One method for estimating abundance of animals is known as line-intercept sampling. When applied to Alaskan wolverines, it is predicted that the proportion of attempts to locate wolverine tracks that should be successful is $p = 0.453$. Suppose that biologists will make 100 attempts to locate wolverine tracks in random locations in Alaska.
a) Show that this sample size is large enough for the sampling distribution of $\hat { p }$ to be approximately normal.
b) What is the mean of the sampling distribution of $\hat { p }$ if the predicted proportion for line-intercept sampling is correct?
c) What is the standard deviation of the sampling distribution of $\hat { p }$ if the predicted proportion for intercept sampling is correct? d) If the proportion predicted for line-intercept sampling is correct, what is the probability that a sample proportion, $\hat { p }$, would differ from $p = 0.453$ by less than $0.05$ ?

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The Answer of One method for estimating abundance of animals is known as line-intercept sampling. When applied to Alaskan wolverines, it is predicted that the proportion of attempts to locate wolverine tracks that should be successful is $p = 0.453$. Suppose that biologists will make 100 attempts to locate wolverine tracks in random locations in Alaska.
a) Show that this sample size is large enough for the sampling distribution of $\hat { p }$ to be approximately normal.
b) What is the mean of the sampling distribution of $\hat { p }$ if the predicted proportion for line-intercept sampling is correct?
c) What is the standard deviation of the sampling distribution of $\hat { p }$ if the predicted proportion for intercept sampling is correct? d) If the proportion predicted for line-intercept sampling is correct, what is the probability that a sample proportion, $\hat { p }$, would differ from $p = 0.453$ by less than $0.05$ ?

One Method for Estimating Abundance of Animals Is Known as Line-Intercept