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 product-sliced apples —was packaged to appeal to children. The researchers showed the packaging to a sample of 328 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 $\bar{x}=3.59$ and $s=1.93$. Suppose the researchers knew that the mean willingness to eat an actual brand of sliced apples (which is not packaged for children) is $\mu=3$. Complete parts a and $\mathbf{b}$ below. 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. State the null and alternative hypotheses. \[ \begin{array}{l} H_{0}: \mu=3 \\ H_{a}: \mu>3 \end{array} \] Find the test statistic. $z=\square$ (Round to two decimal places as needed.)
Solution
Step 1 ：State the null and alternative hypotheses. The null hypothesis \(H_{0}: \mu=3\) and the alternative hypothesis \(H_{a}: \mu>3\).
Step 2 ：Find the test statistic. We are given the sample mean (\(\bar{x}=3.59\)), the sample standard deviation (\(s=1.93\)), the population mean (\(\mu=3\)), and the sample size (n=328).
Step 3 ：We can use these values to calculate the test statistic using the formula for a z-score: \(z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\).
Step 4 ：Substitute the given values into the formula: \(z = \frac{3.59 - 3}{\frac{1.93}{\sqrt{328}}}\).
Step 5 ：The calculated z-score is approximately 5.54. This value represents how many standard deviations the sample mean is away from the population mean.
Step 6 ：Final Answer: The test statistic is \(\boxed{5.54}\).
