Statistical Inference

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Answer Key & Explanations

Statistical Inference · Grade 11 · Worksheet 2

1. 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

2. Emma surveys a random sample of 40 students and finds the mean amount of weekly screen time is 15 hours with a standard deviation of 5 hours. Calculate the 95% confidence interval for the population mean weekly screen time (use z* = 1.96). Answer: (13.45, 16.55) Solution: Identify the given values: sample mean = 15, sample standard deviation = 5, sample size n = 40, z* = 1.96. Calculate the standard error: standard error = s / sqrt(n) = 5 / sqrt(40).
   Full step-by-step solution

   Step 1: Identify the given values: sample mean = 15, sample standard deviation = 5, sample size n = 40, z* = 1.96. Step 2: Calculate the standard error: standard error = s / sqrt(n) = 5 / sqrt(40). Step 3: sqrt(40) is approximately 6.3249, so standard error = 5 / 6.3249 ≈ 0.7906. Step 4: Calculate the margin of error: margin of error = z* * standard error = 1.96 * 0.7906 ≈ 1.5496. Step 5: Find the lower bound: 15 - 1.5496 = 13.4504, round to 13.45. Step 6: Find the upper bound: 15 + 1.5496 = 16.5496, round to 16.55. Step 7: The 95% confidence interval is (13.45, 16.55). The answer is (13.45, 16.55).

3. Emma, a Grade 11 student, is investigating whether the average amount of time students at her high school spend on homework each night is different from the national average of 90 minutes. She surveys a random sample of 40 students and finds that the mean homework time for her sample is 85 minutes, with a standard deviation of 20 minutes. Calculate the test statistic (t-value) for a hypothesis test to determine if the school's average homework time differs from the national average. Round your answer to two decimal places. Answer: -1.58 Solution: Identify the given values. Sample mean (x-bar) = 85 minutes Population mean (mu) = 90 minutes (national average) Sample standard deviation (s) = 20 minutes Sample size (n) = 40 Write the formula for the one-sample t-test statistic.
   Full step-by-step solution

   Step 1: Identify the given values. Sample mean (x-bar) = 85 minutes Population mean (mu) = 90 minutes (national average) Sample standard deviation (s) = 20 minutes Sample size (n) = 40 Step 2: Write the formula for the one-sample t-test statistic. t = (x-bar - mu) / (s / sqrt(n)) Step 3: Calculate the standard error. Standard error = s / sqrt(n) = 20 / sqrt(40) sqrt(40) ≈ 6.3249 Standard error ≈ 20 / 6.3249 ≈ 3.1623 Step 4: Calculate the test statistic. t = (85 - 90) / 3.1623 t = (-5) / 3.1623 t ≈ -1.5811 Step 5: Round to two decimal places. t ≈ -1.58 The test statistic is approximately -1.58.

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. A quality control manager at a factory claims that the mean weight of their cereal boxes is 500 grams. A consumer group suspects the boxes are underfilled and takes a random sample of 40 boxes, finding a sample mean weight of 495 grams with a sample standard deviation of 12 grams. They conduct a hypothesis test with H₀: μ = 500 and H₁: μ < 500 at a significance level of α = 0.05. Calculate the test statistic (t-score) for this hypothesis test, rounding to two decimal places. Answer: -2.64 Solution: Identify the values needed for the t-statistic formula: - Sample mean (x̄) = 495 grams - Population mean under null hypothesis (μ₀) = 500 grams - Sample standard deviation (s) = 12 grams - Sample size (n) = 40 t = (x̄ - μ₀) / (s / sqrt(n)) Standard error = s / sqrt(n) = 12 / sqrt(40) = 12 /…
   Full step-by-step solution

   Step 1: Identify the values needed for the t-statistic formula: - Sample mean (x̄) = 495 grams - Population mean under null hypothesis (μ₀) = 500 grams - Sample standard deviation (s) = 12 grams - Sample size (n) = 40 Step 2: Recall the t-statistic formula: t = (x̄ - μ₀) / (s / sqrt(n)) Step 3: Calculate the standard error: Standard error = s / sqrt(n) = 12 / sqrt(40) = 12 / 6.3246 = 1.897 Step 4: Calculate the t-statistic: t = (495 - 500) / 1.897 = -5 / 1.897 = -2.635 Step 5: Round to two decimal places: t = -2.64 The test statistic is -2.64.

6. 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.