Question

Question 14 3 pts Small Sample Mean Problem. The maker of potato chips uses an automated packaging machine to pack its 20-ounce bag of chips. At the end of every shift 18 bags are selected at random and tested to see if the equipment needs to be readjusted. After one shift, a sample of 18 bags yielded the following data. mean = 20.45 s=.80 n = 18. A 95% confidence interval would have what as the Bound of Error? 20.052 to 20.848 0.398 95% 2.110

Answer 1

Solution :

Given that,

$\bar x$ = 20.45

s =0.80

n =18

Degrees of freedom = df = n - 1 = 18- 1 = 17

a ) At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

  $\alpha$/ 2$\alpha$ = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,17 = 2.110

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.110 * ( 0.80/ $\sqrt$ 18)

=0.398

answer = 0.398