Final answer:
To find the 99% confidence interval for the difference between two population means with unequal variances, use the sample means, sizes, and standard deviations to calculate the interval using the formula and it results in approximately (1.935, 6.065).
Step-by-step explanation:
To find a 99% confidence interval for the difference between two population means with unequal population variances, we use the formula:
x1 - x2 ± z*(√(s1^2/n1 + s2^2/n2)), where x1 and x2 are the sample means, n1 and n2 are the sample sizes, and s1 and s2 are the sample standard deviations.
Given:
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- x1 = 29
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- x2 = 25
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- n1 = 64
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- n2 = 73
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- s1 = 5
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- s2 = 3
To calculate the 99% confidence interval, we'll use the conservative quick rule for degrees of freedom, which uses the smaller of n1 - 1 and n2 - 1. Since the smallest sample size is 64 (n1), the degrees of freedom would be 64 - 1 = 63.
The z-value for a 99% confidence interval is approximately 2.576 (obtained from standard z-tables or statistical software). Now we substitute the values into the formula:
29 - 25 ± 2.576*(√((5^2/64) + (3^2/73)))
Calculate the pooled standard error:
√((5^2/64) + (3^2/73)) ≈ 0.802
So the confidence interval is:
4 ± 2.576 * 0.802 = 4 ± 2.065
Therefore, the 99% confidence interval for the difference between the two population means is approximately (1.935, 6.065).