In a learning experiment, untrained mice are placed in a maze and the time required for each mouse to exit the maze is recorded. The average time for untrained mice to exit the maze is µ = 50 seconds and the standard deviation of their times is ? = 16 seconds. 64 randomly selected untrained mice are placed in the maze and the time necessary to exit the maze is recorded for each.

a) Completely describe the sampling distribution of the sample mean.

b) What is the probability that the sample mean differs from the population mean by more than 3?

c) If one mouse is selected at random, what is the probability that the time necessary to exit the maze differs from the population mean by more than 3?

d) Answering (b) and (c) do you have to make any assumptions? Explain why or why not.

e) If your random sample actually produced a sample mean of 85 seconds, would you consider this unusual? What conclusion might you draw?

a) The sampling distribution of the sample mean is the probability distribution of all possible sample means that could be obtained from repeated sampling. In this experiment, 64 untrained mice are placed in the maze, and the time necessary for each mouse to exit the maze is recorded. The sample mean is calculated by taking the average of these recorded times. Since each mouse is randomly selected, the sample mean can vary from sample to sample. The sampling distribution of the sample mean is expected to be normally distributed.

b) To calculate the probability that the sample mean differs from the population mean by more than 3, we need to use the concept of the standard error of the mean (SE). The standard error of the mean represents the standard deviation of the sampling distribution of the sample mean. In this case, the standard error of the mean can be calculated by dividing the standard deviation of the population by the square root of the sample size. So, SE = σ / √(n), where σ is the standard deviation of the population and n is the sample size.

Once we calculate the standard error, we can use it to determine the probability that the sample mean differs from the population mean by more than 3. We can do this by finding the z-score of the difference of 3 using the formula: z = (x - µ) / SE, where x is the difference we are interested in. Then, we can look up the corresponding probability in the standard normal distribution table.

c) If one mouse is selected at random, the probability that the time necessary to exit the maze differs from the population mean by more than 3 can be calculated using the concept of the standard deviation and the z-score. We divide the difference by the standard deviation to get the z-score, and then we can use the standard normal distribution table to find the corresponding probability.

d) To answer parts (b) and (c) regarding probability calculations, we need to make assumptions about the population being normally distributed. In this case, we assume that the times required for mice to exit the maze are normally distributed. Additionally, we assume that the sampling process is random and independent, which means that each mouse's time is not influenced by any other mouse's time.

e) To determine if a sample mean of 85 seconds is unusual or not, we need to consider the standard error and the distribution of the sample means. We calculate the z-score using the formula mentioned in part (b): z = (x - µ) / SE, where x is the sample mean. By looking up the z-score in the standard normal distribution table, we can find the corresponding probability. If the probability is very low (such as less than 0.05), we can conclude that the sample mean of 85 seconds is unusual and significantly different from the population mean. Conversely, if the probability is high (such as greater than 0.05), we cannot consider the sample mean of 85 seconds as unusual, and it supports the idea that the sample mean is consistent with the population mean.

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