Question

You are investigating whether there are enough little marshmallows in a ceral box. You buy a sample of 10 cereal boxes at random. Your sample average is 100 marshmallows. Marshmallow distribution is Normal with a standard deviation of 15 marshmallows. You want to calculate a 90% confidence interval for the mean number of marshmellows. What is the margin of error? (Round your answer to zero decimals. ex: 15.)

Answer 1

Given that,

= 100

s =15

n = 10

Degrees of freedom = df = n - 1 = 10- 1 = 9

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,9 = 1.833    ( using student t table)

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

= 1.833 * (15 / 10)

E = 8.69

Margin of error = E =9.0