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11. Suppose a random sample of 25 students found a correlation of r = –.53 between...

11. Suppose a random sample of 25 students found a correlation of r = –.53 between number of absences and final grade. Is this enough evidence to conclude that there is a relationship between absences and final grades in general? Use α = .05, two -tailed. Use “Tr” technique as illustrtaed on the PowerPoint lecture. Use of any other formula will be considered incorrect. Include your write up as well. (2 pts) please show all work steps.

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Answer #1

We have to test that there is relationship between absences & final grade. We have to use t-test for significance of correlation coefficient.

Hypothesis -

Null hypothesis - H0 : There is no correlation between absences & final grade, i.e. \rho = 0.

Alternative hypothesis - H1 : There is correlation between absences & final grade, i.e. \rho \neq 0.

Test statistic -

t = \frac{r\sqrt{n-2}}{\sqrt{1-r^{2}}}

Test criterion -

Reject H0 if t \geq t\alpha/2,n-2 or -t \leq -t\alpha/2,n-2 .

Calculations -

We have, r = -0.53, n = 25

So,

t = \frac{r\sqrt{n-2}}{\sqrt{1-r^{2}}}

t = \frac{(-0.53)\sqrt{25-2}}{\sqrt{1-(-0.53)^{2}}}

t = \frac{(-0.53)\sqrt{23}}{\sqrt{1-0.2809}}

t = \frac{(-0.53)(4.7958)}{\sqrt{0.7191}}

t = \frac{-1.8227}{0.8779}

t = -2.0762

Critical value -

t\alpha/2,n-2 = t0.05/2,25-2 = t0.025,23 = 2.06866

Conclusion -

-t > -t\alpha/2,n-2, so we have to accept H0 at 5% level of significance.

Result -

There is sufficient evidence that there is no relationship between absences & final grade.

Numbers in each row of the table are values on a t-distribution with (df) degrees of freedom for selected right-tail (greater

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