Question

Moose Drool Makes Grass More Appetizing Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study \(^{27}\) has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is \(0.183 .\) The standard error for this estimate is 0.016 . (a) Give notation for the quantity being estimated, and define any parameters used. (b) Give notation for the quantity that gives the best estimate, and give its value. (c) Give a \(95 \%\) confidence interval for the quantity being estimated. Interpret the interval in context.

Short answer

The parameter being estimated, indicated as \(\mu\), is the mean level of the toxin in the grass after being treated with moose drool. The best estimate of this parameter is given by the sample mean, denoted as \(\bar{x}\), which was found to be 0.183 ppm. The 95% confidence interval for the mean level of toxin was calculated to be between 0.152 and 0.214 ppm.

Step-by-step solution

Identify and Define Parameters

The parameter being estimated is the mean level of the toxin ergovaline in the grass after being treated with moose drool. In statistical notation, this parameter is usually denoted as \(\mu\).

Distinguish the Best Estimate

The quantity that provides the best estimate in this instance is the sample mean (denoted as \(\bar{x}\)). After the scientists' study, they were able to estimate the mean level of the toxin to be 0.183 ppm.

Calculate a 95% Confidence Interval

The confidence interval can be calculated using the following formula: \[\text{CI} = \(\bar{x}\) \pm Z*\frac{\(SE\)}{\sqrt{n}}\] Where CI designates the confidence interval, Z denotes the Z score (1.96 for a 95% confidence interval), SE represents standard error and n is the sample size. In this problem, we are only given \(\bar{x}\) (0.183) and SE (0.016), without the sample size; the formula simplifies to \[\text{CI} = 0.183 \pm 1.96*(0.016)\]

Compute & Interpret the Confidence Interval

After calculating the above, we get the Confidence Interval between 0.152 and 0.214. This means that we are 95% confident that the true mean level of toxin in the grass treated with moose drool is between 0.152 and 0.214 ppm. This interpretation of the interval reflects the uncertainty of our estimate due to using a sample instead of having data from all grazing situations.