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) [$292.48, $331.32]
C) [$278.09. $345.71]
D) [$266.82, $356.98]

Question

Quizplus · Accepted Answer

We need to use the t-distribution since the sample size is small (<30) and the population standard deviation is unknown. The formula for the confidence interval is:
(sample mean) ± t(α/2, n-1) * (sample standard deviation / sqrt(n))
Where t(α/2, n-1) is the t-value obtained from the t-distribution table with a degree of freedom of n-1 and a confidence level of 90%. From the table, t(0.05, 20) = 1.725.
Plugging in the values, we get:
311.90 ± 1.725 * (51.60 / sqrt(21))
= 311.90 ± 14.42
Therefore, the 90% confidence interval is [$292.48, $331.32].

The Average Natural Gas Bill for a Random Sample of 21