To construct a 99% confidence interval for the true mean change in typing speeds before and after completing the program, we follow these steps:

Step 1: Define the Differences

For each individual, calculate the difference between the typing speeds before and after the program. Let $d_i$ be the difference for each person, where $d_i = \text{Before} - \text{After}$.

Participant	Before	After	Difference ($d$)
1	34	33	34 - 33 = 1
2	30	35	30 - 35 = -5
3	50	36	50 - 36 = 14
4	53	46	53 - 46 = 7
5	33	35	33 - 35 = -2
6	47	43	47 - 43 = 4

So, the differences ($d$) are: 1, -5, 14, 7, -2, 4.

Step 2: Calculate the Sample Mean and Standard Deviation of the Differences

1. Mean of the differences ($\bar{d}$): $\bar{d} = \frac{\sum d_i}{n} = \frac{1 + (-5) + 14 + 7 + (-2) + 4}{6} = \frac{19}{6} \approx 3.17$

2. Standard Deviation of the differences ($s_d$): First, calculate the variance ($s_d^2$) and then take the square root to get the standard deviation: $s_d^2 = \frac{\sum (d_i - \bar{d})^2}{n-1}$ $s_d = \sqrt{\frac{(1 - 3.17)^2 + (-5 - 3.17)^2 + (14 - 3.17)^2 + (7 - 3.17)^2 + (-2 - 3.17)^2 + (4 - 3.17)^2}{5}}$ Let's calculate this step-by-step:

   • $(1 - 3.17)^2 \approx 4.71$
   • $(-5 - 3.17)^2 \approx 66.69$
   • $(14 - 3.17)^2 \approx 118.47$
   • $(7 - 3.17)^2 \approx 14.66$
   • $(-2 - 3.17)^2 \approx 26.47$
   • $(4 - 3.17)^2 \approx 0.69$

   Sum of squared differences: $\sum (d_i - \bar{d})^2 \approx 231.69$ Then, divide by $n - 1$: $s_d^2 \approx \frac{231.69}{5} = 46.34$ $s_d \approx \sqrt{46.34} \approx 6.81$

Step 3: Determine the Critical Value

For a 99% confidence interval with $n = 6$, degrees of freedom ($df$) is $n - 1 = 5$. Using a t-table, the critical value ($t^*$) for 99% confidence and 5 degrees of freedom is approximately 4.032.

Step 4: Calculate the Margin of Error (ME)

$\text{ME} = t^* \times \frac{s_d}{\sqrt{n}} = 4.032 \times \frac{6.81}{\sqrt{6}} \approx 4.032 \times 2.78 \approx 11.2$

Step 5: Construct the Confidence Interval

$\text{Confidence Interval} = \bar{d} \pm \text{ME}$ $= 3.17 \pm 11.2$ $\approx (-8.0, 14.4)$

Final Answer

The 99% confidence interval for the true mean change in typing speeds is approximately: $(-8.0, 14.4)$

This interval means that we are 99% confident that the true mean change in typing speed after using the program falls between -8.0 and 14.4 words per minute.

Would you like further details or have any questions?

Related Questions:

1. How do you determine which t-distribution value to use for a specific confidence interval?
2. What is the impact of sample size on the width of a confidence interval?
3. How would the confidence interval change if we used a 95% confidence level instead?
4. Why is it important to use the differences between paired data for this type of problem?
5. How can we interpret a confidence interval that includes zero?

Tip:

The width of a confidence interval is influenced by the variability of the data, sample size, and the confidence level. A higher confidence level results in a wider interval.