar – X bar) ± Z /2.[sq,rt{(2Y/n1) + (2X/n2)}] ………………………………….(4)

When 2Y and 2X are not known, but can be assumed to be equal [this must be tested and confirmed] to say 2, then (3) becomes:

[{(Y bar – X bar) – (µY - µX)}/{s.sq,rt{(1/n1) + (1/n2)}] ~ tn1 + n2 - 2 …………………...(5),

where s2 = {(n1 - 1) s12 +(n2 - 1)s22}/(n1 + n2 - 2) is the unbiased estimate of 2………(6)

And then, 100(1 - )% Confidence Interval for (µY - µX) is given by:

(Y bar – X bar) ± tn1 + n2 – 2, /2.{s.sq,rt{(1/n1) + (1/n2)}..……………………………….(7)

Now, to work out the solution,

Part (a)

Given n1 = 24, Y bar = 24.8, X bar = 21.1, 2Y = 49, 2X = 64, = 0.01 (1%) and

from Standard Statistical Tables, Z 0.005 = 2.575, by (4)

99% confidence Interval for (µY - µX) is: (24.8 – 21.1) ± 2.575xsq.rt{(49/24) + (64/41)}

= 3.7 ± 2.575x1.898 = 3.7 ± 4.887 ANSWER

Part (b)

Since 2Y and 2X are not known, we apply (7) here.

tn1 + n2 – 2, 0.005 = 2.654 and s2 = {(23x7.12) + (40x8.12)}/63 = 62.8888 or s = 7.930

99% confidence Interval for (µY - µX) is: (24.8 – 21.1) ± 2.654x7.930.sq.rt{(1/24) + (1/41)}

= 3.7 ± 2.654x7.930x0.2570 = 3.7 ± 5.409 ANSWER

Part (c)

As stated under Back-up Theory, justification is that the population variances, 2Y and 2X are equal. By an actual F-test, the hypothesis 2Y = 2X is found to be acceptable.

When the population variances are not known, the above method is equally reliable. ANSWER

Part (d)

99% lower CI for µY is: Y bar – tn1 – 1, 0.01xs1 = 24.8 – 2.807x7.1 = 4.03 ANSWER