Question

A number of students in a Grade 9 maths course at a particular high school were randomly divided into two groups. The experimental group received teaching according to the new curriculum while the control group did not. All the students were given a test of computational skill (out of a total of 5 points) after the course and the results were as follows: for the experimental group, the sample size was 27, the sample mean was 1.19 and the sample variance was 74. For the control group, the sample size was 25, the sample mean was 3.54 and the sample variance was 80. Assume that population variances are equal. Construct a 99% confidence interval for the difference between the population mean test scores of the new and old curriculums. (μ_1-μ_2)

Answer 1

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

1. Experimental group (new curriculum):

   • Sample size (n₁) = 27
   • Sample mean (x̄₁) = 1.19
   • Sample variance (s₁²) = 74

2. 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:

\[ (-8.84, 4.14) \]