Confidence intervals #5 - Questions and Answers

Question 1 Out of a 54 person sample from our fall Math-150 classes, 12.4% of the students are left- handed. Using a TI 84 graphing calculator: Construct an 80% confidence interval for the population proportion, p. Round to tenths of a percent and write a sentence answer. The U.S. has 13.1% of it's population that are left-handed. Did this fall in your confidence interval from part a? Be sure to show all calculator commans used. Answer Step 1 Given information: n = 54 p̂ = 0.124 x = 54 x 0.124 Confidence level = 80% Step 2 The step by step procedure to determine confidence interval for proportion: Press STAT and scroll right to the TESTS menu. Scroll down to A:1-PropZInt and press ENTER. Enter the values, x = 6.696, n = 54 and C-Level = 0.80. Select Calculate and press ENTER. NKS Step 3 The obtained 80% confidence interval for the population proportion, p is (0.0665, 0.1815). Yes the proportion of left handed in U.S i.e 13.1% falls in our confidence interval (0.0665, 0.1815). Question 2 Find t_(α/2), n = 15 for the 98% confidence interval for the mean. Answer Step 1 Provided information that, n = 15 Alpha (α) = 1 - 0.98 = 0.02 α/2 = 0.01 Step 2 t Critical value: t_0.01, n-1 = 2.1447 Question 3 To monitor complex chemical processes, chemical engineers will consider key process indicators, which may be just yield but most often depend on several quantities. Before trying to improve a process, 9 measurements were made on a key performance indicator. 123, 106, 114, 128, 113, 109, 120, 102, 111. Assume that the key performance indicator has a normal distribution. (a) Find the sample variance s² of the sample. (b) Find a 95% confidence interval for σ². (c) Find a 98% confidence interval for σ. Answer Step 1 (a) to find variance value add them up 123+106+114+128+113+109+120+102+111 = 1026 square your answer 1026 * 1026 = 1052676 ...and divide by the number of items. we have 9 items, 1052676/9 = 116964 set this number aside for a moment. take your set of original numbers from step 1, and square them individually this time 123² + 106² + 114² + 128² + 113² + 109² + 120² + 102² + 111² = 117520 subtract the amount in step 2 from the amount in step 3 117520-116964=556 subtract 1 from the number of items in your data set, 9-1=8 divide the number in step 4 by the number in step 5. this gives you the variance 556/8 = 69.5 to find standard deviation value take the square root of your answer from step 6. this gives you the standard deviation √69.5 = 8.3367 Step 2 (b) CONFIDENCE INTERVAL FOR VARIANCE ci = (n - 1)s²/χ²(right) < σ² < (n - 1)s²/χ²(left) where, s² = variance χ²right = (1-confidence level) / 2 χ²left = 1 - χ²right n = sample size since α = 0.05 χ²right = (1-confidence level)/2 = (1 - 0.95)/2 = 0.025 χ²left = 1 - χ²right = 1 - 0.025 = 0.975 the two critical values χ²left, χ²right at 8 df are 17.5345, 2.18 variance (s²) = 69.5 sample size (n) = 9 confidence interval = [8 * 69.5/17.5345 < σ² < 8*69.5/2.18] = [31.7089, 255.081] Step 3 (c) to calculate confidence interval for standard deviation ci = (n - 1)s²/χ²(right) < σ² < (n - 1)s²/χ²(left) where, s = standard deviation χ²right = (1- confidence level)/2 Question 4 The city wishes to estimate the average commute distance for all city employees. They collect a random sample of 97 employees and find a sample mean of x̅ = 7.12 ≈ 7.1 miles. They also assume a population standard deviation of σ = 1.5σ = 1.5 miles. Calculate the z interval to estimate, μ, the average commute distance for all city employees with 95% confidence. Round your answers to 2 decimal places. Find the Critical Value z*: Point Estimate: Margin of Error: Write the interval as (lower bound, upper bound) Answer Step 1 The random variable commute distance for all city employees follow normal distribution. The sample size is 97. The sample mean is 7.1. The population standard deviation is 1.5. We have to find the 95% confidence interval for the population mean. We have to use Z-table. Step 2 From Z-table, the 95% critical value is z* = 1.96. Point estimate: 7.1. ME = z* * σ/√n = 1.96 x 1.5/√97 = 0.30 Margin of error: 0.30. Step 3 CI = x̅ ± ME = 7.1 ± 0.30 = (7.1-0.30, 7.1 + 0.30) = (6.80, 7.40) The 95% confidence interval is (6.80, 7.40). Question 5 Use the confidence interval to find the margin of error and the sample mean. (0.142, 0.300) The margin of error is ?. The sample mean is ?. Answer a) Margin of error, E = (b-a)/2 = (0.300 - 0.142)/2 = 0.079 b) Sample mean, x = (a+b)/2 = (0.142 + 0.300)/2 = 0.221