Skip to content

One sample 2 tail t-test

The Scenario

A botanical garden maintains a rare species of orchid that requires 42% ambient humidity to trigger blooming.
- Too Little: Humidity below 42% causes dehydration and bud drop.
- Too Much: Humidity above 42% causes stagnant moisture and fungal rot.

The garden installs a new misting system. To test its calibration, botanists collect data from 12 sensor zones. The results show a sample mean (\(\bar{x}\)) of 55%. However, the greenhouse is filled with unpredictable micro-climates, some sensors are in direct sunlight, while others are near drafty vents. This environmental chaos results in a sample standard deviation (\(s\)) of 25%.

The Question

Using a 0.05 level of significance, is there sufficient evidence to conclude that the new misting system has significantly deviated from the target humidity of 42%?

1. State the Hypotheses

  • Null Hypothesis (\(H_0\)): \(\mu = 42\) (System is calibrated).
  • Alternative Hypothesis (\(H_a\)): \(\mu \neq 42\) (System is out of calibration).

2. Identify the Test Parameters

  • Sample Mean (\(\bar{x}\)) = \(55\)
  • Hypothesized Mean (\(\mu_0\)) = \(42\)
  • Sample Standard Deviation (\(s\)) = \(25\)
  • Sample Size (\(n\)) = \(12\)
  • Significance Level (\(\alpha\)) = \(0.05\)

3. Calculate Degrees of Freedom (\(df\))

\[df = n - 1 = 12 - 1 = 11\]

4. Determine the Critical t-value (table)

For a two-tailed test with \(df = 11\) and \(\alpha = 0.05\):

\[t_{crit} = \pm 2.201\]

5. Calculate the t-statistic

\[t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{55 - 42}{\frac{25}{\sqrt{12}}} \approx \frac{13}{\frac{25}{3.464}} \approx \frac{13}{7.217} \approx 1.801\]

6. Make the Statistical Decision

Compare the calculated \(t\) to the critical \(t\):

\[|1.801| < 2.201\]

The calculated t-statistic fails to reach the rejection region.

7. Final Conclusion

At the \(0.05\) level of significance, we Fail to Reject the Null Hypothesis. We can't say that the humidity level we found is different from the needed 42%. With only 12 sensors, the high variance (\(25\)) makes the data too uncertain. We cannot mathematically prove the system is broken, even though the average (55) looks quite high.

Is the sample mean different from a specific number \(\mu_{0}\)?

  • What if you tested this with 24 sensors and got the same mean and standard deviation, would that change the result?
  • What about at a 99% confidence level (\(\alpha= .01\))
  • What happens when the sample standard deviation or sample mean decrease?

Let's say you add fans to even out the humidity, and it brings the standard deviation of your measurements down to 10%. You also add 6 more sensors so you have 18 in total, and you want to have at least a confidence level of 99%. You want to know what values you should look for tomorrow when you get the sensor data. What values would your sample mean have to fall between for you to be able to say that the greenhouse has 42% ambient humidity?