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A random sample of six cars from a particular model year had the following fuel consumption figures (in miles per gallon). Find the 99% confidence interval for the true mean fuel consumption for cars of this model yearRounded to 4 decimalsLeft and right end points need to be found.DATA:a. 18.6b. 18.4c. 19.2d. 20.8e. 19.4f. 20.5

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princessesther2011

Answer:

99% confidence interval for the true mean fuel consumption for cars of this model year is a lower limit of 18.0251 miles per gallon and an upper limit of 20.9749 miles per gallon.

Step-by-step explanation:

Confidence interval = mean + or - Error margin (E)

mean = (18.6 + 18.4 + 19.2 + 20.8 + 19.4 + 20.5) ÷ 6 = 116.9 ÷ 6 = 19.5 MPG

sd = sqrt[((18.6-19.5)^2 + (18.4-19.5)^2 + (19.2-19.5)^2 + (20.8-19.5)^2 + (19.4-19.5)^2 + (20.5-19.5)^2) ÷ 6] = sqrt[4.81 ÷ 6] = sqrt[0.802] = 0.896 MPG

n = 6

degree of freedom = n - 1 = 6 - 1 = 5

confidence level = 99%

t-value corresponding to 5 degrees of freedom and 99% confidence level is 4.032

E = t × sd/√n = 4.032 × 0.896/√6 = 1.4749 MPG

Lower limit = mean - E = 19.5 - 1.4749 = 18.0251 mpg

Upper limit = mean + E = 19.5 + 1.4749 = 20.9749 mpg

99% confidence interval is between 18.0251 and 20.9749 mpg

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