A random sample of size n1 = 25, taken from a normal population with a standard deviation σ1 = 5, has a mean ¯x1 = 80. A second random sample of size n2 = 36, taken from a different normal population with a standard deviation σ2 = 3, has a mean ¯x2 = 75. Find a 94% confidence interval for μ1 − μ2.
Question
A random sample of size n1 = 25, taken from a normal population with a standard deviation σ1 = 5, has a mean ¯x1 = 80. A second random sample of size n2 = 36, taken from a different normal population with a standard deviation σ2 = 3, has a mean ¯x2 = 75. Find a 94% confidence interval for μ1 − μ2.
Solution
To find a 94% confidence interval for μ1 − μ2, we can use the formula for the confidence interval for the difference between two means in two independent populations, which is:
(¯x1 - ¯x2) ± Z * sqrt[(σ1^2/n1) + (σ2^2/n2)]
Where:
- ¯x1 and ¯x2 are the sample means
- σ1 and σ2 are the standard deviations
- n1 and n2 are the sample sizes
- Z is the Z-score corresponding to the desired confidence level
Given:
- ¯x1 = 80, σ1 = 5, n1 = 25
- ¯x2 = 75, σ2 = 3, n2 = 36
- Confidence level = 94%
First, we need to find the Z-score that corresponds to a 94% confidence level. The Z-score for a 94% confidence level is approximately 1.88 (you can find this value in a standard Z-table or using a calculator).
Now, we can substitute these values into the formula:
(80 - 75) ± 1.88 * sqrt[(5^2/25) + (3^2/36)] = 5 ± 1.88 * sqrt[(1) + (0.25)] = 5 ± 1.88 * sqrt[1.25] = 5 ± 1.88 * 1.118 = 5 ± 2.10
So, the 94% confidence interval for μ1 − μ2 is (2.90, 7.10).
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