Bivariate Data

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Bivariate Data · Grade 11 · Worksheet 2

1. Matiu surveyed 120 students at his school about their preferred study method (visual aids or reading notes) and their performance in a recent mathematics test. The two-way table below shows the results. Complete the two-way table by calculating the missing values and then determine if there is an association between study method preference and test performance. \begin{tabular}{|c|c|c|c|} \hline & Passed & Failed & Total \\ \hline Visual Aids & 45 & 15 & 60 \\ \hline Reading Notes & ? & ? & 60 \\ \hline Total & 70 & 50 & 120 \\ \hline \end{tabular} What are the missing values in the 'Reading Notes' row, and what do the conditional relative frequencies by row suggest about the relationship between study method and test performance? Answer: Reading Notes Passed: 25, Reading Notes Failed: 35; Visual Aids students have a 75% pass rate, Reading Notes students have a 41.7% pass rate, indicating an association where visual aids are associated with higher pass rates. Solution: Find the missing value for 'Reading Notes Passed'. Total Passed is 70, and Visual Aids Passed is 45, so Reading Notes Passed = 70 - 45 = 25. Find the missing value for 'Reading Notes Failed'.
   Full step-by-step solution

   Step 1: Find the missing value for 'Reading Notes Passed'. Total Passed is 70, and Visual Aids Passed is 45, so Reading Notes Passed = 70 - 45 = 25. Step 2: Find the missing value for 'Reading Notes Failed'. Total Failed is 50, and Visual Aids Failed is 15, so Reading Notes Failed = 50 - 15 = 35. Step 3: Verify row totals. Reading Notes row: 25 + 35 = 60, which matches the given total. Step 4: Calculate conditional relative frequencies for Visual Aids row: Passed = 45/60 = 0.75 (75%), Failed = 15/60 = 0.25 (25%). Step 5: Calculate conditional relative frequencies for Reading Notes row: Passed = 25/60 ≈ 0.4167 (41.7%), Failed = 35/60 ≈ 0.5833 (58.3%). Step 6: Compare the distributions. Visual Aids students have a much higher pass rate (75%) than Reading Notes students (41.7%), indicating an association between study method preference and test performance. The data suggest that students who prefer visual aids are more likely to pass the test. The answer is: Reading Notes Passed = 25, Reading Notes Failed = 35; there is an association, with visual aids linked to higher pass rates.

2. A scatter plot shows the relationship between study hours (x) and test scores (y) for 50 students. The data points form an approximately linear pattern with a correlation coefficient of r = 0.85. The regression line equation is y = 3.2x + 65. If a student studies for 8 hours, what test score does the regression model predict? Answer: 90.6 Solution: y = 3.2x + 65 - x = study hours - y = predicted test score We want the predicted test score for a student who studies 8 hours, so we substitute x = 8 into the equation.
   Full step-by-step solution

   We are given the regression line equation: y = 3.2x + 65 Here, - x = study hours - y = predicted test score We want the predicted test score for a student who studies 8 hours, so we substitute x = 8 into the equation. Step 1: Write the equation with x = 8 y = 3.2 * 8 + 65 Step 2: Multiply 3.2 by 8 3.2 * 8 = 25.6 Step 3: Add 65 to the result 25.6 + 65 = 90.6 So the regression model predicts a test score of 90.6 for 8 hours of study. The correlation coefficient r = 0.85 is not needed for this calculation because the regression equation is already given.

3. Create a scatter plot for Tane's data: (12, 28), (16, 36), (20, 44), (24, 52), (28, 60), (32, 68), (36, 76), (40, 84), (44, 92), (48, 100). Calculate the correlation coefficient r. Answer: 1.0 Solution: Calculate the mean of x-values: (12+16+20+24+28+32+36+40+44+48)/10 = 300/10 = 30 Calculate the mean of y-values: (28+36+44+52+60+68+76+84+92+100)/10 = 640/10 = 64 Calculate the sum of (x-mean_x)(y-mean_y): (12-30)(28-64) + (16-30)(36-64) + (20-30)(44-64) + (24-30)(52-64) + (28-30)(60-64) +…
   Full step-by-step solution

   Step 1: Calculate the mean of x-values: (12+16+20+24+28+32+36+40+44+48)/10 = 300/10 = 30 Step 2: Calculate the mean of y-values: (28+36+44+52+60+68+76+84+92+100)/10 = 640/10 = 64 Step 3: Calculate the sum of (x-mean_x)(y-mean_y): (12-30)(28-64) + (16-30)(36-64) + (20-30)(44-64) + (24-30)(52-64) + (28-30)(60-64) + (32-30)(68-64) + (36-30)(76-64) + (40-30)(84-64) + (44-30)(92-64) + (48-30)(100-64) = (-18)(-36) + (-14)(-28) + (-10)(-20) + (-6)(-12) + (-2)(-4) + (2)(4) + (6)(12) + (10)(20) + (14)(28) + (18)(36) = 648 + 392 + 200 + 72 + 8 + 8 + 72 + 200 + 392 + 648 = 2640 Step 4: Calculate the sum of (x-mean_x)^2: (-18)^2 + (-14)^2 + (-10)^2 + (-6)^2 + (-2)^2 + (2)^2 + (6)^2 + (10)^2 + (14)^2 + (18)^2 = 324 + 196 + 100 + 36 + 4 + 4 + 36 + 100 + 196 + 324 = 1320 Step 5: Calculate the sum of (y-mean_y)^2: (-36)^2 + (-28)^2 + (-20)^2 + (-12)^2 + (-4)^2 + (4)^2 + (12)^2 + (20)^2 + (28)^2 + (36)^2 = 1296 + 784 + 400 + 144 + 16 + 16 + 144 + 400 + 784 + 1296 = 5280 Step 6: Calculate r = sum[(x-mean_x)(y-mean_y)] / sqrt[sum(x-mean_x)^2 * sum(y-mean_y)^2] = 2640 / sqrt(1320 * 5280) = 2640 / sqrt(6969600) = 2640 / 2640 = 1.0 The correlation coefficient is 1.0, indicating a perfect positive linear relationship.

4. A scatter plot shows the relationship between the number of hours students spend on extracurricular activities per week (x) and their GPA (y) for 60 high school students. The data points form an approximately linear pattern with a correlation coefficient of r = -0.72. The least squares regression line equation is ŷ = -0.15x + 3.8. According to this model, what GPA would be predicted for a student who spends 6 hours per week on extracurricular activities? Answer: 2.9 Solution: Identify the regression equation: ŷ = -0.15x + 3.8 Substitute x = 6 (hours spent on extracurricular activities) into the equation: ŷ = -0.15(6) + 3.8 Calculate the multiplication: -0.15 × 6 = -0.9 Add the result to the y-intercept: -0.9 + 3.8 = 2.9 The predicted GPA is 2.9 The answer is 2.9.
   Full step-by-step solution

   Step 1: Identify the regression equation: ŷ = -0.15x + 3.8 Step 2: Substitute x = 6 (hours spent on extracurricular activities) into the equation: ŷ = -0.15(6) + 3.8 Step 3: Calculate the multiplication: -0.15 × 6 = -0.9 Step 4: Add the result to the y-intercept: -0.9 + 3.8 = 2.9 Step 5: The predicted GPA is 2.9 The answer is 2.9.

5. A medical researcher is studying the relationship between daily screen time (in hours) and sleep quality scores (on a 0-100 scale) among high school students. After collecting data from 120 students, the researcher calculates a correlation coefficient of r = -0.72. The researcher wants to test whether this correlation is statistically significant at the α = 0.05 level. Formulate the appropriate null and alternative hypotheses, then determine the critical value for this hypothesis test using a two-tailed approach. Answer: H₀: ρ = 0, H₁: ρ ≠ 0, critical value = ±1.980 Solution: We are testing if the population correlation coefficient ρ (rho) is significantly different from zero. The null hypothesis (H₀) assumes no correlation in the population. H₀: ρ = 0 H₁: ρ ≠ 0 We have a sample correlation r = -0.72 from n = 120 students.
   Full step-by-step solution

   Step 1: State the hypotheses We are testing if the population correlation coefficient ρ (rho) is significantly different from zero. The null hypothesis (H₀) assumes no correlation in the population. The alternative hypothesis (H₁) assumes there is a correlation (could be positive or negative). Since the problem says "two-tailed approach", we are checking for any non-zero correlation. So: H₀: ρ = 0 H₁: ρ ≠ 0 Step 2: Identify the test statistic and distribution We have a sample correlation r = -0.72 from n = 120 students. For testing H₀: ρ = 0, the test statistic follows a t-distribution with degrees of freedom df = n - 2. Here, df = 120 - 2 = 118. Step 3: Determine the critical value We use α = 0.05 for a two-tailed test. This means we split α into two tails: α/2 = 0.025 in each tail. We need the t-critical value from the t-distribution table for df = 118 and two-tailed α = 0.05. Since df = 118 is large (greater than 100), the t-distribution is very close to the standard normal distribution. From standard normal, z for 0.025 in each tail is about ±1.96. But for df = 118, the exact critical t-value is slightly larger. Looking up t-tables: For df = 100, t_{0.025} ≈ 1.984 For df = 120, t_{0.025} ≈ 1.980 For df = 118, we can interpolate or use df = 120 as a close approximation. Thus, the critical t-value for df = 118, α/2 = 0.025 is approximately 1.980. Because it's two-tailed, the critical region is t < -1.980 or t > 1.980. So critical value = ±1.980 Step 4: Conclusion for the hypothesis test setup We have: H₀: ρ = 0 H₁: ρ ≠ 0 Critical value = ±1.980 (Note: The actual test statistic t = r * sqrt((n-2)/(1-r^2)) would be computed from the data, but here we only need the critical value for the decision rule.)

6. Create a scatter plot for the data: (9,14), (11,18), (13,22), (15,26), (17,30), (19,34). Calculate the correlation coefficient r. Answer: 1.0 Solution: Calculate the means: x̄ = (9+11+13+15+17+19)/6 = 84/6 = 14, ȳ = (14+18+22+26+30+34)/6 = 144/6 = 24 Calculate the sums needed for r: Σ(x-x̄)(y-ȳ) = (9-14)(14-24)+(11-14)(18-24)+(13-14)(22-24)+(15-14)(26-24)+(17-14)(30-24)+(19-14)(34-24) = (-5)(-10)+(-3)(-6)+(-1)(-2)+(1)(2)+(3)(6)+(5)(10) =…
   Full step-by-step solution

   Step 1: Calculate the means: x̄ = (9+11+13+15+17+19)/6 = 84/6 = 14, ȳ = (14+18+22+26+30+34)/6 = 144/6 = 24 Step 2: Calculate the sums needed for r: Σ(x-x̄)(y-ȳ) = (9-14)(14-24)+(11-14)(18-24)+(13-14)(22-24)+(15-14)(26-24)+(17-14)(30-24)+(19-14)(34-24) = (-5)(-10)+(-3)(-6)+(-1)(-2)+(1)(2)+(3)(6)+(5)(10) = 50+18+2+2+18+50 = 140 Step 3: Calculate Σ(x-x̄)² = (-5)²+(-3)²+(-1)²+(1)²+(3)²+(5)² = 25+9+1+1+9+25 = 70 Step 4: Calculate Σ(y-ȳ)² = (-10)²+(-6)²+(-2)²+(2)²+(6)²+(10)² = 100+36+4+4+36+100 = 280 Step 5: Calculate r = 140/√(70×280) = 140/√19600 = 140/140 = 1.0 The answer is 1.0.