Question

In the article “Sweetening Statistics - What M&Ms Can Teach Us,” M. Paret and E. Martz discussed several statistical analysis that they performed on bags of M&Ms. The authors took a random sample of 200 small bags of peanut M&Ms and calculated the sample mean weight and sample standard deviation of 52.04 g and 2.81 g, respectively.

(a) Determine the 95% confidence interval for the mean weight of all small bags of peanut M&Ms.

(b) According to the manufacturer, the average weight of all small bags of peanut M&Ms should be 49.3 g. Comment on this specification in view of your answer in part (a).

Answer 1

a)

Sample size = n = 200

Sample mean = $\bar{x}$ = 52.04

Standard deviation = s = 2.81

We have to construct 95% confidence interval.

Formula is

$(\bar{x}-E , \bar{x}+E)$

Here E is a margin of error.

E=

Degrees of freedom = n - 1 = 200 - 1 = 199

Level of significance = 0.05

tc = 1.972   ( Using t table)

$E = \frac{1.972*2.81}{\sqrt{200}}=0.3918$

So confidence interval is ( 52.04 - 0.3918 , 52.04 + 0.3918) = > ( 51.6482 , 52.4318)

b)

Claim:  The manufacturer, the average weight of all small bags of peanut M&Ms should be 49.3 g.

The value 49.3 does not belong to the above interval so the manufacturer, the average weight of all small bags of peanut M&Ms should not 49.3 g.