A health journal conducted a study to see if packaging a healthy food product like junk food would influence children's desire to consume the product. A fictitious brand of a healthy food productlong dashsliced appleslong dashwas packaged to appeal to children. The researchers showed the packaging to a sample of 330 school children and asked each whether he or she was willing to eat the product. Willingness to eat was measured on a 5-point scale, with 1="not willing at all" and 5="very willing." The data are summarized as x = 3.27 and s=2.27. Suppose the researchers knew that the mean willingness to eat an actual brand of sliced apples (which is not packaged for children) is u=3. Complete parts a and b below.

Conduct a test to determine whether the true mean willingness to eat the brand of sliced apples packaged for children exceeded 3. Use α=0.01 to make your conclusion.

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the true mean is higher than 3 at 1% of significance.

Step-by-step explanation:

Data given and notation

[tex]\bar X=3.27[/tex] represent the sample mean

[tex]s=2.27[/tex] represent the sample standard deviation for the sample

[tex]n=330[/tex] sample size

[tex]\mu_o =3[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 higher than 3, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 3[/tex]

Alternative hypothesis:[tex]\mu > 3[/tex]

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:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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:

[tex]t=\frac{3.27-3}{\frac{2.27}{\sqrt{300}}}=2.06[/tex]

P-value

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

[tex]df=n-1=330-1=329[/tex]

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

[tex]p_v =P(t_{(329)}>2.06)=0.02[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the true mean is higher than 3 at 1% of significance.