Question

Packaging of a children’s health food. Can packaging of a healthy food product influence children’s desire to consume the product? This was the question of interest in an article published in the Journal of Consumer Behaviour (Vol. 10, 2011). A fictitious brand of a healthy food product—sliced apples—was packaged to appeal to children (a smiling cartoon apple was on the front of the package). The researchers showed the packaging to a sample of 408 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 follows: \(\bar x = 3.69\) , s = 2.44. Suppose the researchers knew that the mean willingness to eat an actual brand of sliced apples (which is not packaged for children) is \(\mu = 3\).

a. Conduct a test to determine whether the true mean willingness to eat the brand of sliced apples packaged for children exceeded 3. Use\(\alpha = 0.05\)

to make your conclusion.

b. The data (willingness to eat values) are not normally distributed. How does this impact (if at all) the validity of your conclusion in part a? Explain.

Short answer

a. There is evidence to support the claim that the true mean exceeds 3.

b. Since the sample size is too large, the condition to use the z-test is satisfied. Therefore, there is no effect on the conclusion's validity, although the data are not normally distributed.

Step-by-step solution

Given information

Let X represents the willingness of school children to eat an actual brand of sliced apples.

A random sample of size 408 has a mean\(\bar x = 3.69\)and standard deviation of s=2.44.

Need to test the researcher's claim that the true mean willingness to eat the brand of sliced packaged for children exceeded 3.

Defining the null hypothesis and obtaining the test statistic

a.

The null and alternative hypotheses are:

\({H_0}:\mu = 3\)against

The test statistic is:

\(\begin{aligned}z &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{3.69 - 3}}{{\frac{{2.44}}{{\sqrt {408} }}}}\\ &= \frac{{0.69}}{{0.1208}}\\ &= 5.71\end{aligned}\)

Therefore, the test statistic is \(z = 5.71\).

Obtaining the p-value and interpreting the result

The p-value for the right-tailed test is:

\(\begin{aligned}p &= P\left( {Z > 5.71} \right)\\ \approx 0.00\end{aligned}\).

Since the p-value is less than 0.05, reject the null hypothesis\(\mu = 3\).

Therefore, evidence to support the claim that the true mean exceeds 3.

Explaining the assumptions

b.

Since the sample size is too large, the condition to use the z-test is satisfied. Therefore, there is no effect on the conclusion's validity, although the data are not normally distributed.