This is the last question I have to ask for a project at my university, I have until the 20th of August to answer this, but I have been staring at it all week and have found no clues that I can understand in the book provided to me by the university. I'm bordering on desperate now.
Can packaging of a healthy food product influence child´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: x-bar = 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 µ = 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 α = 0.05 to make your conclusion. (2.5 points)
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)? (0.5 points)
So basically I need the answer to this. I have tried everything I can remember and I find nothing.
I'm certain it has something to do with a hypothesis test for the true mean, but I lack the knowledge on how to properly solve it, or if it is the Z-value or the T-value or something else they are asking me to provide.
I tried using this formula: Z=(3.69-3)/(2.44/√408)=5.71. And then I realised that I didn't even use the α =0.05 that part A demands I use.
In order to pass the test this calculation has to yield a Z-score which is "sufficiently large," for which $Z>Z_{1-.05}=Z_{.95}\approx 1.55$. What that means is if we were to suppose, contrary to what we believe, that the true mean of a child's willingness to eat sliced apples from the smiling apple package was only $3$ (the null hypothesis) then the probability of a Z-score larger than 1.55 is only $5$%. The fact that our Z-score was larger than $1.55$ is considered sufficient evidence to rule out that null hypothesis.
If you want to check the table relating significance to Z-score yourself you can take a look at this link: http://www.z-table.com/.