Let's solve the problem **step by step** as asked. --- ## Problem Restatement Given: - Two groups: those who attend classes (Attending) and those who do not (Not attending) - Summary statistics: | | Attend | Not Attend | |-----------------|--------|------------| | Average | 86.25 | 80.68 | | Std Deviation | 14.04 | 22.44 | | Number of students | 98 | 42 | We need to: 1. **(a)** Test if there is a significant difference in average marks at 1% significance level. 2. **(b)** Calculate a 99% confidence interval for the difference. 3. **(c)** Interpret the result. --- # (a) Hypothesis Test (Two-Sample t-Test) ### **Step 1: State Hypotheses** - Null hypothesis \( H_0 \): \(\mu_1 = \mu_2\) (no difference in means) - Alternative hypothesis \( H_a \): \(\mu_1 \neq \mu_2\) (means are different) Where: - \(\mu_1\): Mean for "Attending" - \(\mu_2\): Mean for "Not attending" --- ### **Step 2: Test Statistic** Use the formula for the t-statistic for two independent samples (unequal variances): \[ t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Where: - \(\bar{X}_1 = 86.25\), \(s_1 = 14.04\), \(n_1 = 98\) - \(\bar{X}_2 = 80.68\), \(s_2 = 22.44\), \(n_2 = 42\) --- ### **Step 3: Plug in Values** Calculate standard error (SE): \[ SE = \sqrt{\frac{14.04^2}{98} + \frac{22.44^2}{42}} \] \[ = \sqrt{\frac{197.1216}{98} + \frac{503.5536}{42}} \] \[ = \sqrt{2.012 + 11.989} \] \[ = \sqrt{14.001} \] \[ = 3.742 \] Now, the test statistic: \[ t = \frac{86.25 - 80.68}{3.742} \] \[ t = \frac{5.57}{3.742} \] \[ t = 1.49 \] Wait, let's verify SE again as the numbers seem off: #### Recalculate: First term: \( \frac{14.04^2}{98} = \frac{197.1216}{98} \approx 2.012 \) Second term: \( \frac{22.44^2}{42} = \frac{503.5536}{42} \approx 11.989 \) Sum: \( 2.012 + 11.989 = 14.001 \) \( \sqrt{14.001} = 3.742 \) So, values are correct. --- ### **Step 4: Degrees of Freedom (Welch's approximation)** \[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{ \left(\frac{s_1^2}{n_1}\right)^2 }{ n_1-1 } + \frac{ \left(\frac{s_2^2}{n_2}\right)^2 }{ n_2-1 } } \] \[ = \frac{ (2.012 + 11.989)^2 }{ \frac{(2.012)^2}{97} + \frac{(11.989)^2}{41} } \] \[ = \frac{14.001^2}{ \frac{4.048}{97} + \frac{143.736}{41} } \] \[ = \frac{196.028}{0.04175 + 3.506} \] \[ = \frac{196.028}{3.5478} \] \[ = 55.27 \approx 55 \] --- ### **Step 5: Critical Value** Significance level = 1% (two-tailed), \( \alpha = 0.01 \) For \( df \approx 55 \), \( t_{0.005, 55} \approx 2.67 \) --- ### **Step 6: Decision** Calculated \( t = 1.49 \) < \( 2.67 \) - **Fail to reject \( H_0 \)**. - There is **not enough evidence at the 1% level** to say the means are different. --- ## (b) 99% Confidence Interval for Difference in Means \[ (\bar{X}_1 - \bar{X}_2) \pm t^* \times SE \] Where \( t^* \) is the 99% critical value (\( \approx 2.67 \)). \[ 5.57 \pm 2.67 \times 3.742 \] \[ 5.57 \pm 9.99 \] So the confidence interval is: \[ (-4.42, 15.56) \] --- ## (c) Interpretation - Since the 99% confidence interval **includes 0**, we **cannot conclude** that there is a significant difference in mean marks between students who attend and those who do not. - **Logic**: If 0 is in the interval, the difference could plausibly be zero, so no evidence for a difference at this confidence level. --- # **Summary Table** | Step | Calculation | Result | |------|-------------|--------| | a | t-statistic | 1.49 (fail to reject \( H_0 \)) | | b | 99% CI | (-4.42, 15.56) | | c | Conclusion | No significant difference at 1% level | --- ### **Final Answers** **(a)** At the 1% significance level, there is **not enough evidence** to conclude that attending class causes a difference in average marks. **(b)** The 99% confidence interval for the difference in means is **(-4.42, 15.56)**. **(c)** Since the interval includes 0, there is **no statistically significant difference** in average marks between the two groups at the 1% level. --- Let me know if you need more explanation or step-by-step calculations!

# Detailed Explanation of the Solution ## Problem Restatement We are tasked with determining whether there is a significant difference in average marks between two groups of students: those who attend classes and those who do not. We have the necessary statistical data to perform a hypothesis test and calculate a confidence interval. --- ## (a) Hypothesis Test (Two-Sample t-Test) ### Step 1: State Hypotheses - **Null Hypothesis \( H_0 \)**: There is no difference in average marks between the two groups (\(\mu_1 = \mu_2\)). - **Alternative Hypothesis \( H_a \)**: There is a difference in average marks (\(\mu_1 \neq \mu_2\)). ### Step 2: Test Statistic To compare the means of two independent samples with potentially unequal variances, we use the formula for the t-statistic: \[ t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Where: - \(\bar{X}_1\) and \(\bar{X}_2\) are sample means. - \(s_1\) and \(s_2\) are sample standard deviations. - \(n_1\) and \(n_2\) are sample sizes. ### Step 3: Plug in Values We substitute the given values into the formula to calculate the t-statistic. #### Standard Error Calculation 1. Calculate the variance for each group: - \(s_1^2 = 14.04^2 = 197.1216\) - \(s_2^2 = 22.44^2 = 503.5536\) 2. Calculate the standard error (SE): \[ SE = \sqrt{\frac{197.1216}{98} + \frac{503.5536}{42}} \approx 3.742 \] 3. Calculate the t-statistic: \[ t = \frac{86.25 - 80.68}{3.742} \approx 1.49 \] ### Step 4: Degrees of Freedom We utilize Welch's approximation to determine the degrees of freedom (df): \[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{ \left(\frac{s_1^2}{n_1}\right)^2 }{ n_1-1 } + \frac{ \left(\frac{s_2^2}{n_2}\right)^2 }{ n_2-1 } } \] Calculating this gives us \( df \approx 55 \). ### Step 5: Critical Value Using a t-distribution table, we find the critical value for \( df \approx 55 \) at a 1% significance level (two-tailed). This value is approximately \( t_{0.005, 55} \approx 2.67 \). ### Step 6: Decision We compare the calculated t-statistic with the critical value: - Since \( t = 1.49 < 2.67 \), we **fail to reject \( H_0 \)**. --- ## (b) 99% Confidence Interval for Difference in Means To construct the 99% confidence interval for the difference in means: 1. Calculate the margin of error: \[ ME = t^* \times SE = 2.67 \times 3.742 \approx 9.99 \] 2. Establish the confidence interval: \[ (86.25 - 80.68) \pm 9.99 = 5.57 \pm 9.99 \] Thus, the confidence interval is: \[ (-4.42, 15.56) \] --- ## (c) Interpretation The confidence interval includes zero, indicating that it is plausible that there is no difference in average marks between the two groups. This aligns with our hypothesis test results, where we failed to reject the null hypothesis. Therefore, we conclude that there is no statistically significant difference in average marks at the 1% level. --- # Summary of Results | Step | Calculation | Result | |------|-------------|--------| | a | t-statistic | 1.49 (fail to reject \( H_0 \)) | | b | 99% CI | (-4.42, 15.56) | | c | Conclusion | No significant difference at 1% level | --- ### Final Answers 1. **(a)** At the 1% significance level, there is **not enough evidence** to conclude that attending class causes a difference in average marks. 2. **(b)** The 99% confidence interval for the difference in means is **(-4.42, 15.56)**. 3. **(c)** Since the interval includes 0, there is **no statistically significant difference** in average marks between the two groups at the 1% level.

# Detailed Calculation Steps for the Problem ## Problem Overview We are comparing the average marks of two groups: students who attend classes versus those who do not. The data provided includes means, standard deviations, and sample sizes for both groups. --- ## Given Data - **Group 1 (Attending)** - Average: \( \bar{X}_1 = 86.25 \) - Standard Deviation: \( s_1 = 14.04 \) - Sample Size: \( n_1 = 98 \) - **Group 2 (Not Attending)** - Average: \( \bar{X}_2 = 80.68 \) - Standard Deviation: \( s_2 = 22.44 \) - Sample Size: \( n_2 = 42 \) --- ## (a) Hypothesis Test Calculation ### Step 1: Hypotheses - Null Hypothesis \( H_0 \): \( \mu_1 = \mu_2 \) - Alternative Hypothesis \( H_a \): \( \mu_1 \neq \mu_2 \) ### Step 2: Test Statistic Formula \[ t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] ### Step 3: Calculate Standard Error (SE) 1. **Calculate variance for both groups:** - \( s_1^2 = 14.04^2 = 197.1216 \) - \( s_2^2 = 22.44^2 = 503.5536 \) 2. **Calculate SE:** \[ SE = \sqrt{\frac{197.1216}{98} + \frac{503.5536}{42}} \] - For Group 1: \[ \frac{197.1216}{98} \approx 2.012 \] - For Group 2: \[ \frac{503.5536}{42} \approx 11.989 \] - Combine: \[ SE = \sqrt{2.012 + 11.989} = \sqrt{14.001} \approx 3.742 \] ### Step 4: Calculate t-statistic \[ t = \frac{86.25 - 80.68}{3.742} = \frac{5.57}{3.742} \approx 1.49 \] ### Step 5: Degrees of Freedom Calculation Using Welch's approximation: \[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{ \left(\frac{s_1^2}{n_1}\right)^2 }{ n_1-1 } + \frac{ \left(\frac{s_2^2}{n_2}\right)^2 }{ n_2-1 } } \] 1. **Calculate components:** - \( \frac{s_1^2}{n_1} = \frac{197.1216}{98} \approx 2.012 \) - \( \frac{s_2^2}{n_2} = \frac{503.5536}{42} \approx 11.989 \) 2. **Calculate the df:** \[ df = \frac{(2.012 + 11.989)^2}{\frac{(2.012)^2}{97} + \frac{(11.989)^2}{41}} \] - Numerator: \[ (2.012 + 11.989)^2 \approx 196.028 \] - Denominator: \[ \frac{(2.012)^2}{97} = \frac{4.048144}{97} \approx 0.04175 \] \[ \frac{(11.989)^2}{41} = \frac{143.736321}{41} \approx 3.506 \] \[ 0.04175 + 3.506 \approx 3.5478 \] - Final df: \[ df \approx \frac{196.028}{3.5478} \approx 55.27 \approx 55 \] ### Step 6: Critical Value For a two-tailed test at \( \alpha = 0.01 \) and \( df \approx 55 \): - Critical value \( t_{0.005, 55} \approx 2.67 \) ### Step 7: Conclusion Since \( t = 1.49 < 2.67 \), we **fail to reject \( H_0 \)**. There is insufficient evidence to conclude a significant difference in average marks. --- ## (b) Confidence Interval Calculation ### Step 1: Margin of Error (ME) \[ ME = t^* \times SE \] Using \( t^* \approx 2.67 \): \[ ME = 2.67 \times 3.742 \approx 9.99 \] ### Step 2: Confidence Interval \[ (\bar{X}_1 - \bar{X}_2) \pm ME \] \[ 5.57 \pm 9.99 \] Thus: \[ (-4.42, 15.56) \] --- ## (c) Interpretation The confidence interval \((-4.42, 15.56)\) includes 0, indicating that it is plausible that there is no difference in average marks. This confirms the hypothesis test conclusion. --- # Summary of Results | Step | Calculation | Result | |------|-------------|--------| | a | t-statistic | 1.49 (fail to reject \( H_0 \)) | | b | 99% CI | (-4.42, 15.56) | | c | Conclusion | No significant difference at 1% level | --- ### Final Answers 1. **(a)** At the 1% significance level, there is **not enough evidence** to conclude a difference in average marks. 2. **(b)** The 99% confidence interval for the difference in means is **(-4.42, 15.56)**. 3. **(c)** Since the interval includes 0, there is **no statistically significant difference** in average marks between the two groups at the 1% level.