Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 442442 gram setting. Based on a 2323 bag sample where the mean is 438438 grams and the variance is 196196, is there sufficient evidence at the 0.10.1 level that the bags are underfilled or overfilled? Assume the population distribution is approximately normal.

Step 1 of 5:

State the null and alternative hypotheses.

Step 2 of 5:

Find the value of the test statistic. Round your answer to three decimal places.

Step 3 of 5:

Specify if the test is one-tailed or two-tailed.

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Step 5 of 5:

Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: the bags are neither underfilled nor overfilled.

Alternative hypothesis: H_a: the bags are underfilled or overfilled.

H₀: µ = 442 versus H_a: µ ≠ 442

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 442

Xbar = 438

S = 14

n = 23

df = n – 1 = 22

α = 0.1

Critical value = - 1.7171 and 1.7171

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (438 - 442)/[14/sqrt(23))

t = -1.3702

P-value = 0.1844

(by using t-table)

P-value > α = 0.1

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the bags are not underfilled or overfilled.