The problem asks us to take the given data, where "Hours Studying" is one variable and "Overall Class Performance" is another, and test if there is a statistically significant relationship between the two variables. We will test this using linear regression.

1. Problem Description

The problem asks us to take the given data, where "Hours Studying" is one variable and "Overall Class Performance" is another, and test if there is a statistically significant relationship between the two variables. We will test this using linear regression.

2. Solution Steps

We'll perform a linear regression analysis to determine if there's a statistically significant relationship between the hours spent studying and the overall class performance. We need to calculate the correlation coefficient $r$, and perform a t-test to determine the statistical significance.

The data is:

Hours Studying (x): 0.62, 1.50, 0.34, 0.97, 3.54, 0.69, 1.53, 0.32, 1.94, 1.25, 1.42, 3.07, 3.99, 1.73, 1.90

Overall Class Performance (y): 1.02, 4.62, 1.60, 1.59, 4.67, 1.52, 2.28, 1.68, 2.50, 4.04, 2.63, 3.53, 3.90, 2.75, 2.95

First, we calculate the summary statistics:

$n = 15$

$\sum x_i = 25.71$

$\sum y_i = 41.28$

$\sum x_i^2 = 61.867$

$\sum y_i^2 = 133.2804$

$\sum x_i y_i = 88.3514$

Next, we find the means:

$\bar{x} = \frac{\sum x_i}{n} = \frac{25.71}{15} = 1.714$

$\bar{y} = \frac{\sum y_i}{n} = \frac{41.28}{15} = 2.752$

Now we can calculate the correlation coefficient $r$:

$r = \frac{n\sum x_i y_i - (\sum x_i)(\sum y_i)}{\sqrt{[n\sum x_i^2 - (\sum x_i)^2][n\sum y_i^2 - (\sum y_i)^2]}}$

$r = \frac{15(88.3514) - (25.71)(41.28)}{\sqrt{[15(61.867) - (25.71)^2][15(133.2804) - (41.28)^2]}}$

$r = \frac{1325.271 - 1061.4408}{\sqrt{[928.005 - 661.0041][1999.206 - 1703.9184]}}$

$r = \frac{263.8302}{\sqrt{[267.0009][295.2876]}}$

$r = \frac{263.8302}{\sqrt{78836.645}} = \frac{263.8302}{280.7786} \approx 0.94$

Now we perform a t-test to assess statistical significance:

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

$t = \frac{0.94\sqrt{15-2}}{\sqrt{1 - (0.94)^2}}$

$t = \frac{0.94\sqrt{13}}{\sqrt{1 - 0.8836}}$

$t = \frac{0.94(3.6056)}{\sqrt{0.1164}}$

$t = \frac{3.3893}{0.3412} \approx 9.93$

Degrees of freedom $df = n-2 = 15-2 = 13$

Looking up the t-value of 9.93 with 13 degrees of freedom, we get a very small p-value (p < 0.001). Typically, we reject the null hypothesis if the p-value is less than 0.

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5. Since our p-value is significantly less than 0.05, we reject the null hypothesis.

3. Final Answer

There is a statistically significant positive relationship between hours spent studying and overall class performance. The correlation coefficient is approximately 0.94, and the t-test results in a statistically significant result (p < 0.001).