In a simple random sample of 48 men, the mean wrist circumference of the men is 19.3 centimeters with a standard deviation of 1.2 centimeters. round your answer to the nearest hundredth. which interval is the 95% confidence interval for the mean wrist circumference for all men? 18.99,19.61 18.96, 19.64 12.85, 19.75 19.14, 19.46

Choice B
The 95% confidence interval for the population mean mu is (18.96, 19.64)



Work Shown:

xbar = 19.3
s = 1.2
n = 48
z = 1.960 (use a calculator or table to compute this critical value)

Lower Limit L
L = xbar - z*(s/sqrt(n))
L = 19.3 - 1.960*(1.2/sqrt(48))
L = 18.9605180417166
L = 18.96

Upper Limit U
U = xbar + z*(s/sqrt(n))
U = 19.3 + 1.960*(1.2/sqrt(48))
U = 19.6394819582836
U = 19.64

The confidence interval is therefore (L,U) = (18.96, 19.64)

Ithink you’re answering is b