Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 57 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 17.2 and a standard deviation of 2.6. What is the 98% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to 4 decimal places. [,] (Report and round to 4 decimal places)

Answer 1

Solution :

Given that,

= 17.2

s = 2.6

n = 57

Degrees of freedom = df = n - 1 = 57 - 1 = 56

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,56 =2.395

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

= 2.395 * (2.6 / 57)

= 0.8247

Margin of error = 0.8247

The 98% confidence interval estimate of the population mean is,

- E < < + E

17.2 - 0.8247 < < 17.2 + 0.8247

16.3753 < < 18.0247

(16.3753, 180247 )