Question

Determine the decision criterion for rejecting the null hypothesis in the given hypothesis test. Describe the values of the test statistic that would result in rejection of the null hypothesis. A cereal company claims that the mean weight of the cereal in its packets is at least 14 oz. You wish to test this claim at the 0.02 level of significance. The mean weight for the random sample of 45 cerial packets is 13.8 oz with a standard deviation of 0.3 oz. What criterion would be used for rejecting the null hypothesis, that oz ? Reject H0 if test statistic < 2.05. Reject H0 if test statistic > -2.05. Reject H0 if test statistic < -2.05. Reject H0 if test statistic < -2.33.

Answer 1

We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution

Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is
`z_(crit)=-2.05`

We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"Reject H0 if test statistic < -2.05

And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.

Explanation:

Data given and notation

`\bar X=13.8` represent the sample mea n

`s=0.3` represent the sample standard deviation

`n=45` sample size

`\mu_o =68` represent the value that we want to test

`\alpha=0.02` 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 mean is at least 14 oz, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 14`

Alternative hypothesis:
`\mu < 14`

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=(13.8-14)/((0.3)/(√(45)))=-4.47`

Critical region

We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution

Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is
`z_(crit)=-2.05`

We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"

So then the correct answer for this case would be:

Reject H0 if test statistic < -2.05

And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.