Question

The manager of a gasoline supply corporation wants to estimate the actual amount of gasoline contained in standard 42-gallon barrels purchased from a nationally known manufacturer. The manufacturer’s specifications state that the amount of gasoline is normally distributed with a standard deviation of 0.65 gallon. A random sample of 25 barrels is selected, and the sample mean amount of gasoline per 42-gallon barrel is 42.15 gallons. Construct a 90% confidence interval estimate for the population mean amount of gasoline contained in a standard 42-gallon barrel.

Select one:

a. 42.15 ± 1.645 (0.65) / 5

b. 42 ± 1.645 (0.65) / 5

c. 42.15 ± 1.7109 (0.65) / 25

d. 42 ± 1.7109 (0.65) / 25

Answer #1

solution:

Given data

Sample size (n) = 25

Sample Mean () = 42.15

Standard deviation () = 0.65

For 90% confidence interval , = 1 - CL = 1 - 0.90 = 0.1

Critical value : Zc = Z(/2) = Z(0.05) = 1.645 [ use Z distribution table ]

The confidence interval for population proportion is given by

CI : **±**
Z * (
)

: 42.15 **± 1.645** * ( 0.65 /
)

: 42.15 **± 1.645 * (0.65 / 5)**

**Option
-A** is correct

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