Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 406 gram setting. It is believed that the machine is underfilling the bags. A 24 bag sample had a mean of 398 grams with a standard deviation of 28. A level of significance of 0.05 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer 1

We will perform a one-sample t-test to see whether the machine is underfilling the bags.

Null Hypothesis: μ >= 406

Alternate Hypothesis: μ < 406

The formula to calculate the t-statistic is:

x-bar = 398, s = 28, n = 24

Hence, t = (398 - 406)/5.72

t = -1.399

Let's calculate the t-critical value at 0.05 and df = n - 1 = 23 which is -1.713 (left-tailed)

Hence, as t>tc, we cannot reject the null hypothesis. Therefore, we conclude that the mean weight is not significantly different than 406 grams. The machine is not under-filling the bags.