Question

Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who works for the manufacturer of the cereal, wants to check if the packaging process is working improperly: boxes are being over-filled or under-filled. A sample of 40 cereal boxes of Granola Crunch yields a mean weight of 1.02 pounds of cereal per box. Assume that the weight is normally distributed with a population standard deviation of 0.04 pound. Conduct a hypothesis test at the 5% significance level.

Answer 1

`z=(1.02-1)/((0.04)/(√(40)))=3.162`

`p_v =2*P(z>3.162)=0.0016`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

Explanation:

Data given and notation

`\bar X=1.02` represent the sample mean

`\sigma=0.04` represent the population standard deviation

`n=40` sample size

`\mu_o =1` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is equal to 1 pound or no :

Null hypothesis:
`\mu =1`

Alternative hypothesis:
`\mu \\eq 1`

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(1.02-1)/((0.04)/(√(40)))=3.162`

P-value

Since is a two-sided test the p value would be:

`p_v =2*P(z>3.162)=0.0016`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.