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A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 65 specimens and counts the number of seeds in each. Use her sample results (mean = 77.3, standard deviation = 5.5) to find the 80% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).80% C.I. =Answer should be obtained without any preliminary rounding.

Sagot :

The answer is (76.42 ; 78.17)

Given that:

A botanist is attempting to determine how many seeds are typically present in a particular fruit. She collects 65 samples and counts how many seeds are present in each. Using her sample data, the mean is 77.3 and the SD is 5.5.

Sample size, n = 65

Mean, xbar = 77.3

Standard deviation, s = 5.5

Confidence level, Zcritical at 80% = 1.28

Confidence interval :

Xbar ± Margin of error

Margin of Error = Zcritical * s/sqrt(n)

Margin of Error = 1.28 * 5.5/sqrt(65)

Margin of Error =0.873

Lower boundary = 77.3 - 0.873 =76.42

Upper boundary = 77.3  + 0.873 = 78.17

Therefore, the answer is (76.42 ; 78.17)

To learn more about Standard deviation click here:

brainly.com/question/16555520

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