In 1998, as an advertising campaign the Nabisco Company announced a \"1000 Chips

Question

In 1998, as an advertising campaign the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of their Chips Ahoy! cookies contained at least 1000 chocolate chips(Your professor remembers these, gosh he is old). Dedicated Statistics students at the Air Force Academy purchased some randomly selected bags of cookies and counted the chocolate chips. Some of their data are given below. 1219 1214 1087 1200 1419 1121 1325 1345 1244 1258 1356 1132 1191 1270 1295 1135 What is the population of interest? What is the parameter we are trying to learn something about. Are the assumptions and conditions met for performing a hypothesis test and constructing a confidence interval? What would be appropriate hypotheses for a hypothesis test? Calculate a p-value. This p-value is a probability. What is it the probability of? Explain this in the context of this problem. Do you believe Nabisco's claim? Why or why not? Explain. Construct a 95% confidence interval. Explain what the 95% confidence interval means in the context of this problem.

Explanation / Answer

population of interest  is all the 18-ounce bag of their Chips Ahoy! cookies

we are trying to measure the population mean of cookie chips

as the sample size is less than 40 , we can conduct a ttest as

t = (xbar-mu)/(sd/sqrt(n))

The hypothesis is

H0 : every 18-ounce bag of their Chips Ahoy! cookies does not containe at least 1000 chocolate chips

H1 : every 18-ounce bag of their Chips Ahoy! cookies contained at least 1000 chocolate chips

we calculate the xbar and standard deviation of the sample as

> xbar
[1] 1238.188
> s
[1] 94.282

and n = 16

t = (xbar-mu)/(sd/sqrt(n)) = (1238.18-1000)/(94.28/sqrt(16)) = 10.10

now we check the t table to check the p value for df = n-1 = 16-1 = 15

Please note that this is a directional test ,

and the p value is

The P-Value is < .00001.

The result is significant at p < .05.

Hence we reject the null hypothesis in favor of alternate hypothesis to conclude H1 : every 18-ounce bag of their Chips Ahoy! cookies contained at least 1000 chocolate chips

margin of error is

t*sd/sqrt(n) , we see the value is 2.13

2.13*94.282/sqrt(16) = 50.20

so the 95% ci is

mean+- moe

1238.188 +- 50.20

solving for plus and minus sign

1187.9
1288.3

we are 95% confident that the true value of the mean would lie in this interval

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