The average natural gas bill for a random sample of 21 homes in 19810 zip code during the month of March was $311.90 with a sample standard deviation of $51.60.The 90% confidence interval around this sample mean is _______.
A) [$285.98,$337.82]
B) [$292.46,$331.34]
C) [$278.09.$345.71]
D) [$266.82,$356.98]

Question

Quizplus · Accepted Answer

To find the 90% confidence interval, we use the formula:

Sample Mean ± (t-value)*(standard error of the mean)

where t-value comes from the t-distribution with n-1 degrees of freedom and a 90% confidence level. For a sample size of 21, the degrees of freedom is 20 and the t-value is 1.725.

The standard error of the mean is found by dividing the sample standard deviation by the square root of the sample size:

standard error of the mean = 51.60 / sqrt(21) = 11.22

Plugging in the values, we get:

311.90 ± 1.725*11.22

which simplifies to:

311.90 ± 19.34

Therefore, the 90% confidence interval is [$292.46, $331.34], which is closest to choice B.

The Average Natural Gas Bill for a Random Sample of 21