Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is believed that the machine is underfilling the bags.

A 26 bag sample had a mean of 427 grams with a standard deviation of 15.
Assume the population is normally distributed. A level of significance of 0.05 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

P-value = 0.0261

We conclude that the machine is under-filling the bags.

Explanation:

We are given the following in the question:

Population mean, μ = 433 gram

Sample mean,
`\bar{x}` = 427 grams

Sample size, n = 26

Alpha, α = 0.05

Sample standard deviation, σ = 15 grams

First, we design the null and the alternate hypothesis

`H_(0): \mu = 433\text{ grams}\\H_A: \mu < 433\text{ grams}`

We use one-tailed(left) t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(427 - 433)/((15)/(√(26)) ) = -2.0396`

Now, we calculate the p-value using the standard table.

P-value = 0.0261

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.

We conclude that the machine is under-filling the bags.