A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 38 specimens and counts the number of seeds in each. Use her sample results (mean = 33.9, standard deviation = 6.8) to find the 98% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.  98% C.I. =

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Q: Suppose you select a random sample of N=81, and find X = 50, and s = 9.9. Determine the 99%…

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Q: Find a 95% confidence interval for a mean, µ, where n = 324, x̄ = 2.7, s = .91, and se = .06.

A: Given: n = 324, x̄ = 2.7, s = .91, se = .06.

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A: Given Data: Confidence level - 90%: The expression for the confidence level is,

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Q: find the margin of error for a sample mean, given n=16, s=4, Xbar=17. for a 95% confidence interval.

A: The margin of error for sample mean is given by,

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