Question

A food snack manufacturer samples 15 bags of pretzels off the assembly line and weighed their contents. If the sample mean is 10.2 and the sample standard deviation is 0.25, find the 95% confidence interval of the true mean. Interpret the confidence interval.

Point Estimate: ______

Critical Value:

Margin of Error: ________

Confidence Interval: _______

Interpretation:

Answer 1

Solution :

Given that,

= 10.2

s =0.25

n =15

Degrees of freedom = df = n - 1 =15 - 1 =14

Point Estimate: 10.2

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t _/2,df = t_0.025,14 = 2.145 ( using student t table)

Critical Value:2.145

Margin of error = E = t_/2,df * (s /n)

= 2.145* ( 0.25/ 15)

E=0.1385

The 95% confidence interval estimate of the mean is,

- E < < + E

10.2 - 0.1385 < <10.2 + 0.1385

10.0615 < < 10.3385

(10.0615 , 10.3385 )

The 95% confidence interval estimate of the mean is, (10.0615 , 10.3385 )