Question

A cereal manufacturer produces cereal in boxes having a labeled weight of 15.7 ounces. Suppose that a random sample of 31 boxes have a mean of 16.14 ounces with a standard deviation of 1.18 ounces. Perform a hypothesis test to determine if the true mean weight of this particular cereal exceeds 15.7 ounces. Use α = 0.05.

Answer 1

`t=(16.14-15.7)/((1.18)/(√(31)))=2.076`

`p_v =P(t_((30))>2.076)=0.0233`

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 reject the null hypothesis, and we can conclude that the true mean for the weigths of cereal boxes is higher than 15.7 at 5% of signficance.

Explanation:

Data given and notation

`\bar X=16.14` represent the sample mean

`s=1.18` represent the sample standard deviation

`n=31` sample size

`\mu_o =15.7` represent the value that we want to test

`\alpha=0.05` 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 higher than 15.7 or no, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 15.7`

Alternative hypothesis:
`\mu > 15.7`

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)

Calculate the statistic

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

`t=(16.14-15.7)/((1.18)/(√(31)))=2.076`

P-value

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

`df=n-1=31-1=30`

Since is a one side test the p value would be:

`p_v =P(t_((30))>2.076)=0.0233`

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 toreject the null hypothesis, and we can conclude that the true mean for the weigths of cereal boxes is higher than 15.7 at 5% of signficance.