population mean=42.6

standard deviation=12

n=16
M=34.4. Is this mean a reasonable outcome for a sample of n=16, or is this sample mean very different from what would usually be expected?

I obtained a z -score of -2.73. Thus, that would make the sample mean a reasonless outcome. Am I correct?

For a sample of 16, the snadard deviation of the mean should be reduced by a factor of sqrt 16 = 4 to be 12/4 = 3. If the measured mean is 34.4, which is 8.2 below the expected mean, the z-score is -8.2/3 = -2.73.

I don't know what you mean by a "reasonless" outcome. It is not unreasonable, but it is quite unlikely (0.31% probability that the measured mean will be this low or lower).

I meant REASONABLE, sorry. Thanks!

To determine whether the sample mean of 34.4 is a reasonable outcome for a sample of n=16, we can calculate the z-score and compare it to the standard normal distribution.

The formula to calculate the z-score is:

z = (x - μ) / (σ / √n)

Where:
x is the sample mean
μ is the population mean
σ is the standard deviation
n is the sample size

Let's calculate the z-score:

z = (34.4 - 42.6) / (12 / √16)
z = (34.4 - 42.6) / (12 / 4)
z = -8.2 / 3
z = -2.73

Your calculation of the z-score is correct, which means the sample mean of 34.4 is approximately 2.73 standard deviations below the population mean.

To determine if this sample mean is reasonable, we need to consider how likely it is to obtain a sample mean that far from the population mean. In general, if the z-score is very small (close to 0) or very large (far from 0), it suggests that the sample mean is not representative of the population mean.

In this case, a z-score of -2.73 indicates that the sample mean is significantly lower than the population mean. This means that the sample mean of 34.4 is very different from what would usually be expected, and it is an unusually low outcome.

In conclusion, based on the z-score of -2.73, the sample mean of 34.4 is not a reasonable outcome for a sample of n=16. It is significantly different from what would usually be expected.

To determine whether a sample mean of 34.4 is a reasonable outcome or not, we can use the concept of z-scores.

A z-score measures how many standard deviations a value is from the mean. It tells us whether the value is typical or an outlier relative to the rest of the distribution.

To calculate the z-score, we can use the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.

In this case, the population mean (μ) is 42.6, the standard deviation (σ) is 12, the sample mean (x) is 34.4, and the sample size (n) is 16.

Substituting these values into the formula, we can calculate the z-score:

z = (34.4 - 42.6) / (12 / √16)
= -8.2 / (12 / 4)
= -8.2 / 3
= -2.73

Based on the z-score of -2.73, we can conclude that the sample mean of 34.4 is very different from what would usually be expected. A z-score of -2.73 indicates that the sample mean is about 2.73 standard deviations below the population mean, which suggests that it is a fairly extreme outcome.