Statistical Inference

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Statistical Inference · Grade 11 · Worksheet 1

1. Matiu is a Grade 11 student investigating whether a new online learning platform improves student performance in mathematics. He collects a random sample of 40 students who used the platform for one semester and finds their average final exam score is 76.4. The school's historical average final exam score for all students (without the platform) is 72.0 with a known population standard deviation of 8.2. Matiu wants to conduct a hypothesis test at a significance level of α = 0.05 to determine if the platform leads to higher scores. Calculate the z-test statistic for this hypothesis test, rounded to two decimal places. Answer: 3.39 Solution: Identify the null and alternative hypotheses. H₀: μ = 72.0 (the platform has no effect) H₁: μ > 72.0 (the platform improves scores) Identify the given values.
   Full step-by-step solution

   Step 1: Identify the null and alternative hypotheses. H₀: μ = 72.0 (the platform has no effect) H₁: μ > 72.0 (the platform improves scores) Step 2: Identify the given values. Sample mean (x̄) = 76.4 Population mean (μ) = 72.0 Population standard deviation (σ) = 8.2 Sample size (n) = 40 Significance level (α) = 0.05 Step 3: Calculate the standard error. Standard error = σ / √n = 8.2 / √40 = 8.2 / 6.3249 = 1.2965 Step 4: Calculate the z-test statistic. z = (x̄ - μ) / (σ / √n) = (76.4 - 72.0) / 1.2965 = 4.4 / 1.2965 = 3.394 Step 5: Round to two decimal places. z = 3.39 The calculated z-test statistic is 3.39.

2. A university is studying whether a new study method improves test scores. They randomly select 64 students who use the method and find their average score is 82 with a standard deviation of 12. The historical average score for all students is 78. The researchers conduct a hypothesis test at α = 0.05 significance level to determine if the new method leads to higher scores. Calculate the test statistic for this hypothesis test. Answer: 2.67 Solution: H₀: μ = 78 (the new method has no effect) H₁: μ > 78 (the new method improves scores) Standard error = s/√n = 12/√64 = 12/8 = 1.5 t = (sample mean - population mean) / standard error t = (82 - 78) / 1.5 = 4 / 1.5 = 2.6667 t = 2.67 The test statistic is 2.67.
   Full step-by-step solution

   Step 1: Identify the null and alternative hypotheses H₀: μ = 78 (the new method has no effect) H₁: μ > 78 (the new method improves scores) Step 2: Calculate the standard error Standard error = s/√n = 12/√64 = 12/8 = 1.5 Step 3: Calculate the test statistic t = (sample mean - population mean) / standard error t = (82 - 78) / 1.5 = 4 / 1.5 = 2.6667 Step 4: Round to two decimal places t = 2.67 The test statistic is 2.67.

3. A scatter plot displays the relationship between daily temperature in °C (x) and ice cream sales in dollars (y) for 60 days. The least squares regression line is calculated as ŷ = 15.2x + 120. A residual plot shows points randomly scattered around zero with no discernible pattern. If the temperature reaches 32°C, what total ice cream sales does the regression model predict? Answer: 606.4 Solution: Identify the regression equation: ŷ = 15.2x + 120 Substitute x = 32 into the equation: ŷ = 15.2(32) + 120 Calculate 15.2 × 32: 15.2 × 30 = 456, 15.2 × 2 = 30.4, so 456 + 30.4 = 486.4 Add 120: 486.4 + 120 = 606.4 The predicted ice cream sales are $606.40 The answer is 606.4.
   Full step-by-step solution

   Step 1: Identify the regression equation: ŷ = 15.2x + 120 Step 2: Substitute x = 32 into the equation: ŷ = 15.2(32) + 120 Step 3: Calculate 15.2 × 32: 15.2 × 30 = 456, 15.2 × 2 = 30.4, so 456 + 30.4 = 486.4 Step 4: Add 120: 486.4 + 120 = 606.4 Step 5: The predicted ice cream sales are $606.40 The answer is 606.4.

4. A scatter plot shows the relationship between hours studied (x) and test scores (y) for 50 students. The data points form an approximately linear pattern with a correlation coefficient of r = 0.82. A least squares regression line is drawn through the data with equation ŷ = 2.5x + 65. If a student studies for 8 hours, what test score does the regression model predict? Answer: 85 Solution: ŷ = 2.5x + 65 - 2.5 is the slope - 65 is the y-intercept We want the predicted test score when x = 8 hours. Substitute x = 8 into the equation. ŷ = 2.5 * 8 + 65 Multiply 2.5 by 8.
   Full step-by-step solution

   We are given the regression equation: ŷ = 2.5x + 65 Here: - ŷ is the predicted test score - x is the number of hours studied - 2.5 is the slope - 65 is the y-intercept We want the predicted test score when x = 8 hours. Step 1: Substitute x = 8 into the equation. ŷ = 2.5 * 8 + 65 Step 2: Multiply 2.5 by 8. 2.5 * 8 = 20 Step 3: Add 65 to the result from Step 2. 20 + 65 = 85 Step 4: Interpret the result. The regression model predicts that a student who studies for 8 hours will score 85 on the test. Final answer: 85

5. A scatter plot shows the relationship between advertising spending (in thousands of dollars) and monthly sales revenue (in thousands of dollars) for 30 small businesses. The least squares regression line is calculated as ŷ = 3.2x + 42.5. The residual plot shows points randomly scattered around zero with no clear pattern. If a business spends $15,000 on advertising, what monthly sales revenue does the regression model predict? Answer: 90.5 Solution: The regression equation is ŷ = 3.2x + 42.5, where x is advertising spending in thousands of dollars and ŷ is predicted sales revenue in thousands of dollars.
   Full step-by-step solution

   Step 1: The regression equation is ŷ = 3.2x + 42.5, where x is advertising spending in thousands of dollars and ŷ is predicted sales revenue in thousands of dollars. Step 2: The business spends $15,000 on advertising, which is 15 thousand dollars, so x = 15. Step 3: Substitute x = 15 into the regression equation: ŷ = 3.2(15) + 42.5 Step 4: Calculate 3.2 × 15 = 48 Step 5: Add 42.5: 48 + 42.5 = 90.5 Step 6: The predicted sales revenue is 90.5 thousand dollars, which is $90,500. The answer is 90.5.

6. P(Z > 1.96) = ? Answer: 0.025 Solution: We are given: P(Z > 1.96) where Z is the standard normal random variable (mean 0, variance 1). We want the probability that Z is greater than 1.96. This is the area under the standard normal curve to the right of 1.96.
   Full step-by-step solution

   We are given: P(Z > 1.96) where Z is the standard normal random variable (mean 0, variance 1). Step 1: Understand the problem. We want the probability that Z is greater than 1.96. This is the area under the standard normal curve to the right of 1.96. Step 2: Use symmetry and standard normal tables. Standard normal tables usually give P(Z <= z), i.e., the cumulative distribution function (CDF) from the left up to z. So, P(Z > 1.96) = 1 - P(Z <= 1.96). Step 3: Look up P(Z <= 1.96) in the standard normal table. From the standard normal table, the value for z = 1.96 is 0.9750. That means P(Z <= 1.96) = 0.9750. Step 4: Subtract from 1. P(Z > 1.96) = 1 - 0.9750 = 0.0250. Step 5: Interpret the result. This means there is a 2.5% chance that a standard normal random variable exceeds 1.96. Final answer: 0.025