Question

Case Study Discussion for Module 6

The online discussion for module 6 draws on the case study for chapter 8, "The 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 UnitedStates 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 toanswer the following questions and to conduct the analyses required in each part.


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.)
Construct and interpret a normal probability plot, boxplot, and histogram of the data.
Use the graphs in part (b) to identify outliers, if any.
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.
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.)

Answer 1

(a) Point Estimate for mean = Xbar = 52986/ 42 = 1261.57

(b) I had asked you for the data to answer this question, but haven't heard from you yet, so can not draw graphs.

(c) Since we do not know population standard deviation we have to use t-score, and since the sample is large ( n>30), we can assume the sampling distribution to be approximately normal, and hence can use one-sample-t-interval.

(d) Here Xbar = 1261.57 and s = 117.6,

Margin of Error E = t*s/sqrt(n) = 2.020 *117.6/sqrt(42) = 36.65

So using TI-84 calculator, I have calculated 95% CI as follows:

( 1224.9 , 1298.2)

B)

E)

One-Sample T: No.of Chips

Variable      N    Mean StDev SE Mean       95% CI

No.of Chips 42 1261.6 117.6     18.1 (1224.9, 1298.2)

There is 95% probability that TRUE Mean Number of chips are between 1224.9 and 1298.2.

Hope this helps.

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Thanks.

Hope this helps.

Please rate this highly and award full points.

Thanks.