Week 12 ‐ Discussion Questions STAT 400, Spring 2022, D. Unger Table of Contents Exercise 1 (Hypothesis Test for µ) ............................................... Error! Bookmark not defined. Exercise 2 (Hypothesis Test for µ) ............................................................................................................... 1 Exercise 3 (Hypothesis Test for µ) ............................................................................................................... 2 Exercise 4 (Hypothesis Test for p) ............................................................................................................... 2 Exercise 5 (Hypothesis Test for p) ............................................................................................................... 2 Exercise 1 (Confidence Interval for µ) The overall standard deviation of the diameters of a certain set of ball bearings is = 0.005 mm. The overall mean diameter of the ball bearings must be 4.300 mm. A sample of 81 ball bearings had a sample mean diameter of 4.299 mm. Is there a reason to believe that the actual overall mean diameter of the ball bearings is not 4.300 mm? (a) Perform the appropriate hypothesis test using a 10% level of significance. Solution Step 1: µ = the actual overall mean diameter of the ball bearings H0: µ = 4.300 vs. H1: µ ≠ 4.300 Step 2: 𝑋 = the sample mean diameter of the ball bearings; 𝑋 ~ Normal(µ, σ2/n) with σ known. We will use 𝑍/√. Step 3: Step 4:Step 5: There is significant evidence to suggest that the actual overall mean diameter of the ball bearings is not 4.300 mm. (b) Repeat the hypothesis test in part a, but provide evidence in the form of an appropriate confidence interval instead. Solution Since the 90% confidence interval of (4.29809, 4.29991) does not include the null hypothesis claim of 4.300, then we reject H0 at the α = 0.10 level of significance. (c) Repeat the hypothesis test in part a, but provide evidence in the form of the test statistic and critical/rejection region instead. Solution Since the test statistic value of z = –1.80 < –1.645, then it lies in the rejection region and we reject H0 at the α = 0.10 level of significance. (d) Repeat the hypothesis test in part a, but this time using a 5% level of significance Solution The data, test statistic, and p-value are unchanged. But now we have that 0.0718 > 0.05. Since p-value > α, then we fail to reject H0 at the α = 0.05 level of significance. We conclude that there is not significant evidence to suggest that the actual overall mean diameter of the ball bearings is not 4.300 mm. Exercise 2 (Confidence Interval for µ) A trucking firm believes that its mean weekly loss due to damaged shipments is at most $1800. Half a year (26 weeks) of operation shows a sample mean weekly loss of $1921.54 with a sample standard deviation of $249.39. Perform the appropriate hypothesis test using an α = 0.10 level of significance. Solution Step 1: µ = the overall mean weekly loss due to damaged shipments H0: µ ≤ 1800 vs. H1: µ > 1800 Step 2: 𝑋 = the sample mean weekly loss due to damaged shipments; 𝑋 ~ Normal(µ, σ2/n) with σ unknown. We will use 𝑇/√. Step 3: Step 4: Step 5: There is significant evidence to suggest that overall mean weekly loss due to damaged shipments is greater than $1800. Exercise 3 (Confidence Interval for µ) Metaltech Industries manufactures carbide drill tips used in drilling oil wells. The life of a carbide drill tip is measured by how many feet can be drilled before the tip wears out. Metaltech claims that under typical drilling conditions, the life of a carbide tip follows a normal distribution with mean of at least 32 feet. Suppose some customers disagree with Metaltech’s claims and argue that Metaltech is overstating the mean (i.e. the mean is actually less than 32). Metaltech agrees to examine arandom sample of 25 carbide tips to test its claim against the customers’ claim. If the Metaltech’s claim is rejected, Metaltech has agreed to give customers a price rebate on past purchases. Suppose Metaltech decided to use a 5% level of significance and the observed sample mean is 30.5 feet with the sample variance 16 feet2. Perform the appropriate hypothesis test. Solution Step 1: µ = the population mean lifetime of a drill tip measured in distance (feet) H0: µ ≥ 32 vs. H1: µ < 32 Step 2: 𝑋 = the sample mean lifetime of a drill tip measured in distance (feet); 𝑋 ~ Normal(µ, σ2/n) with σ unknown. We will use 𝑇/√. Step 3: Step 4: Step 5: There is significant evidence to suggest that overall mean lifetime of a drill tip is less than 32 feet. Exercise 4 (Hypothesis Test for p) In a random sample of 160 students from Faber College, 56 believe that resume inflation (i.e., misrepresenting yourself on your resume) is unethical. (a) Find the p-value of the test H0: p = 0.40 versus H1: p < 0.40. Solution Step 1: p = the proportion of all Faber College students who believe that resume inflation is unethical H0: p = 0.40 vs. H1: p < 0.40 Step 2: 𝑝 = the sample proportion; 𝑝 ~ Normal(p, p(1–p)/n). We will use 𝑍 –/. Step 3: 𝑝 0.35 𝑧𝑝 𝑝𝑝1– 𝑝/𝑛0.35 0.400.400.60/1600.050.03871.29 p-value = P[Z < –1.29] = 0.985 (b) Is there significant evidence to suggest that fewer than 2 out of every 5 students think resume inflation is unethical at the 0.10 level of significance? Solution Step 4: Since the p-value is less than α = 0.10, we reject the null hypothesis. Step 5: Yes, there is significant evidence to suggest that fewer than 2 out of every 5 students (p < 0.40) think resume inflation is unethical. Exercise 5 (Hypothesis Test for p) An economist states that 10% of Champaign-Urbana’s labor force is unemployed. A random sample of 400 people in the labor force is obtained, of whom 28 are unemployed. (a) Test whether Champaign-Urbana’s unemployment rate is as the economist claims or is different than the claim at a 1% level of significance using the critical region method for evidence. Solution Step 1: p = the proportion of all Champaign-Urbana residents who are unemployed H0: p = 0.10 vs. H1: p ≠ 0.10 Step 2: 𝑝 = the sample proportion; 𝑝 ~
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