Question

6. Let X equal the weight (in pounds) of a “12-ounce” can of buttermilk biscuits. Assume that the distribution of X is approximately normally distributed. We measure 18 net weights and find out that the sample mean is 11.3 and sample standard deviation is 4. Find a 95% confidence interval for µ. (Hint: use tdistribution)

A. 1.45 B. 1.96 C. 2.10 D. 7.07 E. 5.21

Answer 1

Solution :

Given that,

= 11.3

s = 4

n = 18

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

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,17 =2.10

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

= 2.10 * (4 / 18)

= 1.99

Margin of error = 1.99

The 95% confidence interval estimate of the population mean is,

- E < < + E

11.3 - 1.99 < < 11.3 + 1.99

9.31 < < 13.29

(9.31, 13.29 )