"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $\backslash (2.02!!!$Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $\backslash )1$, but now I wasn't so sure. However, I was determined to find out. I went to $53$of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

• $5$stores $@\backslash (2.02$

•$15$stores $@\backslash )0.25$

• $3$stores $@\backslash (1.29$

• $6$stores $@\backslash )0.35$

• $4$stores $@\backslash (2.27$

• $7$stores $@\backslash )1.50$

• $5$stores $@\backslash (1.89$

• $8$stores $@0.75$.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most$\backslash )1,$then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

Step by step solution

01

Given information

Given the price per box of mac and cheese from $53$grocery stores,

Sample size$=53$

02

Explanation

Hypothesis: The null hypothesis says that the average price of a box of macaroni and cheese is less than $\$1$, while the alternate hypothesis claims that the average price is greater than $\$1$

$H_0:\mu \le 1$

$H_0:\mu >1$

We utilize the students' t-distribution for this test because we wouldn't know the population standard deviation.

$$\begin{gathered} t_n-1=t_{53}-1 \\ =t_{52} \end{gathered}$$

where$\mu$is the population mean and $\overline{x}$is the sample.

The sample mean is,

$$\begin{gathered} \overline{X}=\frac{2.02+2.02+...+0.75}{53} \\ =1.03 \end{gathered}$$

Thet-test statistic is calculated as,

$$\begin{gathered} t=\frac{\overline{X}-\mu }{s/\sqrt{n}} \\ \text{where}\overline{X}\text{isthepopulationmean} \\ n=53 \\ t=\frac{1.03-1}{0.75/\sqrt{53}} \\ =\frac{0.03}{0.103} \\ =0.291 \end{gathered}$$

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