Question

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?

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Esmeralda

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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.

Given:

Population mean:

Sample mean:

Sample variance:

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

where

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

Substituting the given values in the formula, we have:

where

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

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:

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

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