In: Math
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
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 = Z0.025 = 1.96
Margin of error = E = Z/2* ( /n)
= 1.96 * (0.4 / 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