A sample of size n = 42 has sample mean x? = 53.1 and sample standard deviation s = 8.2. Construct an 80% confidence interval for the population mean ?. Round the answers to one decimal place. An 80% confidence interval for the population mean ? is 51.5 < ? < 54.7. If the sample size were n = 31, would the confidence interval be narrower or wider? The confidence interval would be (Choose one)
Added by Derrick F.
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We can use the formula: Margin of Error = Z * (s / sqrt(n)) where Z is the Z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size. For an 80% confidence interval, the Z-score is 1.28 (you can find Show more…
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