A sample is chosen from a population with y = 46, and a treatment is then administered to the sample. After treatment, the sample mean is M = 47 with a sample variance of s2 = 16. Based on this information, what is the value of Cohen's d?

To calculate Cohen's d, we need the population mean, the sample mean, and the pooled standard deviation.

Given:
Population mean: $\mu = 46$
Sample mean: $\bar{X} = 47$
Sample variance: $s^2 = 16$

To calculate the pooled standard deviation, we can use the formula:

$$s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}}$$

where $n_1$ and $n_2$ are the sample sizes, and $s_1$ and $s_2$ are the sample standard deviations.

Since we only have information for one sample, we can use the sample variance as an estimate for the population variance:

$$s^2 \approx \sigma^2$$

Substituting the given values in the formula, we have:

$$s_p = \sqrt{\frac{(n-1)s^2}{n-1}} = s$$

where $n$ is the sample size.

Therefore, the pooled standard deviation is equal to the sample standard deviation:

$$s_p = s = \sqrt{16} = 4$$

Now we can calculate Cohen's d, which is the difference between the sample mean and the population mean, divided by the pooled standard deviation:

$$d = \frac{\bar{X} - \mu}{s_p} = \frac{47 - 46}{4} = \frac{1}{4} = 0.25$$

Answer: The value of Cohen's d is 0.25.