Question

It is supposed that a machine, used for filling plastic bottles with a net volume of 16.0 oz, on average, does not perform according to specifications. An engineer will collect 15 measurements and will reset the machine if there is evidence that the mean fill volume is different from 16 oz. The resulting data from a random sample yield \overline{x}=16.0367x=16.0367 and s=0.0551. Should the engineer reset the machine with a level of significance 5%? Find also the pp-value.

Answer 1

Step-by-step explanation:

The null hypothesis is that the mean fill volume is 16 oz, and the alternative hypothesis is that the mean fill volume is different from 16 oz. The level of significance is 5%, and the sample size is 15. The test statistic is t=0.0367t=0.0367. The p-value is 0.9738.

Since the p-value is greater than the level of significance, we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the mean fill volume is different from 16 oz. The engineer should not reset the machine.

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The test statistic is calculated as follows:

t = (x - μ) / s / sqrt(n)

= (16.0367 - 16) / 0.0551 / sqrt(15)

= 0.0367

The p-value is calculated using the t-distribution with 14 degrees of freedom.

p-value = 2 * t.cdf(-0.0367, 14)

= 0.9738

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