Let's go through each part of the problem step by step. ### i) Null and Alternate HypothesesThe null hypothesis (H0) typically states that there is no effect or no difference. The alternate hypothesis (H1) states that there is an effect or a difference. - **Null Hypothesis (H0):** The mean difference in reading age between the experimental group and the control group is zero. (μD = 0) - **Alternate Hypothesis (H1):** The mean difference in reading age between the experimental group and the control group is not zero. (μD ≠ 0) ### ii) Sample Mean of DTo find the sample mean of D, sum all the differences and divide by the number of pairs. \[ D = [-3, -2, +1, -4, -4, -1] \] \[ \text{Sample Mean} (\bar{D}) = \frac{\sum D}{n} = \frac{(-3) + (-2) + 1 + (-4) + (-4) + (-1)}{6} = \frac{-13}{6} = -2.17 \] ### iii) Sample Standard Deviation of DTo find the sample standard deviation, use the formula: \[ s_D = \sqrt{\frac{\sum (D_i - \bar{D})^2}{n-1}} \] First, calculate each \( (D_i - \bar{D})^2 \): \[ (-3 + 2.17)^2 = 0.6889 \] \[ (-2 + 2.17)^2 = 0.0289 \] \[ (1 + 2.17)^2 = 9.8289 \] \[ (-4 + 2.17)^2 = 3.3769 \] \[ (-4 + 2.17)^2 = 3.3769 \] \[ (-1 + 2.17)^2 = 1.3689 \] Sum these values: \[ 0.6889 + 0.0289 + 9.8289 + 3.3769 + 3.3769 + 1.3689 = 18.6704 \] Then divide by \( n-1 \): \[ s_D = \sqrt{\frac{18.6704}{5}} = \sqrt{3.73408} = 1.93 \] ### iv) Standardized Test StatisticThe standardized test statistic (t) is calculated as: \[ t = \frac{\bar{D} - \mu_D}{s_D / \sqrt{n}} \] Under the null hypothesis, \( \mu_D = 0 \): \[ t = \frac{-2.17 - 0}{1.93 / \sqrt{6}} = \frac{-2.17}{0.787} = -2.76 \] ### v) P-value for the Standardized Test StatisticFor a one-tailed test with \( t = -2.76 \) and \( df = n-1 = 5 \): Using a t-distribution table or calculator, the p-value for \( t = -2.76 \) with 5 degrees of freedom is approximately 0.020. ### vi) ConclusionBased on the p-value (0.020), which is less than the typical significance level of 0.05, we reject the null hypothesis. This suggests that there is a statistically significant difference in the reading age between the experimental group and the control group.

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Let's go through each part of the problem step by step.

i) Null and Alternate Hypotheses

The null hypothesis (H0) typically states that there is no effect or no difference. The alternate hypothesis (H1) states that there is an effect or a difference.

• Null Hypothesis (H0): The mean difference in reading age between the experimental group and the control group is zero. (μD = 0)
• Alternate Hypothesis (H1): The mean difference in reading age between the experimental group and the control group is not zero. (μD ≠ 0)

ii) Sample Mean of D

To find the sample mean of D, sum all the differences and divide by the number of pairs. $D = [-3, -2, +1, -4, -4, -1]$ $\text{Sample Mean} (\bar{D}) = \frac{\sum D}{n} = \frac{(-3) + (-2) + 1 + (-4) + (-4) + (-1)}{6} = \frac{-13}{6} = -2.17$

iii) Sample Standard Deviation of D

To find the sample standard deviation, use the formula: $s_D = \sqrt{\frac{\sum (D_i - \bar{D})^2}{n-1}}$ First, calculate each $(D_i - \bar{D})^2$: $(-3 + 2.17)^2 = 0.6889$ $(-2 + 2.17)^2 = 0.0289$ $(1 + 2.17)^2 = 9.8289$ $(-4 + 2.17)^2 = 3.3769$ $(-4 + 2.17)^2 = 3.3769$ $(-1 + 2.17)^2 = 1.3689$ Sum these values: $0.6889 + 0.0289 + 9.8289 + 3.3769 + 3.3769 + 1.3689 = 18.6704$ Then divide by $n-1$: $s_D = \sqrt{\frac{18.6704}{5}} = \sqrt{3.73408} = 1.93$

iv) Standardized Test Statistic

The standardized test statistic (t) is calculated as: $t = \frac{\bar{D} - \mu_D}{s_D / \sqrt{n}}$ Under the null hypothesis, $\mu_D = 0$: $t = \frac{-2.17 - 0}{1.93 / \sqrt{6}} = \frac{-2.17}{0.787} = -2.76$

v) P-value for the Standardized Test Statistic

For a one-tailed test with $t = -2.76$ and $df = n-1 = 5$: Using a t-distribution table or calculator, the p-value for $t = -2.76$ with 5 degrees of freedom is approximately 0.020.

vi) Conclusion

Based on the p-value (0.020), which is less than the typical significance level of 0.05, we reject the null hypothesis. This suggests that there is a statistically significant difference in the reading age between the experimental group and the control group.