To construct a 99% confidence interval for the difference between the population mean test scores of the new and old curriculums (i.e., \( \mu_1 - \mu_2 \)), we can follow the steps below. Here, we will define the groups as follows:
- Group 1: Experimental group (new curriculum)
- Group 2: Control group (old curriculum)
Step 1: Gather sample statistics
From the problem, we have:
-
Experimental group (Group 1)
- Sample size (\( n_1 \)) = 27
- Sample mean (\( \bar{x_1} \)) = 1.19
- Sample variance (\( s_1^2 \)) = 74
-
Control group (Group 2)
- Sample size (\( n_2 \)) = 25
- Sample mean (\( \bar{x_2} \)) = 3.54
- Sample variance (\( s_2^2 \)) = 80
Step 2: Calculate the pooled standard deviation
Since we assume that the population variances are equal, we can calculate the pooled variance (\( s_p^2 \)):
\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]
Calculating the pooled variance: \[ s_p^2 = \frac{(27 - 1) \times 74 + (25 - 1) \times 80}{27 + 25 - 2} \]
\[ s_p^2 = \frac{26 \times 74 + 24 \times 80}{50} \]
\[ = \frac{1924 + 1920}{50} = \frac{3844}{50} = 76.88 \]
Thus, the pooled standard deviation (\( s_p \)) is: \[ s_p = \sqrt{76.88} \approx 8.76 \]
Step 3: Calculate the standard error of the difference in means
The standard error (SE) for the difference in means is given by:
\[ SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \]
Calculating the SE: \[ SE = 8.76 \sqrt{\frac{1}{27} + \frac{1}{25}} = 8.76 \sqrt{0.037037 + 0.04} = 8.76 \sqrt{0.077037} \approx 8.76 \times 0.277 \]
\[ SE \approx 2.43 \]
Step 4: Find the critical value
For a 99% confidence interval with degrees of freedom (\( df = n_1 + n_2 - 2 = 27 + 25 - 2 = 50 \)), we look up the t-value in the t-table. For \( df = 50 \) and a two-tailed test at the 99% confidence level, the critical t-value (approximately) is: \[ t_{0.005, 50} \approx 2.676 \]
Step 5: Calculate the confidence interval
The confidence interval for the difference in means can be calculated using:
\[ (\bar{x_1} - \bar{x_2}) \pm t_{\alpha/2} \cdot SE \]
First, calculate the difference in sample means: \[ \bar{x_1} - \bar{x_2} = 1.19 - 3.54 = -2.35 \]
Applying the values we have:
\[ -2.35 \pm 2.676 \cdot 2.43 \]
Calculating the margin of error: \[ 2.676 \cdot 2.43 \approx 6.52 \]
Now we can construct the confidence interval: \[ (-2.35 - 6.52, -2.35 + 6.52) = (-8.87, 4.17) \]
Final Result
The 99% confidence interval for the difference between the population mean test scores of the new and old curriculums is:
\[ \boxed{(-8.87, 4.17)} \]