In: Math
A report from the U.S. Department of Agriculture shows that the mean American consumption of carbonated beverages per year is greater than 52 gallons. A random sample of 30 Americans yielded a sample mean of 69 gallons. Assume that the population standard deviation is 20 gallons, and the consumption of carbonated beverages per year follows a normal distribution. Calculate a 95% confidence interval, and write it in a complete sentence related to the scenario.
Solution :
Given that,
sample mean =
= 69
Population standard deviation =
= 20
Sample size = n =30.
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* ( 20 / 30
)
= 7.16
At 95% confidence interval
is,
- E <
<
+ E
69 - 7.16 <
< 69 + 7.16
61.84<
<76.16
lower limit 61.84
upper limit 76.16
sample mean fall between lower and upper limits
( , )