Statistical Inference

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Answer Key & Explanations

Statistical Inference · Grade 11 · Worksheet 1

1. H₀: μ = 120, H₁: μ ≠ 120, sample mean = 124.5, σ = 15, n = 64, α = 0.05; z = ? Answer: 2.4 Solution: Identify the formula for the z-test statistic: z = (sample mean - hypothesized mean) / (σ/√n) Substitute the given values: z = (124.5 - 120) / (15/√64) Calculate the denominator: 15/√64 = 15/8 = 1.875 Calculate the numerator: 124.5 - 120 = 4.5 Divide numerator by denominator: 4.5 / 1.875 = 2.4…
   Full step-by-step solution

   Step 1: Identify the formula for the z-test statistic: z = (sample mean - hypothesized mean) / (σ/√n) Step 2: Substitute the given values: z = (124.5 - 120) / (15/√64) Step 3: Calculate the denominator: 15/√64 = 15/8 = 1.875 Step 4: Calculate the numerator: 124.5 - 120 = 4.5 Step 5: Divide numerator by denominator: 4.5 / 1.875 = 2.4 The answer is 2.4.

2. A nutrition researcher is studying the relationship between daily fruit consumption and blood pressure levels. After collecting data from 180 adults, she calculates a correlation coefficient of r = -0.42 between fruit servings and systolic blood pressure. She wants to test if this correlation is statistically significant at α = 0.05. What is the appropriate null hypothesis for this test? Answer: B. H₀: ρ = 0 Solution: In correlation hypothesis testing, we're testing whether a relationship exists in the population The null hypothesis always states there is no relationship or no effect For correlation tests, we use ρ (rho) to represent the population correlation coefficient The null hypothesis should be H₀: ρ =…
   Full step-by-step solution

   Step 1: In correlation hypothesis testing, we're testing whether a relationship exists in the population Step 2: The null hypothesis always states there is no relationship or no effect Step 3: For correlation tests, we use ρ (rho) to represent the population correlation coefficient Step 4: The null hypothesis should be H₀: ρ = 0, meaning no correlation exists in the population Step 5: The sample correlation r = -0.42 is what we're testing against this null hypothesis Step 6: Therefore, the correct null hypothesis is H₀: ρ = 0 The correct answer is H₀: ρ = 0.

3. Mere is investigating whether the average weekly screen time for Grade 11 students at her school exceeds the national average of 24 hours. She surveys a random sample of 64 students and finds a sample mean of 26.8 hours with a sample standard deviation of 8 hours. Using a significance level of α = 0.05, test the claim that the true mean weekly screen time for students at her school is greater than 24 hours. What is the calculated test statistic? Round your answer to two decimal places. Answer: 2.80 Solution: Identify the null and alternative hypotheses. H₀: μ = 24 (the school's average is equal to the national average) H₁: μ > 24 (the school's average is greater than the national average) Write down the given values.
   Full step-by-step solution

   Step 1: Identify the null and alternative hypotheses. H₀: μ = 24 (the school's average is equal to the national average) H₁: μ > 24 (the school's average is greater than the national average) Step 2: Write down the given values. Sample mean (x̄) = 26.8 Hypothesized population mean (μ₀) = 24 Sample standard deviation (s) = 8 Sample size (n) = 64 Step 3: Calculate the standard error. Standard error = s / sqrt(n) = 8 / sqrt(64) = 8 / 8 = 1 Step 4: Calculate the t-test statistic. t = (x̄ - μ₀) / (s / sqrt(n)) = (26.8 - 24) / 1 = 2.8 / 1 = 2.80 The calculated test statistic is 2.80.

4. A sample of 81 students from a large university has a mean GPA of 3.16 with a standard deviation of 0.36. Construct a 96% confidence interval for the population mean GPA. (Use z* = 2.05) Answer: (3.078, 3.242) Solution: Identify the given values. Sample mean (x̄) = 3.16, sample standard deviation (s) = 0.36, sample size (n) = 81, critical value (z*) = 2.05.
   Full step-by-step solution

   Step 1: Identify the given values. Sample mean (x̄) = 3.16, sample standard deviation (s) = 0.36, sample size (n) = 81, critical value (z*) = 2.05. Step 2: Calculate the standard error: SE = s / sqrt(n) = 0.36 / sqrt(81) = 0.36 / 9 = 0.04. Step 3: Calculate the margin of error: ME = z* × SE = 2.05 × 0.04 = 0.082. Step 4: Find the lower bound: x̄ - ME = 3.16 - 0.082 = 3.078. Step 5: Find the upper bound: x̄ + ME = 3.16 + 0.082 = 3.242. Step 6: The 96% confidence interval is (3.078, 3.242). The answer is (3.078, 3.242).

5. Emma is a researcher investigating whether a new online tutoring platform improves student performance on standardized math tests. She randomly selects 40 Grade 11 students who use the platform for one month and records their improvement scores (post-test score minus pre-test score). The sample mean improvement is 15 points with a sample standard deviation of 10 points. Emma wants to determine if the true mean improvement for all Grade 11 students who use the platform is greater than 12 points. Using a significance level of α = 0.05, calculate the test statistic (t-value) for this hypothesis test. Round your answer to two decimal places. Answer: 1.90 Solution: Identify the null and alternative hypotheses. H0: μ = 12 (the true mean improvement is 12 points) H1: μ > 12 (the true mean improvement is greater than 12 points) Identify the given values.
   Full step-by-step solution

   Step 1: Identify the null and alternative hypotheses. H0: μ = 12 (the true mean improvement is 12 points) H1: μ > 12 (the true mean improvement is greater than 12 points) Step 2: Identify the given values. Sample mean (x̄) = 15 Claimed population mean (μ0) = 12 Sample standard deviation (s) = 10 Sample size (n) = 40 Step 3: Calculate the standard error. Standard error = s / sqrt(n) = 10 / sqrt(40) = 10 / 6.3249 = 1.5811 Step 4: Calculate the test statistic (t-value). t = (x̄ - μ0) / (s / sqrt(n)) = (15 - 12) / 1.5811 = 3 / 1.5811 = 1.8974 Step 5: Round to two decimal places. t = 1.90 (rounded) The answer is 1.90.

6. P(Z > 1.645) = ? Answer: 0.05 Solution: The problem asks for P(Z > 1.645), which is the probability that a standard normal random variable exceeds 1.645. For a standard normal distribution, P(Z > z) = 1 - P(Z ≤ z).
   Full step-by-step solution

   Step 1: The problem asks for P(Z > 1.645), which is the probability that a standard normal random variable exceeds 1.645. Step 2: For a standard normal distribution, P(Z > z) = 1 - P(Z ≤ z). Step 3: Using a standard normal table or calculator, we find that P(Z ≤ 1.645) = 0.95. Step 4: Therefore, P(Z > 1.645) = 1 - 0.95 = 0.05. The answer is 0.05.