Question

A manufacturer of potato 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 of overfilling the bags. A 61 bag sample had a mean of 424 grams with a standard deviation of 26. Assume the population is normally distributed. A level of significance of 0.01 will be used. Specify the type of hypothesis test.

Answer 1

`t=(424-420)/((26)/(√(61)))=1.202`

`p_v =2*P(t_((60))>1.202)=0.234`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.

Explanation:

Data given and notation

`\bar X=424` represent the sample mean

`s=26` represent the sample standard deviation

`n=61` sample size

`\mu_o =420` represent the value that we want to test

`\alpha=0.01` represent the significance level for the hypothesis test.

t 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 different from 420 or not, the system of hypothesis would be:

Null hypothesis:
`\mu = 420`

Alternative hypothesis:
`\mu \\eq 420`

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-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:

`t=(424-420)/((26)/(√(61)))=1.202`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=61-1=60`

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

`p_v =2*P(t_((60))>1.202)=0.234`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.