Question

In: Statistics and Probability

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her...

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)

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Expert Solution


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 )


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