a state legislator wants to determine whether his voters performance rating (0 - 100) has changed from last year to this year. the following table shows the legislators performance from the same ten randomly selected voters for last year and this year. use this data to find the 90% confidence interval for the true difference between the population means. assume that the populations of voters performance ratings are normally distributed for both this year and last year.

\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline rating (last year) & 77 & 89 & 50 & 44 & 54 & 51 & 66 & 42 & 94 & 44 \\
\hline rating (this year) & 50 & 94 & 75 & 55 & 45 & 65 & 94 & 67 & 91 & 46 \\
\hline
\end{tabular}

step 4 of 4: construct the 90% confidence interval. round your answers to one decimal place.

Question

a state legislator wants to determine whether his voters performance rating (0 - 100) has changed from last year to this year. the following table shows the legislators performance from the same ten randomly selected voters for last year and this year. use this data to find the 90% confidence interval for the true difference between the population means. assume that the populations of voters performance ratings are normally distributed for both this year and last year.

\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline rating (last year) & 77 & 89 & 50 & 44 & 54 & 51 & 66 & 42 & 94 & 44 \\
\hline rating (this year) & 50 & 94 & 75 & 55 & 45 & 65 & 94 & 67 & 91 & 46 \\
\hline
\end{tabular}

step 4 of 4: construct the 90% confidence interval. round your answers to one decimal place.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Paired t-Interval", "Mean Difference", "Margin of Error" ], "new_concepts": [], "current_concepts": [ "Paired t-Interval", "Mean Difference", "Margin of Error" ] } ## Calculate paired differences $$ \begin{aligned} d &= \text{Rating (this year)} - \text{Rating (last year)} \ d_1 &= 50 - 77 = -27 \ d_2 &= 94 - 89 = 5 \ d_3 &= 75 - 50 = 25 \ d_4 &= 55 - 44 = 11 \ d_5 &= 45 - 54 = -9 \ d_6 &= 65 - 51 = 14 \ d_7 &= 94 - 66 = 28 \ d_8 &= 67 - 42 = 25 \ d_9 &= 91 - 94 = -3 \ d_{10} &= 46 - 44 = 2 \ \end{aligned} $$ ## Compute sample mean and standard deviation of differences $$ \begin{aligned} n &= 10 \ \bar{d} &= \frac{-27 + 5 + 25 + 11 - 9 + 14 + 28 + 25 - 3 + 2}{10} = \frac{71}{10} = 7.1 \ s_d &= \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} \ \sum (d_i - \bar{d})^2 &= (-27 - 7.1)^2 + (5 - 7.1)^2 + (25 - 7.1)^2 + (11 - 7.1)^2 + (-9 - 7.1)^2 \ &+ (14 - 7.1)^2 + (28 - 7.1)^2 + (25 - 7.1)^2 + (-3 - 7.1)^2 + (2 - 7.1)^2 \ &= (-34.1)^2 + (-2.1)^2 + (17.9)^2 + (3.9)^2 + (-16.1)^2 \ &+ (6.9)^2 + (20.9)^2 + (17.9)^2 + (-10.1)^2 + (-5.1)^2 \ &= 1162.81 + 4.41 + 320.41 + 15.21 + 259.21 + 47.61 + 436.81 + 320.41 + 102.01 + 26.01 \ &= 2694.9 \ s_d &= \sqrt{\frac{2694.9}{9}} = \sqrt{299.4333} \approx 17.3041 \end{aligned} $$ ## Calculate confidence interval $$ \begin{aligned} df &= 9 \ t_{0.05, 9} &\approx 1.833 \ E &= t_{0.05, 9} \times \frac{s_d}{\sqrt{n}} = 1.833 \times \frac{17.3041}{\sqrt{10}} \approx 1.833 \times 5.4720 \approx 10.0302 \ \text{Lower endpoint} &= \bar{d} - E = 7.1 - 10.0302 = -2.9302 \approx -2.9 \ \text{Upper endpoint} &= \bar{d} + E = 7.1 + 10.0302 = 17.1302 \approx 17.1 \end{aligned} $$ Lower endpoint: -2.9 Upper endpoint: 17.1 { "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: