Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15. Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?
What if the size for each sample was increased to 20? Would a sample mean of 115 or more be considered unusual? Why or why not?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to each Z score.
From that, you should be able to decide whether or not it is unusual.
To determine whether a sample mean of 115 or more is unusual, we can use the concept of z-scores and the Central Limit Theorem.
1. For a sample of size 3:
To calculate the z-score, we use the formula:
z = (x - μ) / (σ / √n)
where x = sample mean (115), μ = population mean (100), σ = population standard deviation (15), and n = sample size (3).
Plugging in the values:
z = (115 - 100) / (15 / √3)
z = 15 / (15 / √3)
z = √3
Using a standard normal distribution table or a z-score calculator, we can find that the z-score of √3 is approximately 1.732.
A z-score of 1.732 corresponds to a cumulative probability of approximately 0.958. This means that if the sample size is 3, a sample mean of 115 or more is within the 95.8% percentile of the distribution. Therefore, it would not be considered unusual.
2. For a sample size of 20:
Using the same z-score formula:
z = (x - μ) / (σ / √n)
where n = sample size (20).
Plugging in the values:
z = (115 - 100) / (15 / √20)
z = 15 / (15 / √20)
z = √20
Using a standard normal distribution table or a z-score calculator, we can find that the z-score of √20 is approximately 2.828.
A z-score of 2.828 corresponds to a cumulative probability of approximately 0.997. This means that if the sample size is 20, a sample mean of 115 or more is within the 99.7% percentile of the distribution. Therefore, it would still not be considered unusual.
In summary, for both sample sizes of 3 and 20, a sample mean of 115 or more is not considered unusual since it falls within the higher percentiles of the distribution.
To determine if a sample mean of 115 or more is unusual, we need to calculate the z-score. The z-score measures the number of standard deviations a given value is from the mean.
For the first question, where the sample size is 3, we can calculate the z-score as follows:
z = (sample mean - population mean) / (standard deviation / sqrt(sample size))
z = (115 - 100) / (15 / sqrt(3))
z = 15 / (15 / sqrt(3))
z = sqrt(3)
By referring to a standard normal distribution table or using a statistical software, we can find that the z-score of sqrt(3) corresponds to a probability of approximately 0.91. This means that a sample mean of 115 or more would be unusual for a sample size of 3, as it falls in the upper tail of the distribution.
Now, let's consider the second question, where the sample size is 20. Calculating the z-score:
z = (115 - 100) / (15 / sqrt(20))
z = 15 / (15 / sqrt(20))
z = 15 / (3.3541)
z = 4.4721
Again, referring to a standard normal distribution table or using a statistical software, we can find that the z-score of 4.4721 corresponds to a probability of approximately 1.0. This means that a sample mean of 115 or more would be extremely unusual for a sample size of 20.
In both cases, a sample mean of 115 or more is considered unusual, but with a larger sample size of 20, it becomes even more unusual due to the decrease in sampling variability and increased accuracy in estimating the population mean.