Chapter 14, Problem 12P

### Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570

Chapter
Section

### Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570
Textbook Problem

# As we have noted in previous chapters, even a very small effect can he significant if the sample is large enough. Suppose, for example, that a researcher obtains a correlation of r = 0.60 for a sample of n = 10 participants. a. Is this sample sufficient to conclude that a significant correlation exists in the population? Use a two-tailed test with α = .05. b. If the sample had n = 25 participants, is the correlation significant? Again, use a two-tailed test with α = .05.

a.

To determine
The sample is sufficient for concluding the significant of the correlation.

The sample is not sufficient to conclude the significance of the correlation.

Explanation

Given info:

The correlation coefficient r is 0.60 and sample size is 10.

Calculation:

The correlation coefficient is 0.60.

State the test hypotheses.

Null hypothesis:

H0:ρ=0

Alternative hypothesis:

Ha:ρ0

Test statistics is,

t=rρ(1r2)(n2)

Substitute 0.60 for r, 0 for ρ and 10 for n in the above formula to get the test statistics

t=0.600(10.36)102=0.600.648=0.600.2828=2.125

For 8 degree of freedom at 5% level of significance the critical value is 2.306.

The t value is belong to the critical region fail to reject the null hypothesis the sample correlation is not enough large to reject the null hypothesis.

Conclusion:

The sample is not sufficient to conclude that correlation is significant.

b.

To determine
The sample is sufficient for concluding the significant of the correlation.

The sample is sufficient to conclude the significance of the correlation.

Explanation

Given info:

The correlation coefficient r is 0.60 and sample size is 25.

Calculation:

The correlation coefficient is 0.60.

Set a null hypothesis

H0:ρ=0

Test statistics is,

t=rρ(1r2)(n2)

Substitute 0.60 for r, 0 for ρ and 10 for n in the above formula to get the test statistics

t=0.600(10.36)252=0.600.6423=0.600.1668=3.59

For 23 degree of freedom at 5% level of significance the critical value is 2.069.

The t value is not belong to the critical region so strong evidence to reject the null hypothesis the sample correlation is enough large to reject the null hypothesis.

Thus the correlation is statistically significant.

Conclusion:

The sample is sufficient to conclude that correlation is significant.

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