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A researcher plans to conduct an experiment evaluating the effect of a treatment. A sample of n = 9 participants is selected and each person receives the treatment before being tested on a standardized dexterity task. The treatment is expected to lower scores on the test by an average of 30 points. For the regular population, scores on the dexterity task form a normal distribution with μ = 240 and σ = 30. If the researcher uses a two-tailed test with α = .05, what is the power of the hypothesis test? Again assuming a two-tailed test with α = .05, what is the power of the hypothesis test if the sample size is increased to n = 25?

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Final answer:

To calculate the power of a hypothesis test, we need to find the probability of correctly rejecting the null hypothesis when it is false. In this scenario, with a sample size of 9 and a two-tailed test with α = .05, the power of the test is 0.26%. If the sample size is increased to 25, the power of the test reduces to 0.001%.

Step-by-step explanation:

In this scenario, we are conducting a hypothesis test to evaluate the effect of a treatment on a standardized dexterity task. We are given that the population mean for scores on the task is 240, with a standard deviation of 30. The treatment is expected to lower scores by an average of 30 points.

To find the power of the hypothesis test, we need to calculate the probability of correctly rejecting the null hypothesis when it is false. This can be done using the formula for power, which is given by Power = 1 - .

For a two-tailed test with α = .05 and a sample size of n = 9, we need to find the critical values using a z-table. The critical values will be ± 1.96. The standard error of the mean can be calculated by dividing the standard deviation by the square root of the sample size: σ/√n = 30/√9 = 10.

Using the critical values and the standard error of the mean, we can find the rejection region. The rejection region consists of values that are more extreme than the critical values. In this case, we need to find the probability of observing a mean score that is less than 210 or greater than 270. We can calculate the z-scores for these values using the formula: z = (x - μ) / (σ/√n).

For x = 210, the z-score is (210 - 240) / 10 = -3. For x = 270, the z-score is (270 - 240) / 10 = 3. We can now find the corresponding areas under the standard normal distribution curve using a z-table.

The probability of observing a mean score less than 210 is the area to the left of -3, which is 0.0013. The probability of observing a mean score greater than 270 is the area to the right of 3, which is also 0.0013. The total power of the test is the sum of these probabilities: Power = 0.0013 + 0.0013 = 0.0026, or 0.26%.

If the sample size is increased to n = 25, we can calculate the new standard error of the mean: σ/√n = 30/√25 = 6. The critical values and rejection region will remain the same, but the z-scores will be different. For x = 210, the new z-score is (210 - 240) / 6 = -5. For x = 270, the new z-score is (270 - 240) / 6 = 5. Using a z-table, we find that the probability of observing a mean score less than 210 is 0.000005 and the probability of observing a mean score greater than 270 is also 0.000005. The total power of the test is therefore 0.000005 + 0.000005 = 0.00001, or 0.001%.