16. Gravetter/Wallnau/Forzano, Essentials - Chapter 8 -End-of-chapter question 2

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16. Gravetter/Wallnau/Forzano, Essentials - Chapter 8 -End-of-chapter question 20 Aa Aa Research has shown that 1Q scores have been increasing for years (Flynn, 1984, 1999). The phenomenon is called the Flynn effect and the data indicate that the increase appears to average about 7 points per decade. To examine this effect, a researcher obtains an IQ test with instructions for scoring from 10 years ago and plans to administer the test to a sample of n 25 of today's high school students. Ten years ago, the scores on this IQ test produced a standardized distribution with a mean of = 100 and a standard deviation -15. If there actually has been a 7-point increase in the average IQ during the past 10 years, then find the power of the hypothesis test for each of the following. The researcher uses a two-tailed hypothesis test with = .05 to determine the data indicate a significant change in 1Q over the past 10 years. (Round your answer for z-score value to two decimal places.) The power for the test is the probability of obtaining a z-score than ,which is p The researcher uses a one-tailed hypothesis test with = .05 to determine the data indicate a significant increase in IQ over the past 10 years. (Round your answer for z-score value to two decimal places.) The power for the test is the probability of obtaining a z-score than , which is p = Standard Normal Distribution tools are available to help you find the probabilities. To enable you to select z-scores precisely to two decimal places, only part of the distribution (the left half or the right half) is shown at a time. To use a tool, click the appropriate button. Left Tail Right Tail

Explanation / Answer

Answer to the first question)

For two tailed test for alpha = 0.05

Z critical for alpha = 0.05 , is Z = 1.96

.

if there is 7 point increase

then the new mean = 100+7 = 107

Formula of Z score is as follows:

z = (x - M) / (s / sqrt(n))

where x = 107

M = 100

s = 15

n = 25

.
on plugging these values we get:

Z = (107 -100) / (15/sqrt(25))

Z = 7 /3

Z = 2.33

.

The P value for this Z value will be :

P value = 2 * (1 - P(z < 2.33))
We can get the value of P(z < 2.33) from the Z table

P(z < 2.33) =0.9902

Thus P value = 2* (1-0.9902)

P value = 0.0196

Thus Power = 1 - 0.0196 = 0.9804

.

Thus the Z value will be 2.33 than 1.96 , with Power 0.9804

.

Answer to the second question)

if it is one tailed test, Z = 1.645 for alpha = 0.05

Z for 107 will be same as calculated above 2.33

But P value fro one tailed test will be:

P value = 1 - P(z < 2.33)

P(z < 2.33) = 0.9902

P value = 0.0098

Hence Power = 1 - 0.0098

Poower = 0.9902

Hence the Z value will be 2.33 than 1.645 , with power = 0.9902

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