This is an odd question that only rarely appears on exams. You are asked to determine if the mean of a population falls within a specific interval:
19.16mg ± 1 = 18.16mg to 20.16mg
For this reason, we cannot just use the t-test of an unknown population mean. Instead, we must calculate the 99% confidence interval (which deals with intervals), and then see how that interval compares to the claimed interval of 18.16mg to 20.13mg.
If the entire 99% confidence interval falls within this claimed range, it supports the hypothesis that the mean THC level is within 1 mg of 19.16 mg at the 1% significance level.
Sample mean:
= 19.16 mg
Sample variance:
= 0.0330
Sample size:
= 19
99% Confidence Interval:
Critical t-value for 99% confidence level with 18 degrees of freedom: ![]()
Standard Deviation:
mg
![Rendered by QuickLaTeX.com \[ CI = \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} \]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-a7bdfd9548cc71b67ab30d6c45527eb1_l3.png)
![Rendered by QuickLaTeX.com \[ CI = 19.16 \pm 2.878 \times \frac{0.1817}{\sqrt{19}} \]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-0b9a6d75b96d05ccf9cfaa317af6c6f9_l3.png)
![]()
![]()
Confidence Interval:
mg
We estimate that the mean THC amount falls between 19.0399mg and 19.2801mg, and this type of estimator is correct 99% of the time
Since this interval lies entirely within the range of 18.16 mg to 20.16 mg, we can say there is sufficient evidence (at the 1% significance level) that GH’s product is within 1 mg of 19.16 mg.