Question

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 study27 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.

Answer 1

Answer:

a) $\mu$

b) $\bar{x}$

c) (0.152, 0.214)

Step-by-step explanation:

We are given the following in the question:

Sample mean = 0.183 ppm

Standard error = 0.016

a) quantity being estimated

We have to estimate the population mean.

Notation for population mean:

$\mu$

b) Best estimate for population mean is the sample mean

Notation for sample mean:

$\bar{x}$

The point estimate for population mean is

$\mu = \bar{x} = 0.183$

c) 95% confidence interval

$\bar{x}\pm z_{critical}(\text{Standard Error})$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)$

Thus, the 95% confidence interval is:

(0.152, 0.214)

Interpretation:

After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.

Answer 2

Answer:

95 percent

Step-by-step explanation: