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Name: A researcher is evaluating the influence of a treatment using a sample
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Description: A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of μ = 50 and a standard deviation of σ = 10. The researcher expects a +5-point treatment effect and plans to use a two

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of μ = 50 and a standard deviation of σ = 10. The researcher expects a +5-point treatment effect and plans to use a two-tailed hypothesis test with α = .05.

Standard Normal Distribution

Mean = 0.0

Standard Deviation = 1.0

-2.0-1.00.01.02.0z.5000.50000.00

Compute the power of the test if the researcher uses a sample of n = 4 individuals.

The power for the test is the probability of obtaining a z-score than

, which corresponds to p =

. Thus, the power of the test is

Compute the power of the test if the researcher uses a sample of n = 25 individuals.

The power for the test is the probability of obtaining a z-score than

, which corresponds to p =

. Thus, the power of the test is

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

Given Information:

Population mean $μ = 50$

Population standard deviation $σ = 10$

The researcher expects a +5-point treatment effect.

Significance level $α = 0.05$ (two tailed test)

To compute the power of the test if the researcher uses a sample of n = 4 individuals:

Calculate the parameters of null hypothesis $H_{0}$ is true, sampling distribution:

Mean of sampling distribution: $μ_{M} = μ = 50$

Standard deviation of sampling distribution $σ_{M} = \frac{σ}{\sqrt{n}} = \frac{10}{\sqrt{4}} = 5$

Now, calculate the parameter of $H_{0}$ is false:

$μ = 50 + 5 = 55$

$σ_{M} = 5$

Calculate the critical z-score:

Using standard normal table, critical values for a two tailed test at 0.05 significance level are: -1.96 and +1.96

Raw score (M) = $μ + z σ = 50 + 1.96 (5) = 59.8$

Power = area to the right of the critical value z under $H_{0}$ is false distribution.

Location of critical z in $H_{0}$ is false distribution = $z_{m e a n o f t r e a t m e n t s} - z_{c r i t i c a l}$

$\begin{array}{rcl} z & = & \frac{M - μ}{σ_{M}} \\ = & \frac{59.8 - 55}{5} \\ = & 0.96 \end{array}$

Using standard normal table, value corresponding to 0.96 is 0.83147.

$\begin{array}{rcl} P (z > 0.96) & = & 1 - P (z \leq 0.96) \\ = & 1 - 0.83147 \\ = & 0.16853 \end{array}$

Thus, with a sample of 4 people and 5 points treatment effect, 16.853% of the time the hypothesis test will conclude that there is a significant effect.

Power of the hypothesis is 0.1685

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