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Baucators conducted a study of how students' involvement into social networks = X measured in hours per week affects their test scores Y (in a 100-point scale). They assumed all conditions required for this model and used a sample of n = 20 students. As a result, they derived the fitted line equation in the form: Ý = 70 - 2.85 X with the standard error for the estimated slope equal to SE (6) = 1.5 %3D At the significance level a = 0.05, do they have enough evidence that social networking has NEGATIVE impact on students' test performance? 1. Specify the critical value (or values) required in this case. 2. Formulate Rejection Rule 3. Evaluate the Test Statistic 4. Finally, state your decision.

Baucators conducted a study of how students' involvement into social networks = X measured in hours per week affects their test scores Y (in a 100-point scale). They assumed all conditions required for this model and used a sample of n = 20 students. As a result, they derived the fitted line equation in the form: Ý = 70 - 2.85 X with the standard error for the estimated slope equal to SE (6) = 1.5 %3D At the significance level a = 0.05, do they have enough evidence that social networking has NEGATIVE impact on students' test performance? 1. Specify the critical value (or values) required in this case. 2. Formulate Rejection Rule 3. Evaluate the Test Statistic 4. Finally, state your decision.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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**Study on Social Networking and Test Performance** Educators conducted a study to examine the impact of students' involvement in social networks, denoted as \( X \) (measured in hours per week), on their test scores, denoted as \( Y \) (on a 100-point scale). A sample of \( n = 20 \) students was used, assuming all conditions for the model were met. The resulting fitted line equation is: \[ \hat{Y} = 70 - 2.85 \cdot X \] The standard error for the estimated slope is: \[ SE[b] = 1.5 \] The question posed is whether, at a significance level of \( \alpha = 0.05 \), there is sufficient evidence to claim that social networking negatively impacts students' test performance. To address this, consider the following steps: 1. **Specify the critical value (or values) required in this case.** 2. **Formulate the Rejection Rule.** 3. **Evaluate the Test Statistic.** 4. **Finally, state your decision.** **Solution** [The solution would likely include steps such as calculating the test statistic using the slope and standard error, comparing it to the critical t-value at \( \alpha = 0.05 \) with \( n-2 \) degrees of freedom, and then using this comparison to either reject or fail to reject the null hypothesis.] *Note: The solution section is left incomplete as it requires detailed calculation steps that typically involve statistical tables or software outputs.*
Transcribed Image Text:**Study on Social Networking and Test Performance** Educators conducted a study to examine the impact of students' involvement in social networks, denoted as \( X \) (measured in hours per week), on their test scores, denoted as \( Y \) (on a 100-point scale). A sample of \( n = 20 \) students was used, assuming all conditions for the model were met. The resulting fitted line equation is: \[ \hat{Y} = 70 - 2.85 \cdot X \] The standard error for the estimated slope is: \[ SE[b] = 1.5 \] The question posed is whether, at a significance level of \( \alpha = 0.05 \), there is sufficient evidence to claim that social networking negatively impacts students' test performance. To address this, consider the following steps: 1. **Specify the critical value (or values) required in this case.** 2. **Formulate the Rejection Rule.** 3. **Evaluate the Test Statistic.** 4. **Finally, state your decision.** **Solution** [The solution would likely include steps such as calculating the test statistic using the slope and standard error, comparing it to the critical t-value at \( \alpha = 0.05 \) with \( n-2 \) degrees of freedom, and then using this comparison to either reject or fail to reject the null hypothesis.] *Note: The solution section is left incomplete as it requires detailed calculation steps that typically involve statistical tables or software outputs.*
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