A possible important environmental determinant of lung function in children is the amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups:

A. Group 1 consists of 23 nonsmoking children 5-9 years of age, both of whose parents smoke, who have a mean FEV of 2.1 L and standard deviation of 0.7 L;

B. Group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have a mean FEV of 2.3 L and standard deviation of 0.4 L. Suppose 40 children, both of whose parents smoke, and 50 children, neither of whose parents smoke are recruited for the study. How much power would such a study have using a two-sided test with signicance level = 0.05, assuming that the estimates of the population parameters in the pilot study are correct?

Question

Mathematics

Kate

20 June, 03:42

A possible important environmental determinant of lung function in children is the amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups:

A. Group 1 consists of 23 nonsmoking children 5-9 years of age, both of whose parents smoke, who have a mean FEV of 2.1 L and standard deviation of 0.7 L;

B. Group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have a mean FEV of 2.3 L and standard deviation of 0.4 L. Suppose 40 children, both of whose parents smoke, and 50 children, neither of whose parents smoke are recruited for the study. How much power would such a study have using a two-sided test with signicance level = 0.05, assuming that the estimates of the population parameters in the pilot study are correct?

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Answers (1)

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Darion Archer · Answer 1

At 95% confidence interval, the true mean difference is (-0.547524, 0.147524)

Step-by-step explanation:

Given that

Treatment I: n1=23, mean = 2.1, s1 = 0.7

Treatment II: n2 = 20, mean = 2.3, s1=0.4

The 95% confidence interval for the difference is given as:

{ (mean 1 - mean 2) + or - t0. 0025,35 (√S1^2/n + S2^2/n) }

{ (2.1 - 2. 3) + or - 2.030 (√ (0.7) ^2/23 + (0.4) ^2/20) }

= 0.2 + or - 2.030 (0.171185127)

= 0.2 + or - 0.347505809

= (-0.547524, 0.147524)

We are 95% confident that the limit of - 0.547524 FEV and 0.147524FEV definitely contain the difference between the 2 population means. For the fact that they do contain 0, this confidence interval proves that it is very possible that the two population means are equal, therefore, there is not a significant difference between the two means.