A popular restaurant takes a random sample n = 25 customers and records how long each occupied a table.The found a sample mean of 1.2 hours and a sample standard deviation of 0.3 hours.Find the 95 percent confidence interval for the mean.
A)1.2 ± .118
B)1.2 ± .124
C)1.2 ± .588
D)1.2 ± .609

Question

Quizplus · Accepted Answer

We can use the formula: 
CI = x̄ ± z*(σ/√n)
Where:
x̄ = sample mean = 1.2
σ = sample standard deviation = 0.3
n = sample size = 25
z = z-score for 95% confidence level = 1.96 (from Z-table or calculator)
Plugging in the values, we get:
CI = 1.2 ± 1.96*(0.3/√25)
CI = 1.2 ± 0.124
Therefore, the 95% confidence interval for the mean is 1.2 ± 0.124, which is answer choice B.

A Popular Restaurant Takes a Random Sample N = 25