# Normal Distribution and Chips Ahoy

Case Study: The Chips Ahoy! 1,000 Chips Challenge

Review the case study on page 359 of the textbook.

The data on the number of chocolate chips per bag for 42 bags of Chips Ahoy! cookies were obtained by the students in an introductory statistics class at the United States Air Force Academy in response to the Chips Ahoy! 1,000 Chips Challenge sponsored by Nabisco, the makers of Chips Ahoy! Use the data collected by the students to answer the following questions and to conduct the analyses required in each part.

A. Obtain and interpret a point estimate for the mean number of chocolate chips per bag for all bags of Chips Ahoy! Cookies. (Note: The sum of the data is 52,986.) B .Construct and interpret a normal probability plot, boxplot, and histogram of the data. C. Use the graphs in part (b) to identify outliers, if any. D.Is it reasonable to use the one-mean t-interval procedure to obtain a confidence interval for the mean number of chocolate chips per bag for all bags of Chips Ahoy! cookies? Explain your answer. E. Determine a 95% confidence interval for the mean number of chips per bag for all bags of Chips Ahoy! cookies, and interpret your result in words. (Note: = 1261.6; s = 117.6.)

A. ͞X =∑Xi/n = 52986/42 = 1261.57

BASED ON THE SAMPLE DATA ,WE ESTIMATE THAT MEAN CHIPS PER BAG TO APPROXIMATLY 1261.57 CHIPS PER BAG.

B. GRAPHS ARE ATTACHED

C. I NOTICED MAY BE 1545 OR 1546 ARE OUTLINERS, THAT WHAT IT LOOKED LIKE AT BOX PLOT.

D. I THINK YOU CAN USE ONE MEAN T-INTERVAL PROCEDURE EVEN IF THE STANDARDIZED DEVIATION NOT GIVEN BY USING THE FOLLOWING

STEP 1. FOR A CONFIDENCE LEVEL OF 1-α, USE TABLE IV TO FINDTα/2 WITH DF= N-1 WHERE N IS SAMPLE SIZE WE WANT 95 % CONFIDENCE INTERVAL SO α= 1-0.95 =0.05

FOR N = 42 DF = 42-1= 41

FROM TABLE IV Tα/2=T0.05/2= T0.025=2.020

STEP 2 THE CONFIDENCE INTERVAL FOR µ IS FROM

͞X –T(α/2) * S/√N TO ͞X + T(α/2)*S/√N

͞X = 1261.57

S= ͞X-µ/(N-1/√N) = 1261.57-1000...

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