In: Statistics and Probability
a food snack manufacturer samples 13 bags of pretzels off the assembly line and weighs their contents. If the sample mean is 13.7 oz and the sample standard deviation is 0.05 oz find the 95% confidence interval true mean.
a. (13.4, 14.0)
b. (11.0, 16.0)
c. (11.4, 16.4)
d. (12.4, 15.0)
Solution :
Given that,
= 13.7
s =0.05
n =13 Degrees of freedom = df = n - 1 =13 - 1 = 12
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2= 0.05 / 2 = 0.025
t /2,df = t0.025,12 = 2.179 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.179 * ( 2.17/ 13)
E = 1.3
The 95% confidence interval estimate of the population mean is,
- E < < + E
13.7 - 1.3< <13.7 + 1.3
12.4< < 15.0
( 12.4, 15.0 )