Two sets of data are similar with 95 % confidence level

To assess whether two sets of data are similar with a 95% confidence level, you can perform a hypothesis test for the equality of means. Here’s a step-by-step process using a t-test:

1. Define your null and alternative hypotheses:
2. . — Null Hypothesis (H0): The two sets of data have the same mean.
3. . — Alternative Hypothesis (H1): The two sets of data do not have the same mean.

2. Calculate the means and variances of both data sets.

3. Compute the t-statistic using the formula:

. \[ t = \frac{{\bar{X}_1 — \bar{X}_2}}{{\sqrt{\frac{{S_1^2}}{{n_1}} + \frac{{S_2^2}}{{n_2}}}}}\]

. where:

. — \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means of the two data sets.

. — \(S_1^2\) and \(S_2^2\) are the sample variances of the two data sets.

. — \(n_1\) and \(n_2\) are the sample sizes of the two data sets.

4. Determine the degrees of freedom (df) for the t-distribution. In this case, df is equal to \(n_1 + n_2 — 2\).

5. Find the critical t-value for a 95% confidence level and the calculated degrees of freedom. You can look up this value in a t-table or use statistical software.

6. Compare the calculated t-statistic to the critical t-value. If the calculated t-statistic is greater than the critical t-value, you reject the null hypothesis, indicating that the two data sets are significantly different. If it’s less than the critical t-value, you fail to reject the null hypothesis, suggesting that there is no significant difference between the data sets.

Keep in mind that statistical significance doesn’t necessarily mean practical significance. Even if you find a significant difference, it may not be practically important. Additionally, consider the assumptions of the t-test, such as the normality of data and equal variances, and make sure they hold for your data before conducting the test.

Example

let’s work through an example with two sample sets. Suppose you have two sets of exam scores for two groups of students (Group A and Group B), and you want to determine if the two groups perform similarly at a 95% confidence level.

Here are the sample scores:

Group A: [85, 88, 92, 78, 89, 91, 83, 86, 90, 87]

Group B: [82, 86, 88, 79, 85, 84, 87, 90, 91, 82]

Step 1: Define your hypotheses:

• Null Hypothesis (H0): The mean exam scores for Group A and Group B are the same.
• - Alternative Hypothesis (H1): The mean exam scores for Group A and Group B are not the same.

Step 2: Calculate the means and variances for both groups:

• Group A: Mean (μA) = 87.9, Variance (σA^2) ≈ 14.89
• - Group B: Mean (μB) = 85.3, Variance (σB^2) ≈ 10.61

Step 3: Compute the t-statistic:

\[ t = \frac{{87.9 — 85.3}}{{\sqrt{\frac{{14.89}}{{10}} + \frac{{10.61}}{{10}}}}} \approx 0.972\]

Step 4: Determine the degrees of freedom (df):

df = 10 + 10 — 2 = 18

Step 5: Find the critical t-value for a 95% confidence level with 18 degrees of freedom. You can use a t-table or calculator to find this value. For a two-tailed test at 95% confidence, it’s approximately ±2.101.

Step 6: Compare the calculated t-statistic (0.972) to the critical t-value (±2.101). Since 0.972 is within the range of -2.101 to 2.101, you fail to reject the null hypothesis. This means that at the 95% confidence level, there is no significant difference in the mean exam scores between Group A and Group B.

In this example, we did not find enough evidence to conclude that the two groups perform differently based on the t-test at a 95% confidence level.