Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is believed that the machine is underfilling the bags. A 24 bag sample had a mean of 414 grams with a standard deviation of 15. Assume the population is normally distributed. A level of significance of 0.1 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

Answer 1

We fail to reject the null hypothesis that the bag filling machine works correctly at the 420 gram setting at the level of significance of 0.1. The p-value of the test statistic is 0.0250

Explanation:

We have the following null and alternative hypothesis

`H_(0): \mu = 420` vs
`H_(1): \mu < 420` lower-tail alternative.

For n = 24,
`\bar{x} = 414` and
`s = 15`.

`\bar{X}` is normally distributed with a mean
`\mu` and a standard deviation of
`(15)/(√(24))` (approx). Therefore, we can use as test statistic

`Z = \frac{\bar{X}-420}{15/√(24)}` and the observed value is

`z = (414-420)/(15/√(24)) = -1.9596`

p-value = P(Z < -1.9596) = 0.0250

We can use a table from a book or a programming language to find this probability P(Z < -1.9596).

You can use the instruction pnorm(-1.9596) in the R statistical programming language.

Because the p-value is greater than 0.1 (0.0250 > 0.1) we fail to reject the null hypothesis at the level of significance of 0.1.