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. Dedicated statistics students at the Air Force Academy (no kidding) 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 Find a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies Based on this confidence interval, is the Nabisco Company justified in making this claim? yes

Answer 1

	X	X2
	1219	1485961
	1214	1473796
	1087	1181569
	1200	1440000
	1419	2013561
	1121	1256641
	1325	1755625
	1345	1809025
	1244	1547536
	1258	1582564
	1356	1838736
	1132	1281424
	1191	1418481
	1270	1612900
	1295	1677025
	1135	1288225
Sum =	19811	24663069

The sample mean is computed as follows:

$\bar X = \frac{1}{n} \sum_{i=1}^n X_i = \frac{ 19811}{ 16} = 1238.188$

Also, the sample variance is

$s^2 = \frac{ 1}{n-1}\left(\sum_{i=1}^n X_i^2 - \frac{1}{n}\left(\sum_{i=1}^n X_i\right)^2 \right) = \frac{ 1}{ 16-1}\left( 24663069 - \frac{ 19811^2}{ 16} \right) = 8889.096$

Therefore, the sample standard deviation s is

$s = \sqrt{s^2} = \sqrt{ 8889.096} = 94.282$

The number of degrees of freedom are df = 16 - 1 = 15, and the significance level is α=0.05.

Based on the provided information, the critical t-value for α=0.05 and df = 15 degrees of freedom is t_c = 2.131

The 95% confidence for the population mean μ is computed using the following expression

C1 = (*_*)

Therefore, based on the information provided, the 95 % confidence for the population mean μ is

$CI = \left(1238.188 - \frac{ 2.131 \times 94.282}{\sqrt{ 16}} , 1238.188 + \frac{ 2.131 \times 94.282}{\sqrt{ 16}} \right)$

CI = (1187.949, 1288.427) which completes the calculation.

Since the objective is to make at least 1000 chips and lower limit of CI is more than 1000

Hence, claim is correct