Math  /  Data & Statistics

QuestionPart 1 of 4 Points: 0 of 1 Save
The accompanying data set includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are n=36,xˉ=12.192oz,s=0.099ozn=36, \bar{x}=12.192 \mathrm{oz}, \mathrm{s}=0.099 \mathrm{oz}. Assume that a simple random sample has been selected. Use a 0.01 significance level to test the claim that cans of Coke have a mean volume of 12.00 ounces. Does it appear that consumers are being cheated?
Click the icon to view the data set of regular Coke can volumes.
Identify the null and alternative hypotheses. H0\mathrm{H}_{0} : \square \square \square H1\mathrm{H}_{1} : \square \square (Type integers or decimals. Do not round.)

Studdy Solution

STEP 1

What is this asking? We're checking if the *average* amount of Coke in a can is actually 12 ounces, and if not, are customers getting less than what they paid for? Watch out! Don't mix up the *sample mean* and the *population mean*!
We're testing what's true for *all* Coke cans, not just the ones in our sample.

STEP 2

1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Make a decision

STEP 3

We want to see if the *true average* volume is different from **12 ounces**.
Our **null hypothesis** H0H_0 is that the average is exactly **12 ounces**.
Mathematically, we write H0:μ=12H_0: \mu = 12.

STEP 4

The **alternative hypothesis** H1H_1 is what we're trying to prove.
Here, we're checking if customers are being cheated, meaning the average volume is *less than* **12 ounces**.
So, H1:μ<12H_1: \mu < 12.

STEP 5

We'll use a *t-test* since we don't know the *true population standard deviation*.
The formula for the *t-statistic* is: t=xˉμsnt = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} where xˉ\bar{x} is the **sample mean**, μ\mu is the **population mean** we're testing (from our null hypothesis), ss is the **sample standard deviation**, and nn is the **sample size**.

STEP 6

Let's plug in our values: xˉ=12.192\bar{x} = \textbf{12.192}, μ=12\mu = \textbf{12}, s=0.099s = \textbf{0.099}, and n=36n = \textbf{36}. t=12.192120.09936t = \frac{12.192 - 12}{\frac{0.099}{\sqrt{36}}}

STEP 7

First, let's simplify the numerator: 12.19212=0.19212.192 - 12 = \textbf{0.192}.
Then, the denominator: 0.09936=0.0996=0.0165\frac{0.099}{\sqrt{36}} = \frac{0.099}{6} = \textbf{0.0165}.

STEP 8

Now, divide the numerator by the denominator: t=0.1920.016511.636t = \frac{0.192}{0.0165} \approx \textbf{11.636} Our **t-statistic** is approximately **11.636**.

STEP 9

The *p-value* tells us the probability of getting our results (or more extreme results) if the null hypothesis is true.
Since our alternative hypothesis is μ<12\mu < 12 (a *one-tailed test*), we're looking for the probability of getting a *t-statistic* as large as **11.636** or *larger*.

STEP 10

With a **t-statistic** of **11.636** and degrees of freedom df=n1=361=35df = n - 1 = 36 - 1 = 35, using a *t-table* or calculator, we find a *p-value* extremely close to **0**.

STEP 11

Our **significance level** is α=0.01\alpha = \textbf{0.01}.
If our *p-value* is *less than* α\alpha, we **reject** the null hypothesis.

STEP 12

Since our *p-value* (approximately **0**) is *much less than* **0.01**, we **reject** the null hypothesis.

STEP 13

We reject the null hypothesis that the average volume is 12 ounces.
There's *strong evidence* to suggest that the average volume is *less than* 12 ounces.
It *does* appear that consumers are being cheated!

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