A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of u= 40 and a standard deviation of σ= 12. The researcher expects a 6-point treatment effect and plans to use a two-tailed hypothesis test with α= .05.A. Compute the power of the test if the researcher uses a sample of n=9 individuals (see example 8.6)B. Compute the power of the test if the researcher uses a sample of n=16 individuals Hon Uf power TOr å specific research situation.We start with a normal shaped population with a mean of µ = 80 and a standard deviationof o = 10. A researcher plans to select a sample of n = 25 individuals from this populationand administer a treatment to each individual. It is expected that the treatment will have an8-point effect; that is, the treatment will add 8 points to each individual's score.The top half of Figure 8.10 shows the original population and two possible outcomes:EXAMPLE 8.61. If the null hypothesis is true and there is no treatment effect.2. If the researcher's expectation is correct and there is an 8-point effect.The left-hand side of the figure shows what should happen according to the null hypoth-esis. In this case, the treatment has no effect and the population mean is still pu = 80. On theright-hand side of the figure we show what would happen if the treatment has an 8-pointeffect. If the treatment adds 8 points to each person's score, the population mean after treat-ment will increase to u = 88.The bottom half of Figure 8.10 shows the distribution of sample means for n = 25 foreach of the two outcomes. According to the null hypothesis, the sample means are centeredat u =80 (bottom left). With an 8-point treatment effect, the sample means are centered atu = 88 (bottom right). Both distributions have a standard error of1010= 2σVnV25Notice that the distribution on the left shows all of the possible sample means if the nullhypothesis is true. This is the distribution we use to locate the critical region for the hypothesis hypothesis (the POWC tthe test) is nearly 100%for an 8-point treatmenteffect.+1.96-1.96e crictheseis close to 100% if there is an 8-point treatment effect.To calculate the exact value for the power of the test we must determine whythe distribution on the right-hand side is shaded. Thus, we must locate the exacfor the critical region, then find the probability value in the unit normal table. Fortebution on the left-hand side, the critical boundary of z = +1.96 corresponds toalthat is above the mean by 1.96 standard deviations. This distance is equal to1.960 M= 1.96(2)3.92 pointsThus, the critical boundary of z = +1.96 corresponds to a sample mean of M3.92 = 83.92. Any sample mean greater than M = 83.92 is in the critical region

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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 u= 40 and a standard deviation of σ= 12. The researcher expects a 6-point treatment effect and plans to use a two-tailed hypothesis test with α= .05. A. Compute the power of the test if the researcher uses a sample of n=9 individuals (see example 8.6) B. Compute the power of the test if the researcher uses a sample of n=16 individuals

Hon Uf power TOr å specific research situation.
We start with a normal shaped population with a mean of µ = 80 and a standard deviation
of o = 10. A researcher plans to select a sample of n = 25 individuals from this population
and administer a treatment to each individual. It is expected that the treatment will have an
8-point effect; that is, the treatment will add 8 points to each individual's score.
The top half of Figure 8.10 shows the original population and two possible outcomes:
EXAMPLE 8.6
1. If the null hypothesis is true and there is no treatment effect.
2. If the researcher's expectation is correct and there is an 8-point effect.
The left-hand side of the figure shows what should happen according to the null hypoth-
esis. In this case, the treatment has no effect and the population mean is still pu = 80. On the
right-hand side of the figure we show what would happen if the treatment has an 8-point
effect. If the treatment adds 8 points to each person's score, the population mean after treat-
ment will increase to u = 88.
The bottom half of Figure 8.10 shows the distribution of sample means for n = 25 for
each of the two outcomes. According to the null hypothesis, the sample means are centered
at u =
80 (bottom left). With an 8-point treatment effect, the sample means are centered at
u = 88 (bottom right). Both distributions have a standard error of
10
10
= 2
σ
Vn
V25
Notice that the distribution on the left shows all of the possible sample means if the null
hypothesis is true. This is the distribution we use to locate the critical region for the hypothesis

hypothesis (the POWC t
the test) is nearly 100%
for an 8-point treatment
effect.
+1.96
-1.96
e cric
these
is close to 100% if there is an 8-point treatment effect.
To calculate the exact value for the power of the test we must determine why
the distribution on the right-hand side is shaded. Thus, we must locate the exac
for the critical region, then find the probability value in the unit normal table. Forte
bution on the left-hand side, the critical boundary of z = +1.96 corresponds toal
that is above the mean by 1.96 standard deviations. This distance is equal to
1.960 M
= 1.96(2)
3.92 points
Thus, the critical boundary of z = +1.96 corresponds to a sample mean of M
3.92 = 83.92. Any sample mean greater than M = 83.92 is in the critical region

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.

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$μ = 40 σ = 12 α = 0.05$

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