Question

In a study of the nutritional qualities of fast foods, the amount of fat was measured for a random sample of 36 hamburgers from a particular restaurant chain. The sample mean and sample standard deviation were found to be 50 grams and 4 grams, respectively. Compute and interpret a 95% confidence interval for the true mean fat content in hamburgers served in these restaurants.

Answer 1

Solution :

Given that,

= 50

s =4

n =36 Degrees of freedom = df = n - 1 =36 - 1 =35

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,35 = 2.030 ( using student t table)

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

= 2.030 * ( 4/ 36)

= 1.3533

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

- E < < + E

50 -1.3533 < <50 + 1.3533

48.6467< < 48.6467

( 48.6467, 48.6467)