8.2.3.1.3 - Example: Coffee

In the population of Americans who drink coffee, the average daily consumption is 3 cups per day. A university wants to know if their students tend to drink more coffee than the national average. They ask a random sample of 50 students how many cups of coffee they drink each day and found \(\overline{x}=3.8\) and \(s=1.5\). Do they have evidence that their students drink more than the national average?

1. Check assumption and write hypotheses

Amount of coffee consumed is a quantitative variable. We are given that random sampling methods were employed. Because \(n \ge 30\), we can approximate the sampling distribution using a t distribution. 

This is a right-tailed test because we want to know if the mean in the sample is greater than the national average.

\(H_{0}:\mu= 3\)
\(H_{a}:\mu>3\)

2. Calculate the test statistic
Test Statistic: One Group Mean

\(t=\frac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}\)

\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size

\(t=\frac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{3.8-3}{\frac{1.5}{\sqrt{50}}}=3.771\)

Our \(t\) test statistic is 3.771

3. Determine the p-value

\(df=n-1=50-1=49\)

Distribution Plot of Density vs X - T, DF=49

Using Minitab Express, we can find that \(P(t > 3.771) =0.0002191\)

p-value = 0.0002191

If \(\mu=3\), then the probability of taking a random sample of \(n=50\) and finding \(\overline{x} \geq 3\) is 0.0002191

4. Make a decision

\(p\leq.05\), therefore we reject the null hypothesis.

5. State a "real world" conclusion

There is evidence to state the mean number of coffees consumed in the population of all students at this university is greater than 3.