Statistical Inference

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Statistical Inference · Grade 11 · Worksheet 3

1. Mere is investigating whether the average daily screen time for teenagers in her city is different from the national average of 5.5 hours. She surveys a random sample of 64 teenagers and finds that their mean daily screen time is 6.2 hours with a standard deviation of 1.6 hours. Using a 95% confidence level, calculate the margin of error for her estimate and interpret what this confidence interval suggests about the population mean. Answer: Margin of error = 0.392 hours. The 95% confidence interval is (5.808, 6.592). Since the national average of 5.5 hours is below this interval, there is evidence that the true mean for her city is higher than the national average. Solution: Identify the given information. Sample size n = 64 Sample mean x-bar = 6.2 hours Sample standard deviation s = 1.6 hours Confidence level = 95% Determine the critical t-value.
   Full step-by-step solution

   Step 1: Identify the given information. Sample size n = 64 Sample mean x-bar = 6.2 hours Sample standard deviation s = 1.6 hours Confidence level = 95% Step 2: Determine the critical t-value. For a 95% confidence level and degrees of freedom df = n - 1 = 63, the critical t-value (from t-distribution table) is approximately 1.96 (since n is large, it is close to the z-value). Step 3: Calculate the standard error. Standard error = s / sqrt(n) = 1.6 / sqrt(64) = 1.6 / 8 = 0.2 hours Step 4: Calculate the margin of error. Margin of error = critical t-value * standard error = 1.96 * 0.2 = 0.392 hours Step 5: Construct the 95% confidence interval. Lower bound = x-bar - margin of error = 6.2 - 0.392 = 5.808 hours Upper bound = x-bar + margin of error = 6.2 + 0.392 = 6.592 hours Step 6: Interpretation. We are 95% confident that the true mean daily screen time for all teenagers in Mere's city lies between 5.808 and 6.592 hours. Since the national average of 5.5 hours is below this interval, there is statistical evidence that the average screen time in her city is higher than the national average. The margin of error is 0.392 hours.

2. A university is studying whether a new study technique improves test scores. They randomly select 64 students and find the mean improvement is 8.2 points with a standard deviation of 12 points. Using a significance level of α = 0.05, test the hypothesis H₀: μ = 0 versus H₁: μ > 0. What is the calculated test statistic value? Round your answer to two decimal places. Answer: 5.47 Solution: Sample size n = 64 Sample mean x̄ = 8.2 Sample standard deviation s = 12 Null hypothesis mean μ₀ = 0 t = (x̄ - μ₀) / (s/√n) s/√n = 12/√64 = 12/8 = 1.5 t = (8.2 - 0) / 1.5 = 8.2 / 1.5 = 5.4666...
   Full step-by-step solution

   Step 1: Identify the given values Sample size n = 64 Sample mean x̄ = 8.2 Sample standard deviation s = 12 Null hypothesis mean μ₀ = 0 Step 2: Use the one-sample t-test formula t = (x̄ - μ₀) / (s/√n) Step 3: Calculate the standard error s/√n = 12/√64 = 12/8 = 1.5 Step 4: Calculate the test statistic t = (8.2 - 0) / 1.5 = 8.2 / 1.5 = 5.4666... Step 5: Round to two decimal places t = 5.47 The calculated test statistic is 5.47.

3. P(Z > 1.28) = ? Answer: 0.1003 Solution: We need to find P(Z > 1.28), which is the area under the standard normal curve to the right of z = 1.28. Standard normal tables typically give P(Z < z), the area to the left of a z-value.
   Full step-by-step solution

   Step 1: We need to find P(Z > 1.28), which is the area under the standard normal curve to the right of z = 1.28. Step 2: Standard normal tables typically give P(Z < z), the area to the left of a z-value. Step 3: Using a standard normal table, we find P(Z < 1.28) = 0.8997. Step 4: Since the total area under the curve is 1, P(Z > 1.28) = 1 - P(Z < 1.28) = 1 - 0.8997 = 0.1003. Step 5: The answer is 0.1003.

4. H₀: μ = 100, H₁: μ ≠ 100, sample mean = 104.5, σ = 15, n = 64, α = 0.05; z = ? Answer: 2.4 Solution: Write the formula for the z-test statistic: z = (sample mean - hypothesized mean) / (σ / √n) Substitute the given values: z = (104.5 - 100) / (15 / √64) Calculate the denominator: √64 = 8, so 15 / 8 = 1.875 Calculate the numerator: 104.5 - 100 = 4.5 Divide the numerator by the denominator: 4.5 /…
   Full step-by-step solution

   Step 1: Write the formula for the z-test statistic: z = (sample mean - hypothesized mean) / (σ / √n) Step 2: Substitute the given values: z = (104.5 - 100) / (15 / √64) Step 3: Calculate the denominator: √64 = 8, so 15 / 8 = 1.875 Step 4: Calculate the numerator: 104.5 - 100 = 4.5 Step 5: Divide the numerator by the denominator: 4.5 / 1.875 = 2.4 The z-test statistic is 2.4.

5. Mere is a Grade 11 student investigating whether a new online tutoring platform improves mathematics test scores at her school. She randomly selects 49 students who used the platform for one month. Their mean test score is 74 with a standard deviation of 14. The historical mean score for all students at the school is 70. Based on this sample, calculate the margin of error for a 95% confidence interval for the true population mean score. Use the critical value z* = 1.96 and round your answer to one decimal place. Answer: 3.9 Solution: Identify the given values. Sample mean = 74, sample standard deviation s = 14, sample size n = 49, critical value z* = 1.96. The margin of error for a confidence interval is calculated as E = z* * (s / sqrt(n)).
   Full step-by-step solution

   Step 1: Identify the given values. Sample mean = 74, sample standard deviation s = 14, sample size n = 49, critical value z* = 1.96. Step 2: The margin of error for a confidence interval is calculated as E = z* * (s / sqrt(n)). Step 3: Calculate the standard error: s / sqrt(n) = 14 / sqrt(49) = 14 / 7 = 2. Step 4: Multiply the critical value by the standard error: E = 1.96 * 2 = 3.92. Step 5: Round to one decimal place: 3.9. The answer is 3.9.

6. Tane is investigating whether the average daily screen time for Grade 11 students in his city exceeds the recommended maximum of 2 hours. He surveys a random sample of 49 students and finds a sample mean of 2.4 hours with a sample standard deviation of 0.7 hours. Construct a 95% confidence interval for the population mean daily screen time. Then, based on your interval, determine whether there is evidence that the true mean exceeds 2 hours. Use z* = 1.96 for 95% confidence. Answer: Yes, the entire interval is above 2 hours, so there is evidence that the true mean exceeds 2 hours. Solution: Identify the given values. Sample mean x̄ = 2.4 hours, sample standard deviation s = 0.7 hours, sample size n = 49, confidence level 95% (z* = 1.96). Calculate the standard error.
   Full step-by-step solution

   Step 1: Identify the given values. Sample mean x̄ = 2.4 hours, sample standard deviation s = 0.7 hours, sample size n = 49, confidence level 95% (z* = 1.96). Step 2: Calculate the standard error. Standard error = s / sqrt(n) = 0.7 / sqrt(49) = 0.7 / 7 = 0.1. Step 3: Calculate the margin of error. Margin of error = z* * standard error = 1.96 * 0.1 = 0.196. Step 4: Construct the confidence interval. Lower bound = x̄ - margin of error = 2.4 - 0.196 = 2.204. Upper bound = x̄ + margin of error = 2.4 + 0.196 = 2.596. Step 5: Interpret the interval. We are 95% confident that the true mean daily screen time for all Grade 11 students in the city is between 2.204 hours and 2.596 hours. Step 6: Compare to the recommended 2 hours. The entire confidence interval (2.204, 2.596) is above 2 hours. Therefore, there is evidence that the true mean exceeds 2 hours. The answer is: Yes, the entire interval is above 2 hours, so there is evidence that the true mean exceeds 2 hours.

7. H₀: μ = 100, H₁: μ ≠ 100, z = -2.45, α = 0.02; Reject H₀? Answer: B. yes Solution: This is a two-tailed test because H₁ uses ≠. For α = 0.02, the significance level is split between both tails: α/2 = 0.01 in each tail. Find the critical z-values for a two-tailed test with α = 0.02.
   Full step-by-step solution

   Step 1: This is a two-tailed test because H₁ uses ≠. Step 2: For α = 0.02, the significance level is split between both tails: α/2 = 0.01 in each tail. Step 3: Find the critical z-values for a two-tailed test with α = 0.02. The critical values are z = ±2.33 (from standard normal table for 0.01 in each tail). Step 4: Compare the test statistic z = -2.45 to the critical values. Since -2.45 < -2.33, the test statistic falls in the rejection region. Step 5: Therefore, we reject the null hypothesis H₀. The answer is Yes.