Question

The manufacturers of Good-O use two different types of machines to fill their 25 kg packs of dried dog food. On the basis of random samples of size 15 and 18 from output from machines 1 and 2 respectively, the mean and standard deviation of the weight of the packs of dog food produced were found to be 29.38 kg and 0.305 kg for machine 1 and 23.194 kg and 0.122 kg for machine 2. Hence, under the usual assumptions, determine a 95% confidence interval for the difference between the average weight of the output of machine 1 and machine 2. Use machine 1 minus machine 2, stating the upper limit of the interval correct to three decimal places.

Answer #1

n1 = 15

n2 = 18

1 =29.38

2 = 23.194

s1 = 0.305

s2 = 0.122

Pooled Standard Deviation is given by:

=.0.05

ndf = 15 + 18 - 2 = 31

From Table,critical values of t = 2.0395

Confidence Interval:

(29.38 - 23.194) (2.0395 X 0.0783)

= 6.186 0.1597

= (6.026 ,6.346)

Confidence Interval:

**6.026 <
< 6.346**

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