Question

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A report from the U.S. Department of Agriculture shows that the mean American consumption of carbonated...

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.

Solutions

Expert Solution

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

( , )


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