Question

II. A study was done to test whether the amount of narcissism (Y) on number of friends (X) someone has. A sample of 10 participants was used. Both variables are numeric. The only other data that you have is that r = .4 and SSY = 60. Assume an alpha level of .05.

9. What statistical test should you use to evaluate the hypothesis above? one-way independent ANOVA simple linear regression Pearson Product Moment Correlation 10. What are the null and alternative hypotheses in words?

Null hypothesis:

Alternative hypothesis: Th

11. What is the critical value?

12. Knowing that r = .4 and SSY = 60, calculate the appropriate test statistic (r or F).

14. Can you reject the null? No

Problem II. A study was done to investigate if time management skills is a good predictor of college grade point average (GPA). A sample of 57 participants was used. Assume an alpha level of .05.

1.1 State the null and alternative hypotheses (0.5 point) Null hypothesis: There is no relationship between time management skills and GPA Alternative hypothesis: There is a relationship between time management skills and GPA

1.2 What is the correct statistical test to evaluate the hypothesis? (0.5 point) Simple linear regression

1.3 Calculate the appropriate test statistic given that SSy = 20.6 and r = .5. (4 points) 1.4 What is the standard error? (1 point) standard error = 1.5 Find critical F (0.5 points) 1.6 What is your conclusion? Specifically, state if there is any significance and what is your decision regarding the null hypothesis (0.5 points)

Answer 1

We are asked to test, whether the amount of narcissism (Y) depends on the number of friends (X) someone has. Here, we are interested to determine whether a causal relationship between X and Y.

Given: n = 10, r = 0.40, SSY = 60

Now, the Pearson Product Moment Correlation measures the strength and direction of the linear relationship between two continuous variable - X and Y. However, using 'r' we only arrive at the conclusion that a relationship exists, but says nothing as to whether - X depends on Y or whether Y depends on X.

But, Simple linear regression, on the other hand, is capable of establishing this causal relationship - where, Y is the dependent variable and X denotes the independent variable.

Hence, the correct statistical tool, here, would be a simple linear regression.

10. The null and alternative hypotheses in words:

H₀: The fitted regression model is the same as the intercept only model (i.e. there is no causal relationship between X and Y)

H_a: The fitted regression model is different from the intercept only model (i.e. there is a significant causal relationship between X and Y)

11. The test statistic F to test the above hypothesis is given by:

with critical region

where Between Mean square and Within / error mean square

And Total Sum of Squares SSY = SSB + SSW

11. The critical value of F for (1,8) degrees of freedom at % level is obtained from the F table as:

We get F_critical = 5.3177

12. Computing the test statistic:

Now, introducing the goodness of fit measure r² - which gives the amount of explained variation in y by x, obtained by squaring the correlation coefficient r, given by,

r² = SSB / SSY

i.e (0.40)² = SSB / 60

i.e SSB = 60 x 0.16

= 9.6

and SSW = SSY - SSB = 60 - 9.6 = 50.4

For n = 10 observations, the F statistic is obtained as:

F = 1.52

14. Since the test statistic F = 1.52 < 5.32 does not lie in the rejection region, we fail to reject H₀.

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Problem II. A study was done to investigate if time management skills is a good predictor of college grade point average (GPA). A sample of 57 participants was used. Assume an alpha level of .05.

1.1  Null hypothesis: There is no causal relationship between time management skills and GPA

Alternative hypothesis: There is a causal relationship between time management skills and GPA

1.2 The correct statistical test to evaluate the hypothesis:

Simple linear regression (as mentioned above)

1.3 The appropriate test statistic given that SSY = 20.6 and r = .5,

r² = SSB / SSY

i.e (0.5)² = SSB / 20.6

i.e SSB = 20.6 * 0.25

= 5.15

and SSW = SSY - SSB = 20.6 - 5.15 = 15.45

For n = 57 observations, the F statistic is obtained as:

F = 18.33

1.4 The standard error

SE = 0.530

1.5 Critical value for (1,55) degrees of freedom can be obtained using the excel function:

F_Critical = 4.016

1.6 Since the test statistic F = 18.33 > 4.016 lies in the rejection region, we may reject H₀ at 5% level of significance.We may conclude that there exists a significant causal relationship between time management skills and GPA