Statistical Inference

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Answer Key & Explanations

Statistical Inference · Grade 11 · Worksheet 3

1. A sample of 225 students at a large university has a mean GPA of 3.12 with a standard deviation of 0.48. Construct a 95% confidence interval for the population mean GPA. (Use z* = 1.96) Answer: (3.057, 3.183) Solution: Identify the given values: sample mean (x-bar) = 3.12, sample standard deviation (s) = 0.48, sample size (n) = 225, confidence level = 95%, critical value (z*) = 1.96.
   Full step-by-step solution

   Step 1: Identify the given values: sample mean (x-bar) = 3.12, sample standard deviation (s) = 0.48, sample size (n) = 225, confidence level = 95%, critical value (z*) = 1.96. Step 2: Calculate the standard error: SE = s / sqrt(n) = 0.48 / sqrt(225) = 0.48 / 15 = 0.032. Step 3: Calculate the margin of error: ME = z* * SE = 1.96 * 0.032 = 0.06272, which rounds to 0.063. Step 4: Construct the confidence interval: lower bound = 3.12 - 0.063 = 3.057, upper bound = 3.12 + 0.063 = 3.183. Step 5: The 95% confidence interval for the population mean GPA is (3.057, 3.183). The answer is (3.057, 3.183).

2. Olivia, a researcher at a health institute, wants to estimate the average number of hours per week that Grade 11 students spend on extracurricular activities. She surveys a random sample of 49 students and finds that the sample mean is 11 hours with a sample standard deviation of 3 hours. Based on this sample, calculate a 95% confidence interval for the population mean number of hours per week spent on extracurricular activities by all Grade 11 students. Then, interpret what this interval tells us about the population mean. Answer: The 95% confidence interval is approximately 10.16 to 11.84 hours. We are 95% confident that the true mean number of hours per week spent on extracurricular activities by all Grade 11 students is between 10.16 and 11.84 hours. Solution: Identify the given values. Sample mean = 11 hours, sample standard deviation = 3 hours, sample size n = 49. Confidence level = 95%, so alpha = 0.05.
   Full step-by-step solution

   Step 1: Identify the given values. Sample mean = 11 hours, sample standard deviation = 3 hours, sample size n = 49. Confidence level = 95%, so alpha = 0.05. Since the population standard deviation is unknown, we use the t-distribution. Step 2: Find the critical t-value. For a 95% confidence interval with n - 1 = 48 degrees of freedom, the critical t-value (from t-table) is approximately 2.011. (For large df close to 50, t_0.025 is about 2.009, but we'll use 2.011 for 48 df.) Step 3: Calculate the standard error. Standard error = s / sqrt(n) = 3 / sqrt(49) = 3 / 7 = 0.42857 hours. Step 4: Calculate the margin of error. Margin of error = t_critical * standard error = 2.011 * 0.42857 = approximately 0.8619 hours. Step 5: Construct the confidence interval. Lower bound = sample mean - margin of error = 11 - 0.8619 = 10.1381 hours. Upper bound = sample mean + margin of error = 11 + 0.8619 = 11.8619 hours. Rounded to two decimal places: 10.14 to 11.86 hours. (Alternatively, using t = 2.009 for df=50 gives 11 ± 0.861, so 10.14 to 11.86.) Step 6: Interpretation. We are 95% confident that the true mean number of hours per week that all Grade 11 students spend on extracurricular activities lies between approximately 10.14 and 11.86 hours. This means that if we repeated this sampling process many times, about 95% of the intervals constructed would contain the true population mean. The answer is: The 95% confidence interval is approximately 10.14 to 11.86 hours.

3. A sample of 144 students has a mean GPA of 3.12 with a standard deviation of 0.48. Construct a 99% confidence interval for the population mean GPA. (Use z* = 2.576) Answer: (3.017, 3.223) Solution: Identify the given values: sample mean = 3.12, sample standard deviation = 0.48, sample size n = 144, critical value z* = 2.576 for 99% confidence.
   Full step-by-step solution

   Step 1: Identify the given values: sample mean = 3.12, sample standard deviation = 0.48, sample size n = 144, critical value z* = 2.576 for 99% confidence. Step 2: Calculate the standard error: SE = s / sqrt(n) = 0.48 / sqrt(144) = 0.48 / 12 = 0.04. Step 3: Calculate the margin of error: ME = z* × SE = 2.576 × 0.04 = 0.10304. Step 4: Construct the confidence interval: lower bound = sample mean - ME = 3.12 - 0.10304 = 3.01696 ≈ 3.017; upper bound = sample mean + ME = 3.12 + 0.10304 = 3.22304 ≈ 3.223. Step 5: The 99% confidence interval for the population mean GPA is (3.017, 3.223). The answer is (3.017, 3.223).

4. A scatter plot displays the relationship between daily exercise time (minutes) and resting heart rate (bpm) for 40 adults. The least squares regression line is ŷ = -0.15x + 72. A residual plot shows random scatter around zero with no discernible pattern. According to this model, what resting heart rate would be predicted for an adult who exercises 60 minutes daily? Answer: 63 Solution: Identify the regression equation: ŷ = -0.15x + 72 Identify the given x-value (exercise time): x = 60 minutes Substitute x = 60 into the regression equation: ŷ = -0.15(60) + 72 Calculate -0.15 × 60 = -9 Add the result to 72: -9 + 72 = 63 The predicted resting heart rate is 63 bpm The answer is 63.
   Full step-by-step solution

   Step 1: Identify the regression equation: ŷ = -0.15x + 72 Step 2: Identify the given x-value (exercise time): x = 60 minutes Step 3: Substitute x = 60 into the regression equation: ŷ = -0.15(60) + 72 Step 4: Calculate -0.15 × 60 = -9 Step 5: Add the result to 72: -9 + 72 = 63 Step 6: The predicted resting heart rate is 63 bpm The answer is 63.

5. A random sample of 40 light bulbs from a factory has a mean lifetime of 850 hours with a standard deviation of 45 hours. Construct a 95% confidence interval for the population mean lifetime, using z* = 1.96. What is the margin of error? Answer: 13.94 hours Solution: Identify the given values: sample mean = 850, sample standard deviation = 45, sample size n = 40, critical value z* = 1.96. The margin of error (E) is calculated as E = z* * (standard deviation / sqrt(n)).
   Full step-by-step solution

   Step 1: Identify the given values: sample mean = 850, sample standard deviation = 45, sample size n = 40, critical value z* = 1.96. Step 2: The margin of error (E) is calculated as E = z* * (standard deviation / sqrt(n)). Step 3: Compute sqrt(40). sqrt(40) = 6.3249 (approximately). Step 4: Divide the standard deviation by sqrt(n): 45 / 6.3249 = 7.114 (approximately). Step 5: Multiply by z*: 1.96 * 7.114 = 13.94 (rounded to two decimal places). Step 6: The 95% confidence interval is (850 - 13.94, 850 + 13.94) = (836.06, 863.94). The margin of error is 13.94 hours.

6. A sample of 144 students from a large university has a mean weekly study time of 18.5 hours with a standard deviation of 9.6 hours. Construct a 95% confidence interval for the population mean weekly study time (using z* = 1.96) and state the margin of error. Answer: Margin of error = 1.568 hours; 95% CI = (16.932, 20.068) hours Solution: Identify the given values: sample mean x̄ = 18.5, sample standard deviation s = 9.6, sample size n = 144, confidence level = 95%, critical value z* = 1.96.
   Full step-by-step solution

   Step 1: Identify the given values: sample mean x̄ = 18.5, sample standard deviation s = 9.6, sample size n = 144, confidence level = 95%, critical value z* = 1.96. Step 2: Calculate the standard error: SE = s / sqrt(n) = 9.6 / sqrt(144) = 9.6 / 12 = 0.8. Step 3: Calculate the margin of error: ME = z* × SE = 1.96 × 0.8 = 1.568. Step 4: Construct the confidence interval: lower bound = x̄ - ME = 18.5 - 1.568 = 16.932; upper bound = x̄ + ME = 18.5 + 1.568 = 20.068. Step 5: Interpret: We are 95% confident that the true mean weekly study time for all students at this university is between 16.932 hours and 20.068 hours. The margin of error is ±1.568 hours. The answer is: Margin of error = 1.568 hours; 95% CI = (16.932, 20.068) hours.

7. H₀: μ = 100, H₁: μ ≠ 100, z = 2.45, α = 0.02; Reject H₀? Answer: B. yes Solution: Identify this is a two-tailed test because H₁: μ ≠ 100 For α = 0.02, the critical values are z = ±2.33 (since 0.02/2 = 0.01 in each tail) Compare the test statistic z = 2.45 to the critical value 2.33 Since 2.45 > 2.33, the test statistic falls in the rejection region
   Full step-by-step solution

   Step 1: Identify this is a two-tailed test because H₁: μ ≠ 100 Step 2: For α = 0.02, the critical values are z = ±2.33 (since 0.02/2 = 0.01 in each tail) Step 3: Compare the test statistic z = 2.45 to the critical value 2.33 Step 4: Since 2.45 > 2.33, the test statistic falls in the rejection region Step 5: Therefore, we reject the null hypothesis H₀ Final answer: yes