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Statistical Analysis Solutions

(1) This document contains the solutions to 13 multiple choice questions. (2) Question 13 uses a chi-square test to determine if computer use is independent of gender. The chi-square statistic is calculated to be 12.316, which exceeds the critical value of 9.2104. Therefore, the null hypothesis that computer use is independent of gender is rejected. (3) Question 16 uses a two-sample t-test to compare the mean home prices in Brisbane and Canberra. With a 90% confidence level, the difference between the mean prices in Brisbane and Canberra is estimated to be between $7631 to $15569 higher in Brisbane.

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87 views13 pages

Statistical Analysis Solutions

(1) This document contains the solutions to 13 multiple choice questions. (2) Question 13 uses a chi-square test to determine if computer use is independent of gender. The chi-square statistic is calculated to be 12.316, which exceeds the critical value of 9.2104. Therefore, the null hypothesis that computer use is independent of gender is rejected. (3) Question 16 uses a two-sample t-test to compare the mean home prices in Brisbane and Canberra. With a 90% confidence level, the difference between the mean prices in Brisbane and Canberra is estimated to be between $7631 to $15569 higher in Brisbane.

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Look Dnt Wrry
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© Attribution Non-Commercial (BY-NC)
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Download as PDF, TXT or read online on Scribd
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Solutions to selected questions

Question 1
Question 2
Question 3
Question 4
Question 5
Question 6

Question 7
Question 8
Question 9

Let Melbourne be group 1

(1) H0: µ1 – µ2 = 0
Ha: µ1 – µ2 > 0

( x1 − x 2 ) − ( µ1 − µ 2 )
(2) t=
s1 2 (n1 − 1) + s 2 2 (n2 − 1) 1 1
+
n1 + n2 − 2 n1 n2

(3) α = 0.01
(4) For a one-tailed test and df = 8 + 9 – 2 = 15, t .01, 15 = 2.602. If the observed value of t is
greater than 2.602, the decision is to reject the null hypothesis.

(5) Melbourne Sydney


n1 = 8 n2 = 9
x1 = 47 x 2 = 44
s1 = 3 s2 = 3

(47 − 44) − (0)


(6) t= = 2.06
7(3) 2 + 8(3) 2 1 1
+
15 8 9

(7) Since t = 2.06 < t .01, 15 = 2.602, the decision is to fail to reject the null hypothesis.

(8) There is no significant difference in rental rates between Melbourne and Sydney.
Question 10

H 0 : µ = $82 600

H a : µ < $82 600

x = $78 974, n = 18, σ = $7810, α = .01

zcrit = −2.33Reject Ho if z <-2.33


78 974 − 82 600
z= = −1.97
7810 / 18

Do not reject H 0 .

Question 12
Question 13

H o : Computer use is independent of Gender.

H a : Computer use is not independent of Gender.

8440(9932) (8440)(5128)
e 11 = = 5047.33 e 12 = = 2605.99
16608 16608

(8440)(1548) (8168)(9932)
e 13 = = 786.68 e 21 = = 4884.67
16608 16608

(8168)(5128) (8168)(1548)
e 22 = = 2522.01 e 23 = = 761.32
16608 16608

Computer use

Gender Home Work Library Total

Male 5064 (5047.33) 2653 (2605.99) 723 (786.68) 8440

Female 4868 (4884.67) 2475 (2522.01) 825 (761.32) 8168

Total 9932 5128 1548 16608

(5064 − 5047.33) 2
χ2 = +
5047.33
(2653 − 2605.99) 2
+
2605.99
(723 − 786.68) 2
+
786.68
(4868 − 4884.62) 2
4884.62
(2475 − 2522.01) 2
+ +
2522.01
(825 − 761.32) 2
761.32
χ = 12.316
2

α = .01, df = (c – 1)( r– 1) = (3 – 1)(2 – 1) = 2

χ2 .01,2 = 9.2104

Since the observed χ2 = 12.316 > χ2 .01,2 = 9.2104, the decision is to reject the null hypothesis.

Computer use is not independent of gender.


Question 16

Brisbane Canberra
n 1 = 21 n 2 = 26
x1 = 507 600 x2 = 496 000
s 1 = 9200 s 2 = 7000

df = 21 + 26 – 2
90% level of confidence, α/2 = 0.05, t .05, 45 = 1.68

s1 2 (n1 − 1) + s 2 2 (n2 − 1) 1 1
( x1 − x 2 ) ± t +
n1 + n2 − 2 n1 n2

(9200) 2 (20) + (7000) 2 (25) 1 1


= (507600 – 496000) ± 1.68 +
21 + 26 − 2 21 26
= 11 600 ± 3969
7631 < µ 1 – µ 2 < 15569We can be 90% confident that the mean price of mid range
homes in Brisbane is between $7631 to $15569 higher than in Canberra

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