The 95% confidence interval of the mean for = 13.0, s = 1.6, and N = 21 is _________.
A) 12.01 to 13.99
B) 12.27 to 13.73
C) 11.05 to 15.05
D) 12.95 to 14.95

Question

The 95% confidence interval of the mean for  = 13.0, s = 1.6, and N = 21 is _________.
A) 12.01 to 13.99 
B) 12.27 to 13.73 
C) 11.05 to 15.05 
D) 12.95 to 14.95

Quizplus · Accepted Answer

The formula for the 95% confidence interval of the mean is:

CI = X̄ ± (t_critical)(s/√N)

where X̄ is the sample mean, s is the sample standard deviation, N is the sample size, and t_critical is the t-value from the t-distribution with N-1 degrees of freedom at the 95% confidence level.

Using the given values, we have:

X̄ = 13.0
s = 1.6
N = 21
t_critical = 2.079 (from a t-table with 20 degrees of freedom and a 95% confidence level)

Plugging these values into the formula, we get:

CI = 13.0 ± (2.079)(1.6/√21)
CI = 13.0 ± 0.73
CI = (12.27, 13.73)

Therefore, the 95% confidence interval of the mean for the data is 12.27 to 13.73, and the answer is B.

The 95% Confidence Interval of the Mean for =