Darned old Foofy!
Foofy computes the x̅ (MEAN) from data that her professor says is a random sample from population Q. she correctly computes that this mean has a z-score of +41 on the sampling distribution for population Q. Foofy claims she has proven that this could not be a random sample from population Q. Do you agree or disagree?
Why?
Thanks
To determine whether Foofy's claim is valid or not, we need to understand the concept of z-scores and their relationship to random sampling.
A z-score represents the number of standard deviations an individual data point is away from the mean of a distribution. It helps us understand the relative position of a data point within a distribution. In the context of random sampling, a z-score can also indicate how likely it is to obtain a particular sample mean under the assumption of random sampling from a population.
Foofy claims that a mean with a z-score of +41 on the sampling distribution for population Q is evidence that the sample could not be random. However, this claim is incorrect. Here's why:
1. The z-score of +41 is extremely high. In a normal distribution, such an extreme z-score would correspond to an exceptionally rare event. Therefore, if the sample mean truly had a z-score of +41, it would suggest that the sample mean is highly unlikely to occur by chance alone.
2. Random sampling from a population typically results in sample means that follow a normal distribution.
3. The z-scores of sample means from a random sample follow a standard normal distribution with a mean of 0 and a standard deviation of 1 (also known as the standard normal distribution or z-distribution).
Given these points, it is highly improbable for a sample mean to have a z-score as extreme as +41 under the assumption of random sampling. Therefore, Foofy's claim is reasonable in challenging the validity of this particular sample being a random sample from population Q.
To further analyze this situation, it would be necessary to look into the data and the sampling process in more detail. It is possible that there might be errors in the data or issues with the sampling methodology that led to such an extreme result. Additionally, it is important to consider whether the assumption of random sampling was actually met.
In summary, the extremely high z-score of +41 suggests that the sample mean is highly unlikely to occur by chance alone, giving reasonable doubt to the claim that the sample is a random sample from population Q. However, further analysis would be needed to determine the exact nature of the issue.