Asst5_2011_SOLUTION

Asst5_2011_SOLUTION - COMM 291 JanApr 2011 ASSIGNMENT 5...

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COMM 291 – Jan–Apr 2011 ASSIGNMENT 5 – SOLUTIONS Solution to Question 1 (6 marks) a) The populations being examined are 1) houses with one or more fireplaces and currently for sale and 2) houses without fireplaces and currently for sal e (in the same particular region). b) H 0 : μ 1 2 = 0 H a : 1 2 ≠ 0 Pooled version: Since ( s 1 = 59,827) < 2×( s 2 = 34,131), it is appropriate to assume that σ 1 = 2 and to pool s 1 and s 2 into s p . d.f. = n 1 + n 2 –2 = 98, and p-value = 2×Pr( t df > | t |) = 2×Pr(t 98 > 8.04) < .0001 Regular (unpooled) version: …is also acceptable. d.f. = min( n 1 –1, n 2 –1) = 49 (most conservative choice); p-value = 2×Pr(t 49 > 8.04) < .0001; or d.f. = 78 as estimated by Data Analysis Toolpak; p-value = 2×Pr(t 78 > 8.04) < .0001 Regardless of the approach, the p-value is negligible and the evidence is extremely strong that the listing prices of houses with fireplaces and of those without are different. (Not necessary: Those with fireplaces are apparently more expensive.) For LIVING AREA: The null and alternative hypotheses are the same as for Price. t-stat = -6.4154; Two-tailed P-value < .0001 Conclusion: There is very strong evidence that the living area of houses with fireplaces and those without are different. (Not necessary: Those with fireplaces are larger.) For AGE: t-stat = 3.6136; Two-tailed P-value = .0005 Conclusion: There is very strong evidence that the age of houses with fireplaces and those without are different. (Not necessary: Those with fireplaces are newer.) 2 2 2 2 1 1 2 2 1 2 ( 1) ( 1) 49 59827 49 34131 48704 2 98 189323 111006 8.04 1 1 48704 50 50 p n s n s s n n t - + - × + × = = = + - - = = + ( 29 ( 29 1 2 0 2 2 2 2 1 2 1 2 189323 111006 0 8.04 59827 34141 50 50 x x t s s n n - - ∆ - - = = = + +
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This note was uploaded on 09/01/2011 for the course COMM 291 taught by Professor E.fowler during the Winter '10 term at UBC.

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Asst5_2011_SOLUTION - COMM 291 JanApr 2011 ASSIGNMENT 5...

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