Question

The weights of adult gray tree frogs are known to be normally distributed with a population standard deviation of 0.12 ounces. 34 mature gray tree frogs were weighed. The mean weight of the sample was ?̅= 0.25 oz. Compute the margin of error (E) to construct a 95% confidence interval for the population mean for a sample of size 34. Round, if necessary, to two (2) decimal places. E =

Answer #1

= Solution :

Given that,

Point estimate = sample mean =
= 0.25

Population standard deviation =
= 0.12

Sample size n =34

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2
* (
/n)

= 1.96 * ( 0.12 / 34
)

E= 0.04

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