Fifteen SmartCars were randomly selected and the highway mileage of each was noted The analysis yielded a mean of 47 miles per gallon and a sample standard deviation of 5 miles per gallon Which of the following would represent a 90% confidence interval for the average highway mileage of all SmartCars? a. 47 plusminus (1.753*(5 + 3.8730) b. 47 plusminus (1.345*(5 + 3.8730) c. 47 plusminus (1.765*(5 + 3.8730) d. 47 plusminus (1.645*(5 + 3.8730)

a. 47 plusminus (1.753*(5 ÷ 3.8730)

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 5

n = number of samples = 15

z represents the test statistic

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the test statistic score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 15 - 1 = 14

Since confidence level = 90% = 0.90, α = 1 - CL = 1 – 0.90 = 0.1

α/2 = 0.1/2 = 0.05

the area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 1 - 0.05 = 0.95

Looking at the t distribution table,

z = 1.753

Margin of error = 1.753(5 ÷ 3.8730)

Confidence interval = 47 ± 1.753(5 ÷ 3.8730)

michael scott & dwight schrute

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