Question

Detecting gypsy moths The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. Each month, an SRS of 50 traps is inspected, the number of moths in each trap is recorded, and the mean number of moths is calculated. Based on years of data, the distribution of moth counts is discrete and strongly skewed with a mean of 0.5 and a standard deviation of 0.7.

a. Explain why it is reasonable to use a Normal distribution to approximate the sampling distribution of ${\mathrm{x}}^{-}\overline{x}$for SRSs of size 50 .

b. Estimate the probability that the mean number of moths in a sample of size 50 is greater than or equal to 0.6.

c. In a recent month, the mean number of moths in an SRS of size 50 was $x^{-}=0.6$. $\overline{x}=0.6$. Based on this result, is there convincing evidence that the moth population is getting larger in this state? Explain your reasoning.

Short answer

(a) the sample size is 50 , which is more than 30 and therefore the sampling distribution of the sample means $\overline{x}$is about normal, by the central limit theorem.

(b) The associative probability using table =0.1562

(c) it is plausible that the sample mean number of months is this high by chance alone.

Step-by-step solution

Part (a) Step 1: Given Information

Given,

$$\begin{gathered} \mu =0.5 \\ \sigma =0.7 \\ n=50 \end{gathered}$$

Part (a) Step 2: Simplification

The central limit theorems say that the sampling distribution of the sample means is about normal, when the sample is sufficiently large. When a sample size of 10 or more is used, it is considered to be appropriately large.

By the central limit theorem, the sampling distribution of the sample means $\overline{x}$ is nearly normal in this example because the sample size is 50, which is greater than 30.

Part (b) Step 1: Given Information

Given,

$$\begin{gathered} \mu =0.5 \\ \sigma =0.7 \\ n=50 \end{gathered}$$

Formula used:

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ z=\frac{x-\mu }{\sigma } \end{gathered}$$

Part (b) Step 2: Simplification

The mean of the sampling distribution of the sample mean $\overline{x}$is equal to the population mean:

${\mu }_{\overline{x}}=\mu =0.5$

The sampling distribution of the sample mean's standard deviation $\overline{x}$is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.7}{\sqrt{50}} \\ =0.0990 \end{gathered}$$

The sample size is 30 or more. We know that the sampling distribution of the sample mean$\overline{x}$is approximately normal because of the central limit theorem.

The z-score is a measure of how well something works.

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.6-0.5}{0.0990} \\ =1.01 \end{gathered}$$

The associative probability using table

$$\begin{gathered} P(\overline{x}\ge 0.6)=P(z>1.01) \\ =P(Z<-1.01) \\ =0.1562 \end{gathered}$$

Part (c) Step 1: Given information

Given,

$$\begin{gathered} \mu =0.5 \\ \sigma =0.7 \\ n=50 \end{gathered}$$

Part (c) Step 2: Explanation

No

Since this probability is not insignificant, it is possible that the sample mean number of months is thus large just by coincidence.