Statistics Bba2 Worksheet # 4 (solutions)

UNIVERSITY OF TECHNOLOGY, JAMAICA DEPARTMENT OF SCIENCE AND MATHEMATICS STATISTICS 2 (STA 2004) H 0 : µ = 50

1.

H1 : µ ≠ 50 Two – tailed test, the significance level = 0.06 (6 %) Test statistic:

Z cal =

x − µ 49 − 50 6 = = −1× = −1.2 σ 5 5 n 36

Conclusion: Since −1.88 < Z cal < 1.88 , we accept H 0 , ie. µ = 50 at the 6% level of significance. H 0 : µ = 25 minutes

2.

H1 : µ ≠ 25 minutes Two – tailed test, the significance level = 5 % Test statistic:

Z cal =

x − µ 25.9 − 25 40 σ = 1.5 = 0.9 × 1.5 = 3.79 n 40

Conclusion: Since Z cal > 1.96 , we reject H 0 and accept H1 , ie. µ ≠ 25 minutes at the 5% level of significance. H 0 : µ ≤ 16 ounces

3.

H1 : µ > 16 ounces One – tailed test, the significance level = 1 % Test statistic:

Z cal =

x − µ 16.05 − 16 50 = = 0.05 × = 11.79 σ 0.03 0.03 n 50

Conclusion: Since Z cal > 2.33 , we reject H 0 and accept H1 , ie. µ > 16 ounces at the 1% level of significance. 4. Let µ1 represent the mean length of time Norbrook residents owned their homes and µ 2 represent the mean length of time Beverly Hills residents owned their homes H 0 : µ1 ≥ µ2 H1 : µ1 < µ2 One – tailed test, the significance level = 5 % Test statistic:

Z cal =

x1 − x2 2

2

s1 s2 + n1 n2

=

7.6 − 8.1 (2.3) 2 (2.9) 2 + 40 55

= −0.94

Conclusion: Since Z cal > − 1.65 , we accept H 0 , ie. µ1 ≥ µ2 at the 5 % level of significance.

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5. Please try this question. H 0 : p ≤ 0.60

6.

H1 : p > 0.60 One – tailed test, the significance level = 1 % . Now p = 0.6 and p =

Test statistic:

Z cal =

p− p = p (1 − p) n

210 = 0.7 300

0.7 − 0.6 0.1 = = 3.53 0.6(1 − 0.6) 0.0008 300

Conclusion: Since Z cal > 2.33 , we reject H 0 and accept H1 , ie. p > 0.60 ounces at the 1% level of significance. H 0 : p = 0.5

7.

H1 : p ≠ 0.5 Two – tailed test, the significance level = 5 % . Now p = 0.5 and p =

Test statistic:

Z cal =

87 = 0.435 200

p− p 0.435 − 0.5 = = −1.84 p (1 − p) 0.5(1 − 0.5) n 200

Conclusion: Since −1.96 < Z cal < 1.96 , we accept H 0 , ie. p = 0.5 at the 5 % level of significance. H 0 : p1 = p2

8.

(There is no difference between these two sections in terms of house ownership)

H1 : p1 ≠ p2 Two – tailed test, the significance level = 6 % . Test statistic:

Z cal =

p1 − p2 pc (1 − pc ) pc (1 − pc ) + n1 n2

pc =

200 + 120 320 = = 0.71 250 + 200 450

p1 =

200 = 0.8 250

p2 =

120 = 0.6 200

Z cal =

0.8 − 0.6 0.2 = = 4.65 0.71(1 − 0.71) 0.71(1 − 0.71) 0.043 + 250 200

Conclusion: Since Z cal > 1.88 , we reject H 0 and accept H1 , ie. There is a difference between these two sections in terms of house ownership at the 6 % level of significance.

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