We are given a dataset of pre-test scores ($X_1$) and post-test scores ($X_2$) for a group of students. We want to determine if there was a statistically significant increase in scores from the pre-test to the post-test at a significance level of $\alpha = 0.05$, and if so, by how much did the scores increase. We need to conduct a paired t-test.

1. Problem Description

We are given a dataset of pre-test scores ($X_1$) and post-test scores ($X_2$) for a group of students. We want to determine if there was a statistically significant increase in scores from the pre-test to the post-test at a significance level of $\alpha = 0.05$, and if so, by how much did the scores increase. We need to conduct a paired t-test.

2. Solution Steps

First, calculate the difference $D = X_2 - X_1$ for each student. Then calculate the mean of the differences $\bar{D}$ and the standard deviation of the differences $s_D$. After that, calculate the t-statistic and finally the p-value.

| Pretest ($X_1$) | Posttest ($X_2$) | D = $X_2 - X_1$ |

|---|---|---|

| 90 | 88 | -2 |

| 60 | 67 | 7 |

| 95 | 99 | 4 |

| 93 | 94 | 1 |

| 94 | 100 | 6 |

| 68 | 64 | -4 |

| 88 | 91 | 3 |

| 91 | 95 | 4 |

| 93 | 95 | 2 |

| 84 | 89 | 5 |

| 76 | 82 | 6 |

| 86 | 92 | 6 |

| 83 | 83 | 0 |

| 81 | 85 | 4 |

| 89 | 93 | 4 |

| 65 | 69 | 4 |

| 91 | 90 | -1 |

| 92 | 100 | 8 |

| 87 | 95 | 8 |

| 92 | 96 | 4 |

$n = 20$

Calculate the mean difference $\bar{D}$:

$\bar{D} = \frac{\sum D}{n} = \frac{-2 + 7 + 4 + 1 + 6 - 4 + 3 + 4 + 2 + 5 + 6 + 6 + 0 + 4 + 4 + 4 - 1 + 8 + 8 + 4}{20} = \frac{73}{20} = 3.65$

Calculate the standard deviation of the differences $s_D$:

First calculate $\sum D^2 = (-2)^2 + 7^2 + 4^2 + 1^2 + 6^2 + (-4)^2 + 3^2 + 4^2 + 2^2 + 5^2 + 6^2 + 6^2 + 0^2 + 4^2 + 4^2 + 4^2 + (-1)^2 + 8^2 + 8^2 + 4^2 = 4 + 49 + 16 + 1 + 36 + 16 + 9 + 16 + 4 + 25 + 36 + 36 + 0 + 16 + 16 + 16 + 1 + 64 + 64 + 16 = 439$

$s_D = \sqrt{\frac{\sum D^2 - n\bar{D}^2}{n-1}} = \sqrt{\frac{439 - 20(3.65)^2}{20-1}} = \sqrt{\frac{439 - 20(13.3225)}{19}} = \sqrt{\frac{439 - 266.45}{19}} = \sqrt{\frac{172.55}{19}} = \sqrt{9.0815789} \approx 3.0135658$

Calculate the t-statistic:

$t = \frac{\bar{D}}{s_D / \sqrt{n}} = \frac{3.65}{3.0135658 / \sqrt{20}} = \frac{3.65}{3.0135658 / 4.4721359} = \frac{3.65}{0.6737} \approx 5.418$

Degrees of freedom $df = n - 1 = 20 - 1 = 19$

Significance level $\alpha = 0.05$

Since we are testing for an improvement, this is a one-tailed test. Looking up the t-critical value for a one-tailed test with $df = 19$ and $\alpha = 0.05$, we find $t_{critical} = 1.729$.

Since our calculated t-statistic ($5.418$) is greater than the t-critical value ($1.729$), we reject the null hypothesis that there is no improvement. Therefore, the scores significantly increased.

Alternatively, we can calculate the p-value. Using a t-distribution calculator with $t = 5.418$ and $df = 19$, we find the one-tailed p-value to be approximately $0.00001$. Since this p-value is less than $\alpha = 0.05$, we reject the null hypothesis and conclude that the scores significantly increased.

The estimated increase in scores is the mean difference $\bar{D} = 3.65$.

3. Final Answer

Yes, scores increased. The scores increased by 3.65 on average.