95% CI for Mean Difference: Multimedia vs Traditional Math

Based on your provided image, the handwritten calculations indicate the following sample statistics for two groups of $n_1 = n_2 = 18$ students: Group 1 has a mean $\bar{x}_1 = 6.62$ and standard deviation $s_1 = 1.15$; Group 2 has a mean $\bar{x}_2 = 5.75$ and standard deviation $s_2 = 0.86$.

Answer

The 95% confidence interval for the true difference between the population means ($\mu_1 - \mu_2$) is approximately $[0.203, 1.537]$. This interval suggests that the multimedia instruction group performed significantly better than the traditional group, as the interval does not contain zero.

Explanation

1. Identify Known Quantities From the image and problem description, we extract the following data: Group 1 (Multimedia): $n_1 = 18, \bar{x}_1 = 6.62, s_1 = 1.15$ Group 2 (Traditional): $n_2 = 18, \bar{x}_2 = 5.75, s_2 = 0.86$ Confidence Level: $95\%$ ($\alpha = 0.05$)

2. Determine the Pooled Variance Since we assume independent normal populations and the sample sizes are small ($n < 30$), we use a t-distribution. Because it is not stated that variances are equal, we could calculate unpooled, but in most high-school/introductory contexts, we assume equal variances if not specified otherwise. We calculate the pooled standard deviation ($s_p$): $s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}}$ The pooled variance is a weighted average of the individual sample variances used when population variances are assumed equal.

   Substituting the values: $s_p = \sqrt{\frac{(17 \times 1.3225) + (17 \times 0.7396)}{34}} = \sqrt{\frac{22.48 + 12.57}{34}} \approx 1.015$ The pooled standard deviation combines the variability of both groups into a single estimate.

3. Find the Critical T-Value ⚠️ This step is required on exams: Use the degrees of freedom $df = n_1 + n_2 - 2$. $df = 18 + 18 - 2 = 34$ For a 95% confidence interval ($\alpha/2 = 0.025$), we look up $t_{0.025, 34}$ in a t-table: $t^* \approx 2.032$ T-values represent how many standard errors we must move away from the mean to capture the specified percentage of the distribution.

4. Calculate the Standard Error and Margin of Error The standard error of the difference ($SE$) is calculated as: $SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$ $SE = 1.015 \sqrt{\frac{1}{18} + \frac{1}{18}} = 1.015 \times 0.333 \approx 0.338$ The standard error measures the expected fluctuation in the difference between sample means.

   The Margin of Error ($ME$) is: $ME = t^* \times SE = 2.032 \times 0.338 \approx 0.687$ The margin of error defines the radius of the interval around the point estimate.

5. Construct the Confidence Interval The formula for the confidence interval is: $(\bar{x}_1 - \bar{x}_2) \pm ME$ $6.62 - 5.75 = 0.87$ $(0.87 - 0.687, 0.87 + 0.687)$ Calculating the bounds gives $[0.183, 1.557]$. Note: Using the unpooled Welch-Satterthwaite method yields a similar $t^* \approx 2.04$ and $SE \approx 0.339$, resulting in $[0.178, 1.562]$. We will provide the standard pooled result common in textbooks.

Final Answer

The 95% confidence interval for the difference of the population means $\mu_1 - \mu_2$ is: $\boxed{[0.18, 1.56]}$

Common Mistakes

• Incorrect Degrees of Freedom: Students often use $n-1$ (17) instead of $n_1 + n_2 - 2$ (34) when dealing with two independent samples.
• Using Z instead of T: Because the population standard deviations ($\sigma$) are unknown and sample sizes are small ($n < 30$), the t-distribution is mandatory. Using $z^* = 1.96$ will result in an interval that is too narrow.