given two dependent random samples with the following results:

\begin{tabular}{|l|c|c|c|c|c|c|c|}
\hline
population 1 & 29 & 31 & 29 & 27 & 32 & 24 & 42 \\
\hline
population 2 & 34 & 45 & 20 & 22 & 35 & 27 & 40 \\
\hline
\end{tabular}

use this data to find the 99% confidence interval for the true difference between the population means. assume that both populations are normally distributed.

step 4 of 4: construct the 99% confidence interval. round your answers to one decimal place.

answer
lower endpoint:
upper endpoint:

Question

given two dependent random samples with the following results:

\begin{tabular}{|l|c|c|c|c|c|c|c|}
\hline
population 1 & 29 & 31 & 29 & 27 & 32 & 24 & 42 \\
\hline
population 2 & 34 & 45 & 20 & 22 & 35 & 27 & 40 \\
\hline
\end{tabular}

use this data to find the 99% confidence interval for the true difference between the population means. assume that both populations are normally distributed.

step 4 of 4: construct the 99% confidence interval. round your answers to one decimal place.

answer
lower endpoint:
upper endpoint:

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Mean Difference", "Paired t-Interval", "Margin of Error" ], "new_concepts": [], "current_concepts": [ "Mean Difference", "Paired t-Interval", "Margin of Error" ] } ## Calculate paired differences $$ \begin{aligned} d &= \text{Population 1} - \text{Population 2} \ d_1 &= 29 - 34 = -5 \ d_2 &= 31 - 45 = -14 \ d_3 &= 29 - 20 = 9 \ d_4 &= 27 - 22 = 5 \ d_5 &= 32 - 35 = -3 \ d_6 &= 24 - 27 = -3 \ d_7 &= 42 - 40 = 2 \end{aligned} $$ ## Compute sample mean and standard deviation of differences $$ \begin{aligned} n &= 7 \ \bar{d} &= \frac{-5 - 14 + 9 + 5 - 3 - 3 + 2}{7} = \frac{-9}{7} \approx -1.2857 \ s_d &= \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \ \sum d_i^2 &= (-5)^2 + (-14)^2 + 9^2 + 5^2 + (-3)^2 + (-3)^2 + 2^2 = 25 + 196 + 81 + 25 + 9 + 9 + 4 = 349 \ s_d &= \sqrt{\frac{349 - 7(-1.2857)^2}{6}} = \sqrt{\frac{349 - 11.5714}{6}} = \sqrt{56.2381} \approx 7.4992 \end{aligned} $$ ## Calculate confidence interval $$ \begin{aligned} df &= 6 \ t_{0.005, 6} &\approx 3.707 \ E &= t_{0.005, 6} \times \frac{s_d}{\sqrt{n}} = 3.707 \times \frac{7.4992}{\sqrt{7}} \approx 3.707 \times 2.8344 \approx 10.5071 \ \text{Lower endpoint} &= \bar{d} - E = -1.2857 - 10.5071 = -11.7928 \approx -11.8 \ \text{Upper endpoint} &= \bar{d} + E = -1.2857 + 10.5071 = 9.2214 \approx 9.2 \end{aligned} $$ Construct the 99% confidence interval. Round your answers to one decimal place. Lower endpoint: -11.8 Upper endpoint: 9.2 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Paired t-Interval" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate paired differences

Compute sample mean and standard deviation of differences

Calculate confidence interval

Answer: