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. 42.15 ± 1.7109 (0.65) / 25

c. 42 ± 1.7109 (0.65) / 25

d. 42 ± 1.645 (0.65) / 5

Answer #1

solution

Given that,

= 42.15

s =0.65

n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t 0.05,24 = 1.7109 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 1.7109* ( 0.65/ 25)

The 90% confidence interval is

42.15 ± 1.7109* ( 0.65/ 25)

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