Statistical Inference

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Statistical Inference · Grade 11 · Worksheet 2

1. Tane is a marine biologist studying the average length of a specific species of fish in a large lake. He catches a random sample of 49 fish and measures their lengths. The sample mean length is 42.5 cm with a sample standard deviation of 9.8 cm. Tane wants to estimate the true mean length of this fish species in the entire lake. What is the margin of error for a 95% confidence interval for the population mean? (Use t* = 2.01 for 48 degrees of freedom.) Answer: 2.81 cm Solution: Identify the given values. Sample mean (x-bar) = 42.5 cm Sample standard deviation (s) = 9.8 cm Sample size (n) = 49 Degrees of freedom = n - 1 = 48 Critical t-value for 95% confidence (t*) = 2.01 Calculate the standard error.
   Full step-by-step solution

   Step 1: Identify the given values. Sample mean (x-bar) = 42.5 cm Sample standard deviation (s) = 9.8 cm Sample size (n) = 49 Degrees of freedom = n - 1 = 48 Critical t-value for 95% confidence (t*) = 2.01 Step 2: Calculate the standard error. Standard error = s / sqrt(n) = 9.8 / sqrt(49) = 9.8 / 7 = 1.4 cm Step 3: Calculate the margin of error. Margin of error = t* * (s / sqrt(n)) = 2.01 * 1.4 = 2.814 cm Step 4: Round to two decimal places. Margin of error = 2.81 cm Final answer: The margin of error is 2.81 cm. This means Tane can be 95% confident that the true mean length of the fish in the lake is within 42.5 cm plus or minus 2.81 cm, or between 39.69 cm and 45.31 cm.

2. A scatter plot displays the relationship between daily exercise time (x, in minutes) and resting heart rate (y, in beats per minute) for 60 adults. The least squares regression line is calculated as ŷ = -0.35x + 72. A residual plot shows points randomly scattered around zero with no clear pattern. If an adult exercises for 45 minutes daily, what resting heart rate does the regression model predict? Answer: 56.25 Solution: Identify the regression equation: ŷ = -0.35x + 72. Identify the given x-value (daily exercise time): x = 45 minutes. Substitute x = 45 into the regression equation: ŷ = -0.35 * (45) + 72.
   Full step-by-step solution

   Step 1: Identify the regression equation: ŷ = -0.35x + 72. Step 2: Identify the given x-value (daily exercise time): x = 45 minutes. Step 3: Substitute x = 45 into the regression equation: ŷ = -0.35 * (45) + 72. Step 4: Perform the multiplication: -0.35 * 45 = -15.75. Step 5: Perform the addition: -15.75 + 72 = 56.25. Step 6: The predicted resting heart rate is 56.25 beats per minute. The answer is 56.25.

3. Kaia is an environmental scientist studying the health of a native forest. She collects a random sample of 49 trees and measures their trunk diameters (in cm). The sample mean diameter is 31 cm with a sample standard deviation of 7 cm. Kaia wants to estimate the true mean trunk diameter of all trees in the forest with 95% confidence. Calculate the margin of error for this estimate, and then state the 95% confidence interval for the population mean trunk diameter. Answer: Margin of error = 1.96 * (7 / sqrt(49)) = 1.96 * 1 = 1.96 cm; Confidence interval = (29.04 cm, 32.96 cm) Solution: Identify the given values. Sample size n = 49 Sample mean x-bar = 31 cm Sample standard deviation s = 7 cm Confidence level = 95%, so the critical z-value (for a normal distribution) is z* = 1.96.
   Full step-by-step solution

   Step 1: Identify the given values. Sample size n = 49 Sample mean x-bar = 31 cm Sample standard deviation s = 7 cm Confidence level = 95%, so the critical z-value (for a normal distribution) is z* = 1.96. Step 2: Calculate the standard error of the mean. Standard error = s / sqrt(n) = 7 / sqrt(49) = 7 / 7 = 1 cm. Step 3: Calculate the margin of error. Margin of error = z* * standard error = 1.96 * 1 = 1.96 cm. Step 4: Construct the 95% confidence interval. Lower bound = sample mean - margin of error = 31 - 1.96 = 29.04 cm. Upper bound = sample mean + margin of error = 31 + 1.96 = 32.96 cm. Step 5: Interpret the result. The 95% confidence interval for the true mean trunk diameter of all trees in the forest is (29.04 cm, 32.96 cm). This means we are 95% confident that the population mean lies within this interval. The answer is: Margin of error = 1.96 cm; Confidence interval = (29.04 cm, 32.96 cm).

4. Hana is investigating whether the average daily screen time of Grade 11 students at her school exceeds the national average of 4.5 hours. She randomly selects 49 students and records their daily screen time. The sample mean is 5.1 hours with a sample standard deviation of 1.4 hours. Based on this sample, calculate the margin of error for a 95% confidence interval for the population mean daily screen time of all Grade 11 students at her school. Assume the critical t-value for 48 degrees of freedom at the 95% confidence level is approximately 2.01. Answer: 0.402 hours Solution: Sample size n = 49 Sample mean x_bar = 5.1 hours Sample standard deviation s = 1.4 hours Critical t-value for 95% confidence (df = 48) t_c = 2.01 Standard error = s / sqrt(n) = 1.4 / sqrt(49) = 1.4 / 7 = 0.2 Margin of error = t_c * (s / sqrt(n)) = 2.01 * 0.2 = 0.402 The margin of error is 0.402…
   Full step-by-step solution

   Step 1: Identify the given values: Sample size n = 49 Sample mean x_bar = 5.1 hours Sample standard deviation s = 1.4 hours Critical t-value for 95% confidence (df = 48) t_c = 2.01 Step 2: Calculate the standard error: Standard error = s / sqrt(n) = 1.4 / sqrt(49) = 1.4 / 7 = 0.2 Step 3: Calculate the margin of error: Margin of error = t_c * (s / sqrt(n)) = 2.01 * 0.2 = 0.402 Step 4: Interpret the result: The margin of error is 0.402 hours. This means we can be 95% confident that the true population mean daily screen time for all Grade 11 students at the school is within 5.1 ± 0.402 hours, i.e., between 4.698 and 5.502 hours. The answer is 0.402 hours.

5. H₀: μ = 100, H₁: μ ≠ 100, z = 2.58, α = 0.01; Reject H₀? Answer: A. yes Solution: Identify the type of test. H₁: μ ≠ 100 indicates a two-tailed test. Determine the critical z-value for α = 0.01.
   Full step-by-step solution

   Step 1: Identify the type of test. H₁: μ ≠ 100 indicates a two-tailed test. Step 2: Determine the critical z-value for α = 0.01. For a two-tailed test, the area in each tail is α/2 = 0.005. Step 3: The critical z-value for a two-tailed test with α = 0.01 is ±2.576. Step 4: Compare the test statistic (z = 2.58) to the critical value (2.576). Since 2.58 > 2.576, the test statistic falls in the rejection region. Step 5: Decision: Reject the null hypothesis H₀. The answer is yes.

6. A university is studying whether a new study technique improves exam scores. They randomly select 36 students who use the technique and find their mean score is 82 with a standard deviation of 12. The historical mean score for all students is 78. Using a significance level of α = 0.05, test the hypothesis that the new technique improves scores. Calculate the test statistic and determine if there is sufficient evidence to support the claim. Answer: 2.00 Solution: H₀: μ = 78 (no improvement) H₁: μ > 78 (technique improves scores) For one-sample t-test: t = (x̄ - μ) / (s/√n) x̄ = 82, μ = 78, s = 12, n = 36 t = (82 - 78) / (12/√36) 12/√36 = 12/6 = 2 t = 4 / 2 = 2.00 For one-tailed test with α = 0.05 and df = 35, critical t ≈ 1.69 Since 2.00 > 1.69, we…
   Full step-by-step solution

   Step 1: State the hypotheses H₀: μ = 78 (no improvement) H₁: μ > 78 (technique improves scores) Step 2: Identify the test statistic formula For one-sample t-test: t = (x̄ - μ) / (s/√n) Step 3: Plug in the values x̄ = 82, μ = 78, s = 12, n = 36 t = (82 - 78) / (12/√36) Step 4: Calculate the denominator 12/√36 = 12/6 = 2 Step 5: Calculate the test statistic t = 4 / 2 = 2.00 Step 6: Compare to critical value For one-tailed test with α = 0.05 and df = 35, critical t ≈ 1.69 Since 2.00 > 1.69, we reject H₀ The test statistic is 2.00, and there is sufficient evidence to support the claim that the technique improves scores.