Confidence IntervalsThe scores of 10-year-old children on an IQ test has a standard deviation of 12. Let’s suppose for a sample of 36 children, the mean IQ comes out to be 75. What is the interval in which the population mean will belong if you want to be 90% confident of your estimation?(74.45, 75.55)(71.63, 78.37)(70.87, 79.14)(71.71, 78.29)
Question
Confidence IntervalsThe scores of 10-year-old children on an IQ test has a standard deviation of 12. Let’s suppose for a sample of 36 children, the mean IQ comes out to be 75. What is the interval in which the population mean will belong if you want to be 90% confident of your estimation?(74.45, 75.55)(71.63, 78.37)(70.87, 79.14)(71.71, 78.29)
Solution
To solve this problem, we need to use the formula for a confidence interval:
CI = X ± Z * (σ/√n)
where:
- CI is the confidence interval
- X is the sample mean
- Z is the Z-score (which corresponds to the desired level of confidence)
- σ is the standard deviation of the population
- n is the size of the sample
Given in the problem, we have:
- X = 75
- σ = 12
- n = 36
We need to find the Z-score for a 90% confidence level. The Z-score for a 90% confidence level is 1.645 (you can find this value in a Z-table or use a calculator that provides this value).
Substituting these values into the formula, we get:
CI = 75 ± 1.645 * (12/√36)
Solving the equation gives us:
CI = 75 ± 3.29
So, the confidence interval is (75 - 3.29, 75 + 3.29) = (71.71, 78.29)
Therefore, the correct answer is (71.71, 78.29).
Similar Questions
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