SOLUTION: 2) The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test. Use a 0.05 significance level to test the cla

Question 1170375: 2) The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test. Use a 0.05 significance level to test the claim that there is a linear correlation between hours studied and test score.
Hours 5 10 4 6 10 9
Scores 64 86 69 86 59 87
1) Null and Alternative Hypothesis

2) Calculator Work

3) Test Statistic, P-Value and Correlation Coefficient r, r=0.2242

4) Conclusion about the null hypothesis
Method 1:
Method 2:

5) Final conclusion that addresses the original claim

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1) Null and Alternative Hypothesis:**
* **Null Hypothesis (H0):** There is no linear correlation between hours studied and test scores (ρ = 0).
* **Alternative Hypothesis (H1):** There is a linear correlation between hours studied and test scores (ρ ≠ 0).
**2) Calculator Work:**
You've mentioned that the correlation coefficient, r, is 0.2242. To perform the hypothesis test, we'll need to calculate the test statistic and p-value. Most calculators with statistical functions can do this. Here's a general outline of how to do this on a graphing calculator:
* Enter the hours studied data into List 1 (L1) and the test scores into List 2 (L2).
* Perform a linear regression t-test. The calculator will provide the t-statistic and the p-value.
* We are given r=0.2242.
* The number of pairs n=6.
**3) Test Statistic, P-Value, and Correlation Coefficient r:**
* **Correlation Coefficient (r):** 0.2242 (given)
* **Test Statistic (t):**
* $t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}}$
* $t = \frac{0.2242 \sqrt{6-2}}{\sqrt{1-0.2242^2}}$
* $t = \frac{0.2242 \sqrt{4}}{\sqrt{1-0.05026564}}$
* $t = \frac{0.2242 * 2}{\sqrt{0.94973436}}$
* $t = \frac{0.4484}{0.97454316}$
* t = 0.4601
* **P-Value:**
* Degrees of freedom (df) = n - 2 = 6 - 2 = 4
* Using a t-distribution table or calculator with df = 4 and t = 0.4601, we find the two-tailed p-value.
* The p-value is approximately 0.667.
**4) Conclusion about the Null Hypothesis:**
* **Method 1: Comparing P-value to Significance Level:**
* The p-value (0.667) is greater than the significance level (0.05).
* Therefore, we fail to reject the null hypothesis.
* **Method 2: Comparing Test Statistic to Critical Value:**
* For a two-tailed test with df = 4 and α = 0.05, the critical t-values are ±2.776.
* The test statistic (0.4601) is within the range of -2.776 and 2.776.
* Therefore, we fail to reject the null hypothesis.
**5) Final Conclusion that Addresses the Original Claim:**
* At the 0.05 significance level, there is not sufficient evidence to support the claim that there is a linear correlation between hours studied and test scores.
* In other words, based off of the small sample size, we can not conclude that the hours studied has a linear correlation with the test scores.

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