Question

A candy manufacturer claims that the average weight of their individually wrapped candy is 3 grams. Sixty-four candy pieces were randomly selected and the average weight of those pieces was 2.8 grams with a standard deviation of 0.5 gram. You are interested in determining if the mean weight of the company's candy is less than 3 grams. Let µ represent the true mean weight of all the manufacturer's individually wrapped candies. If the p-value is 0.0007, what is the correct decision? Let alpha = 0.05.

Answer 1

`t=(2.8-3)/((0.5)/(√(64)))=-3.2`

`p_v =P(t_((63))<-3.2)\approx 0.0007`

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 reject the null hypothesis, so we can conclude that the true mean is significantly lower than 3 grams at 5% of signficance.

Explanation:

Data given and notation

`\bar X=2.8` represent the sample mean

`s=0.5` represent the sample standard deviation for the sample

`n=64` sample size

`\mu_o =3` 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 lower than 3 grams, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 3`

Alternative hypothesis:
`\mu < 3`

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=(2.8-3)/((0.5)/(√(64)))=-3.2`

P-value

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

`df=n-1=64-1=63`

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

`p_v =P(t_((63))<-3.2)\approx 0.0007`

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 reject the null hypothesis, so we can conclude that the true mean is significantly lower than 3 grams at 5% of signficance.