MGMT2262 Tutorial Sheet 6 Solutions

1) Ho: p £ 35%

Ha: p > 35%

Reject Ho if Z > 1.645

p-hat = 84/200 = 0.42

Z = (0.42 – 0.35)/sqrt(0.35*0.65/200) = 2.08

Reject Ho

Conclude that the percentage of Ipod users in the 18-34 age range has significantly increased.

2) P-value = P(Z > 2.08) = 0.5 – 0.4812 = 0.0188. Since this p-value is greater than 1%, we would have not have rejected Ho.

3) Lower limit = 0.42 – 1.96sqrt(0.42*0.58/200) = 0.42 – 0.0684 = 0.3516

Upper limit = 0.42 + 0.0684 = 0.4884

With 95% confidence, the percentage of Ipod users in the 18-34 age bracket ranges from 35.16% to 48.84%.

4) Sample size = n = (1.96/0.03)²(0.42)(0.58) = 1039.8 which we round to 1040.

5) E = 1.96sqrt(0.42*0.58/500) = 0.0433 or 4.33%

6) Ho: s £ 0.75

Ha: s > 0.75

Reject Ho if c² > 36.415

c² = 24(0.9)²/(0.75)² = 34.56

Do not reject Ho

Conclude that there is not a significant increase in variability for this study.

7) From the Chi-square table, we see the 5% critical value is 36.415 and the 10% critical value is 33.1962. Since the test statistic falls between these values, the p-value is between 5% and 10%.

8) Lower limit for s² = 24(0.9)²/39.6941 = 0.4935

Upper limit for s² = 24(0.9)²/12.4012 = 1.5676

Lower limit for s = sqrt(0.4935) = 0.7025

Upper limit for s = sqrt(1.5676) = 1.2520

Under a two-tail test the null and alternative hypotheses would be:

Ho: s = 0.75

Ha: s ¹ 0.75

Under this scenario, we would not reject Ho at a 5% level of significance since the hypothesis standard deviation of 0.75 falls inside the 95% confidence interval.