Unformatted text preview: (42.8 – 33.4) +
2.432(.947) = 9.4 2.30 = 11.7. No upper or lower bound can, standing alone, indicate significant differences. For
example, in this problem, for all the reader knows, a lower limit may be less than zero. 34. The data is summarized in the table. Brand
1
2 a. Start with a probability statement n
4
4 mean
13.9
12.2 s
1.22
1.01 So a (1)100%
confidence interval is given . b. The pooled variance is given by
deviation is , so the pooled standard
The critical value for this confidence interval is given by t.025;6 = 2.447, and the interval is given by . The confidence interval indicates that there is no significant difference between the population means.
c. The difference between the two intervals is a matter of the method for calculating the standard error and the degrees of
freedom. The standard error is , virtually identical to the pooled standard error. The degrees of freedom are . So 5 degrees of freedom will be used to find the critical value t.025;5 = 2.571, giving the 95% confidence interval
. This interval is .20 (5%) units wider than the pooled interval in part (b). This is entirely due to the one extra
degree of freedom resulting in a smaller critical value. This shows...
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 Spring '04
 Chung
 Normal Distribution, Standard Error, 4 week, 0.1%, actual size, 1 min, 5 degrees

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