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A study wants to examine the relationship between student anxiety for an exam and the number of hours studied. The data is as follows: 

Student Anxiety Scores 5 10 5 11 12 4 3 2 6 1 

Study Hours 1 6 2 8 5 1 4 6 5 2 

1.Why is a correlation the most appropriate statistic?

2.What is the null and alternate hypothesis? 

3.What is the correlation between student anxiety scores and number of study hours? Select alpha and interpret your findings. Make sure to note whether it is significant or not and what the effect size is. 

4.How would you interpret this?

5.What is the probability of a type I error? What does this mean?

6.How would you use this same information but set it up in a way that allows you to conduct a t-test? An ANOVA? 

|(a) We have two sets of data (two variables) and both are measured on the ratio scale of measurement. We are interested to see if there is any correlation |

|between the anxiety scores and study hours. In particular, we want to see if more anxiety before an exam results in more hours of study. Therefore correlation|

|(coefficient) is the most appropriate statistic. |

| |

|(b) The hypotheses are |

|Ho: Anxiety scores and Study hours are not correlated |

|Ha: Anxiety scores and Study hours are correlated |

| |

|(c) |

| |

| |

|x |

|y |

|x^2 |

|y^2 |

|xy |

| |

| |

| |

|5 |

|1 |

|25 |

|1 |

|5 |

| |

| |

| |

|10 |

|6 |

|100 |

|36 |

|60 |

| |

| |

| |

|5 |

|2 |

|25 |

|4 |

|10 |

| |

| |

| |

|11 |

|8 |

|121 |

|64 |

|88 |

| |

| |

| |

|12 |

|5 |

|144 |

|25 |

|60 |

| |

| |

| |

|4 |

|1 |

|16 |

|1 |

|4 |

| |

| |

| |

|3 |

|4 |

|9 |

|16 |

|12 |

| |

| |

| |

|2 |

|6 |

|4 |

|36 |

|12 |

| |

| |

| |

|6 |

|5 |

|36 |

|25 |

|30 |

| |

| |

| |

|1 |

|2 |

|1 |

|4 |

|2 |

| |

|n = |

|10 |

| |

| |

| |

| |

| |

| |

|Sums = |

| |

|59 |

|40 |

|481 |

|212 |

|283 |

| |

| |

|SS(x) = Σ(x^2) - [(Σx)^2 /n] = 481 - (59^2 /10) = 132.9 |

|SS(y) = Σ(y^2) - [(Σy)^2 /n] = 212 - (40^2 /10) = 52.0 |

|SS(xy) = Σ(xy) - [(Σx)(Σy)/n] = 283 - (59 * 40/10) = 47.0 |

|Correlation coefficient, r = SSxy / √[SS(x) * SS(y)] = 47/√(132.9 * 52) = 0.5654 |

|Let us use a 5% level of significance (α = 0.05) |

|n = 10 |

|r = 0.5654 |

|From the table of critical values for Pearson correlation, we find critical r = 0.632 |

|Since 0.5654 < 0.632, the correlation is not significant. |

|One measure of effect size is the coefficient of determination r^2 = 0.5654^2 = 0.3197. This means only about 31.97% of the variation in Study hours is due to|

|the variation in Anxiety scores. |

| |

|(d) The above results mean that Anxiety scores and Study hours are not significantly correlated with only about 32% of the observed variation in the response |

|variable (Study hours) accounted for by the predictor variable (Anxiety score). |

| |

|(e) Probability of Type I error = α = 0.05. This means 5% of the time, our decision to reject Ho will be wrong. In other words, 5% of the time, we will find |

|correlation between Anxiety scores and Study hours, where in reality there is none. |

| |

|(f) |

|A t- test can be used to test if the two variables are correlated or not by using the t- statistic with n – 2 degrees of freedom. Here are the steps for that |

|… |

|Hypotheses: |

|Ho: There is no significant correlation, that is ρ = 0 |

|Ha: There is significant correlation, that is ρ ≠ 0 |

|Decision Rule: |

|t (Two-tailed), α = 0.05 |

|Degrees of freedom = 10 - 2 = 8 |

|Lower Critical t- score = -2.306 |

|Upper Critical t- score = 2.306 |

|Reject Ho if |t| > 2.306 |

|Test Statistic: |

|SE = √{(1 - r^2)/DOF} = √((1 - 0.5654^2)/8) = 0.2916 |

|t = r/SE = 0.5654/0.2916 = 1.939 |

|p- value = 0.0885 |

|Decision (in terms of the hypotheses): |

|Since 1.939 < 2.306 we fail to reject Ho |

|Conclusion (in terms of the problem): |

|The result is not significant. There is no sufficient evidence that the variables are correlated. |

| |

|Since we have only one predictor variable, an ANOVA will report an F value equal to t^2 but the same p- value. This is shown below … |

|[pic] |

|SST = SS(y) = Σ(y - y-bar)^2 = 52 |

|SSR = Σ[(y' - y-bar)^2] = 16.6215 OR SSR = [SS(xy)]^2 / SS(x) = 16.6215 |

|SSE = SST - SSR OR Σ[(y - y')^2] = 35.3785 |

|Df (Total) = n - 1 = 10 – 1 = 9 |

|Df (R) = 1 |

|Df (E) = Df (Total) - Df (R) = 8 |

|MSR = SSR/Df (R) = 16.6215 |

|MSE = SSE/Df (E) = 4.4223 |

|F = MSR/MSE = 3.76 |

|α = 0.05 |

|Critical F- score = 5.3177 |

|p- value = 0.0885 |

|SE for the slope of the x variable = √[MSE/SS(x)] = 0.1824 |

|t- score for the slope of the x variable = slope/SE = 1.939 |

|ANOVA table |

| |

| |

| |

| |

| |

| |

|Source |

|SS |

|df |

|MS |

|F |

|p-value |

| |

|Regression |

|16.6215 |

|1 |

|16.6215 |

|3.76 |

|.0885 |

| |

|Residual |

|35.3785 |

|8 |

|4.4223 |

| |

| |

| |

|Total |

|52.0000 |

|9 |

|  |

|  |

|  |

| |

| |

|[Note that F = 3.76 and t = 1.939, and that 3.76 = 1.939^2. The p- value reported is the same as for t- test, which is 0.0885. Effect size = SSR/SST = |

|16.6215/52 = 0.3197 (31.97% as before).] |

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