To construct a 99% confidence interval for the difference between the population mean test scores of the experimental group (new curriculum) and the control group (old curriculum), we will use the formula for the confidence interval for the difference between two means when the population variances are assumed to be equal.
Step 1: Gather the necessary statistics
-
Experimental group (new curriculum):
- Sample size (n₁) = 27
- Sample mean (x̄₁) = 1.19
- Sample variance (s₁²) = 74
-
Control group (old curriculum):
- Sample size (n₂) = 25
- Sample mean (x̄₂) = 3.54
- Sample variance (s₂²) = 80
Step 2: Calculate the pooled variance
The pooled variance (s_p²) is calculated as follows:
\[ s_p² = \frac{(n₁ - 1)s₁² + (n₂ - 1)s₂²}{n₁ + n₂ - 2} \]
Substituting in the values:
\[ s_p² = \frac{(27 - 1) \times 74 + (25 - 1) \times 80}{27 + 25 - 2} = \frac{26 \times 74 + 24 \times 80}{50} \]
Calculating each term:
\[ = \frac{1924 + 1920}{50} = \frac{3844}{50} = 76.88 \]
Step 3: Calculate the standard error (SE) of the difference between means
The standard error (SE) of the difference between the two means is calculated using the pooled variance:
\[ SE = \sqrt{s_p² \left( \frac{1}{n₁} + \frac{1}{n₂} \right)} = \sqrt{76.88 \left( \frac{1}{27} + \frac{1}{25} \right)} \]
Calculating the fractions:
\[ \frac{1}{27} \approx 0.03704, \quad \frac{1}{25} = 0.04 \]
Now we sum them:
\[ \frac{1}{27} + \frac{1}{25} \approx 0.03704 + 0.04 = 0.07704 \]
Now calculate the SE:
\[ SE = \sqrt{76.88 \times 0.07704} \approx \sqrt{5.92} \approx 2.43 \]
Step 4: Find the critical value (t*) for a 99% confidence interval
For a 99% confidence interval and degrees of freedom (df):
\[ df = n₁ + n₂ - 2 = 27 + 25 - 2 = 50 \]
Using a t-table or calculator, the critical t-value for 99% CI with 50 degrees of freedom is approximately 2.676.
Step 5: Calculate the confidence interval
Now calculate the confidence interval for the difference \((μ_1 - μ_2)\):
\[ CI = (x̄₁ - x̄₂) \pm t^* \cdot SE \]
First, calculate \(x̄₁ - x̄₂\):
\[ x̄₁ - x̄₂ = 1.19 - 3.54 = -2.35 \]
Now construct the CI:
\[ CI = -2.35 \pm 2.676 \cdot 2.43 \]
Calculating the margin of error:
\[ ME = 2.676 \cdot 2.43 \approx 6.49 \]
Thus, the confidence interval is:
\[ CI = -2.35 \pm 6.49 \]
Calculating the upper and lower bounds:
Lower bound:
\[ -2.35 - 6.49 \approx -8.84 \]
Upper bound:
\[ -2.35 + 6.49 \approx 4.14 \]
Final Result
The 99% confidence interval for the difference between the population mean test scores of the new and old curricula is:
\[ (-8.84, 4.14) \]