Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 58 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 14.9 and a standard deviation of 1.3. 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,

= 14.9

s = 1.3

n = 58

Degrees of freedom = df = n - 1 = 58 - 1 = 57

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,57 =2.394

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

= 2.394 * ( 1.3 / 58 )

= 0.4086

Margin of error = 0.4086

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

- E < < + E

14.9 - 0.4086 < < 14.9 + 0.4086

14.4914 < < 15.3086

(14.4914, 15.3086 )