A biologist is sampling oranges to determine the amount of juice in each orange. She tested...

A biologist is sampling oranges to determine the amount of juice in each orange. She tested 50 oranges chosen at random. The average amount of juice was = 3.1 ounces with a standard deviation of σ = 0.4 ounces. Find a 95% confidence interval for the population mean number of ounces of juice in an orange.

elect one: a. 2.85 < μ < 3.35

b. 3.04 < μ < 3.16

c. 2.99 < μ < 3.21

d. 3.14 < μ < 3.21

Solutions

Expert Solution

Solution :

Given that,

= 3.1

= 0.4

n = 50

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( / $\sqrt$ n)

= 1.96 * (0.4 / $\sqrt$ 50)

= 0.11

At 95% confidence interval estimate of the population mean is,

- E < < + E

3.1 - 0.11 < < 3.1 +0.11

2.99 < < 3.21

The 95% confidence interval for the population mean is 2.99 < μ < 3.21

milcah answered 4 months ago

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