Question

1. The U.S. Department of Agriculture reported a random sample of 36 Americans consumed an average of 69 gallons of carbonated beverages per year. Assume that the population standard deviation is 20 gallons. Find the 95% confidence interval for the number of gallons of carbonated beverages consumed per year by Americans.(round the final answer to 3 decimal places.)

Answer 1

Hence, the confidence interval is
`(63.517,74.483)` gallons.

Given :

Sample size
`=36`

Sample mean
`=69` gallons

Standard deviation
`=20` gallons

So, the standard error
`=(20)/(√(36))=20/6=3.33`

Confidence level
`=95\%`

To find :

The
`95\%` confidence interval for the number of gallons of carbonated beverages consumed per year by Americans.

Explanation :

`\because` Confidence interval
`=(\text{Sample mean}-\text{margin error},\text{Sample mean}+\text{margin error})`

Margin error
`=Z_{(\alpha)/(2)}*(\sigma)/(√(n))`

Here,
`\alpha=1-95\%=1-0.95`

`\Rightarrow \alpha=0.05\Rightarrow (\alpha)/(2)=0.025`

`Z_{(\alpha)/(2)}=Z_(0.025)=1.645` from the
`z-`table.

So, Margin error
`=Z_{(\alpha)/(2)}*(\sigma)/(√(n))`

`=1.645*(20)/(6)=1.645*3.33`

`=5.483`

Therefore, the confidence interval is:

`(69-5.483,69+5.483)`

`=(63.517,74.483)` gallons.