Decision Probability

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Decision Probability · Grade 11 · Worksheet 2

1. A medical researcher is testing whether a new physical therapy protocol improves mobility scores for patients recovering from knee surgery. The null hypothesis states there is no improvement (H₀: μ = 0), while the alternative hypothesis claims there is improvement (H₁: μ > 0). After conducting a one-sample t-test with 36 patients, the researcher obtains a test statistic of t = 2.42. Using a significance level of α = 0.01 and knowing the critical value for a one-tailed test with 35 degrees of freedom is approximately 2.438, what decision should the researcher make about the therapy's effectiveness? Answer: D. Fail to reject H₀ and conclude insufficient evidence Solution: Identify the test statistic (t = 2.42) and critical value (t_critical = 2.438) Since this is a one-tailed test with H₁: μ > 0, we reject H₀ only if t > t_critical Compare the values: 2.42 < 2.438 Since the test statistic does not exceed the critical value, we fail to reject the null hypothesis…
   Full step-by-step solution

   Step 1: Identify the test statistic (t = 2.42) and critical value (t_critical = 2.438) Step 2: Since this is a one-tailed test with H₁: μ > 0, we reject H₀ only if t > t_critical Step 3: Compare the values: 2.42 < 2.438 Step 4: Since the test statistic does not exceed the critical value, we fail to reject the null hypothesis Step 5: This means there is insufficient evidence at the α = 0.01 level to conclude the therapy improves mobility scores The correct answer is Fail to reject H₀ and conclude insufficient evidence.

2. A city is evaluating whether to implement a new traffic light system at a dangerous intersection. The current system has an average of 4.2 accidents per month. After installing the new system for 6 months, the city recorded the following monthly accident counts: 3, 2, 4, 1, 3, 2. The city wants to test if the new system significantly reduces accidents using a one-tailed t-test with α = 0.05. The null hypothesis is H₀: μ = 4.2 (no improvement), and the alternative is H₁: μ < 4.2. What is the calculated t-statistic, and should the city reject the null hypothesis? Answer: C. t = -3.87, Reject H₀ Solution: Sample data: 3, 2, 4, 1, 3, 2 Sum = 3 + 2 + 4 + 1 + 3 + 2 = 15 Sample mean (x̄) = 15/6 = 2.5 Deviations from mean: (3-2.5)=0.5, (2-2.5)=-0.5, (4-2.5)=1.5, (1-2.5)=-1.5, (3-2.5)=0.5, (2-2.5)=-0.5 Squared deviations: 0.25, 0.25, 2.25, 2.25, 0.25, 0.25 Sum of squared deviations =…
   Full step-by-step solution

   Step 1: Calculate the sample mean Sample data: 3, 2, 4, 1, 3, 2 Sum = 3 + 2 + 4 + 1 + 3 + 2 = 15 Sample mean (x̄) = 15/6 = 2.5 Step 2: Calculate the sample standard deviation Deviations from mean: (3-2.5)=0.5, (2-2.5)=-0.5, (4-2.5)=1.5, (1-2.5)=-1.5, (3-2.5)=0.5, (2-2.5)=-0.5 Squared deviations: 0.25, 0.25, 2.25, 2.25, 0.25, 0.25 Sum of squared deviations = 0.25+0.25+2.25+2.25+0.25+0.25 = 5.5 Sample variance = 5.5/(6-1) = 5.5/5 = 1.1 Sample standard deviation (s) = sqrt(1.1) = 1.049 Step 3: Calculate the t-statistic t = (x̄ - μ) / (s/√n) = (2.5 - 4.2) / (1.049/√6) = -1.7 / (1.049/2.449) = -1.7 / 0.428 = -3.87 Step 4: Determine the critical value and make decision Degrees of freedom = n - 1 = 6 - 1 = 5 For a one-tailed test with α = 0.05 and df = 5, the critical t-value is -2.015 Since -3.87 < -2.015, we reject the null hypothesis The correct answer is t = -3.87, Reject H₀.

3. A medical researcher is testing whether a new therapy reduces anxiety levels in patients. She conducts a hypothesis test with H₀: μ = 50 (no change in anxiety score) and H₁: μ < 50 (anxiety decreases). With a sample of 36 patients, she calculates a test statistic of t = -2.45. Using a significance level of α = 0.01 and knowing the critical value for a one-tailed test with 35 degrees of freedom is approximately -2.438, should she reject the null hypothesis? Answer: B. yes Solution: Identify the test statistic: t = -2.45 Identify the critical value: -2.438 Compare the test statistic to the critical value: Since this is a left-tailed test (H₁: μ < 50), we reject H₀ if t < critical value Check the comparison: -2.45 < -2.438 Since the test statistic is less than the critical…
   Full step-by-step solution

   Step 1: Identify the test statistic: t = -2.45 Step 2: Identify the critical value: -2.438 Step 3: Compare the test statistic to the critical value: Since this is a left-tailed test (H₁: μ < 50), we reject H₀ if t < critical value Step 4: Check the comparison: -2.45 < -2.438 Step 5: Since the test statistic is less than the critical value, we reject the null hypothesis Step 6: Conclusion: Yes, she should reject the null hypothesis

4. Matiu is a farmer deciding between two irrigation systems for his orchard. System A costs $2,400 to install and has a 70% chance of increasing his annual profit by $4,800 and a 30% chance of increasing it by only $1,200 due to variable rainfall. System B costs $1,800 to install and has a 55% chance of increasing his annual profit by $3,600 and a 45% chance of increasing it by $2,400. Based on the expected net profit (profit increase minus installation cost) for one year, which irrigation system should Matiu choose, and what is the expected net profit of that system? Answer: System B with expected net profit of $1,260 Solution: Calculate the expected profit increase for System A. Expected increase = (0.70 * 4800) + (0.30 * 1200) = 3360 + 360 = 3720. Expected net profit = 3720 - 2400 = 1320.
   Full step-by-step solution

   Step 1: Calculate the expected profit increase for System A. Expected increase = (0.70 * 4800) + (0.30 * 1200) = 3360 + 360 = 3720. Expected net profit = 3720 - 2400 = 1320. Step 2: Calculate the expected profit increase for System B. Expected increase = (0.55 * 3600) + (0.45 * 2400) = 1980 + 1080 = 3060. Expected net profit = 3060 - 1800 = 1260. Step 3: Compare expected net profits: System A gives $1,320, System B gives $1,260. Since $1,320 > $1,260, System A has a higher expected net profit. Step 4: The better decision is System A with an expected net profit of $1,320. Final answer: System A with expected net profit of $1,320.