Question 1177447: The Hinton Valley Bottling Company distributes cola in bottles labeled 32oz. An inspector randomly chooses 50 of these bottles, measures their contents, and obtains a sample mean of 31.4oz. and a sample standard deviation of 1.75oz. Using a 0.01 level of significance, test the inspector’s claim that the company is cheating the consumers. Find the p-value for this test.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! We're going to conduct a hypothesis test for a mean.
Let $\mu$ be the true mean volume of cola in the bottles.
1. **Hypotheses:**
* Null hypothesis: $\mu = 32$ (The company is not cheating.)
* Alternative hypothesis: $\mu < 32$ (The company is cheating.)
2. **Test statistic:**
We use the t-statistic, since the population standard deviation is unknown:
```
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
```
where $\bar{x} = 31.4$ is the sample mean, $\mu_0 = 32$ is the hypothesized mean, $s = 1.75$ is the sample standard deviation, and $n = 50$ is the sample size. This gives us
```
t = \frac{31.4 - 32}{1.75 / \sqrt{50}} \approx -2.42.
```
3. **P-value:**
The p-value is the probability of observing a sample mean as extreme as $\bar{x} = 31.4$, assuming the null hypothesis is true. Since this is a left-tailed test, the p-value is
```
P(T < -2.42) \approx 0.0096,
```
where $T$ has a t-distribution with $n - 1 = 49$ degrees of freedom.
4. **Conclusion:**
Since the p-value (0.0096) is less than $\alpha = 0.01$, we reject the null hypothesis. There is enough evidence to conclude that the company is cheating the consumers.
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