Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below. Person Person A B C D E F G H I J Test A 94 89 87 107 75 96 103 108 116 121 Test B 96 89 86 113 73 100 101 107 123 123 Test Statistic: Critical Value: 99% Confidence interval:ÂÂ
Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below.
Person
Person A B C D E F G H I J
Test A 94 89 87 107 75 96 103 108 116 121
Test B 96 89 86 113 73 100 101 107 123 123
Test Statistic:
Critical Value:ÂÂ
99% Confidence interval:ÂÂ
The Correct Answer and Explanation is :
To analyze this data, we’ll compare the scores from Test A and Test B to see if there is a significant difference between them. We will use a paired sample t-test, which is appropriate since both sets of scores come from the same individuals. We’ll calculate the t-statistic, the critical value for a 99% confidence level, and the confidence interval.
Step 1: Calculate the differences between Test A and Test B scores.
For each person, subtract the Test A score from the Test B score:
| Person | Test A | Test B | Difference (B – A) |
|---|---|---|---|
| A | 94 | 96 | 2 |
| B | 89 | 89 | 0 |
| C | 87 | 86 | -1 |
| D | 107 | 113 | 6 |
| E | 75 | 73 | -2 |
| F | 96 | 100 | 4 |
| G | 103 | 101 | -2 |
| H | 108 | 107 | -1 |
| I | 116 | 123 | 7 |
| J | 121 | 123 | 2 |
Step 2: Calculate the mean and standard deviation of the differences.
Mean of differences (D̄) = (2 + 0 – 1 + 6 – 2 + 4 – 2 – 1 + 7 + 2) / 10 = 1.5
Standard deviation of differences (SD) = √[(Σ(difference – D̄)²) / (n-1)]
Step 3: Perform a t-test.
The formula for the t-statistic is:
[
t = \frac{D̄}{\frac{SD}{\sqrt{n}}}
]
Where:
- ( D̄ ) is the mean difference (1.5).
- ( SD ) is the standard deviation of the differences.
- ( n ) is the number of pairs (10).
Step 4: Determine the critical value for a 99% confidence level.
For 9 degrees of freedom (n-1), using a t-table, the critical value at a 99% confidence level is approximately 3.249.
Step 5: Calculate the 99% confidence interval.
The 99% confidence interval for the mean difference is calculated as:
[
CI = D̄ \pm t_{\text{critical}} \times \frac{SD}{\sqrt{n}}
]
Conclusion:
- If the absolute value of the t-statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference between the two tests.
- The confidence interval will give us a range of values within which we can be 99% confident the true mean difference lies.
I’ll now calculate the exact values for the standard deviation, t-statistic, and confidence interval. Let me compute that.
Here are the results:
- t-statistic: 1.45
- Critical value (for 99% confidence level with 9 degrees of freedom): 3.25
- 99% Confidence interval: (-1.87, 4.87)
Explanation:
- t-statistic (1.45): This value represents the ratio of the mean difference between the two tests to the standard error of the difference. It is used to determine whether the observed difference is statistically significant.
- Critical value (3.25): This value comes from the t-distribution table for a 99% confidence level and 9 degrees of freedom. If the absolute value of the t-statistic exceeds this critical value, we would reject the null hypothesis.
- Confidence interval (-1.87, 4.87): This interval represents the range within which we are 99% confident that the true mean difference lies. Since this interval includes zero, it suggests that there is no significant difference between the two tests at the 99% confidence level.
Conclusion:
Since the absolute value of the t-statistic (1.45) is less than the critical value (3.25), we fail to reject the null hypothesis. This means there is no statistically significant difference between the two tests at the 99% confidence level.