Calculating Confidence Intervals: Steps and Examples for Students

LECTURE 6 Q The quantities needed for computation of the 95% CI (confidence interval) for true average preferred height are σ=2.0, n=31, and X=80.0. The resulting interval is? Given: * σ = Population standard deviation = 2.0 * n = Sample size = 31 * X = Sample mean = 80.0 * To compute a 95% confidence interval, we need the z-score corresponding to 97.5 percentile, which is z = 1.96 Now the formula for a confidence interval is: X ± z*(σ/√n) Plugging in the values: X = 80.0 z = 1.96 σ = 2.0 n = 31 80.0 ± 1.96 * (2.0 / √31) 80.0 ± 1.96 * (2.0 / 5.57) 80.0 ± 0.7 The resulting 95% confidence interval is: (79.3, 80.7) Therefore, the 95% confidence interval for the true average preferred height is (79.3, 80.7). Q1 (a) A team of efficiency experts intends to use the mean of a random sample of size n = 150 to estimate the average mechanical aptitude of assembly-line workers in a large industry (as measured by a certain standardized test). If, based on experience, the efficiency experts can assume that σ = 6.2 for such data, what can they assert with probability 0.99 x = 20? (b) If a random sample of size n = 20 from a normal population with the variance σ2 = 225 has the mean x ̄ = 64.3, construct a 95% confidence interval for the population mean μ. (a) Given the following information:

σ (population standard deviation) = 6.2 n (sample size) = 150 Confidence level is 0.99 To calculate the margin of error (x), we use the z-score corresponding to the 99% confidence level. Since it is a two-tailed test with α = 0.01, the z score is 2.58. The margin of error formula is: x = z * (σ/√n) Plugging in the values: x = 2.58 * (6.2/√150) x = 2.58 * (6.2/12.25) x = 2.58 * 0.51 x ≈ 1.3 Therefore, with 99% confidence, the efficiency experts can assert that the sample mean will be within ±1.3 units of the true population mean. (b) Given: n = 20 x̄ = 64.3 σ2 = 225 σ = √225 = 15 95% confidence level → z* = 1.96 Margin of error E = z* * (σ/√n) = 1.96 * (15/√20) = 1.96 * (15/4.47) ≈ 6.57 95% Confidence Interval = (x̄ - E, x̄ + E) = (64.3 - 6.57, 64.3 + 6.57) = (57.73, 70.87) Therefore, the 95% confidence interval for the population mean μ is (57.73, 70.87). Q2

A CI is desired for the true average stray-load loss Î¼ (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that strayload loss is normally distributed with stv = 3.0. (a) Compute 95% CI for μ when n=25,¯x=58.3 (b) Compute 95% CI for μ when n=100,¯x=58.3 (c) Compute 99% CI for μ when n=100,¯x=58.3 (d) Compute 82% CI for μ when n=100,¯x=58.3 Given information: σ (standard deviation) = 3.0 μ (true average stray-load loss) (a) n = 25, x̄ = 58.3, 95% CI z-score = 1.96 Margin of error (E) = 1.96 * (3.0/√25) = 1.176 95% CI = (x̄ - E, x̄ + E) = (57.124, 59.476) (b) n = 100, x̄ = 58.3, 95% CI z-score is still 1.96 E = 1.96 * (3.0/√100) = 0.588 95% CI = (57.712, 58.888) (c) n = 100, x̄ = 58.3, 99% CI z-score = 2.58 for 99% CI E = 2.58 * (3.0/√100) = 0.776 99% CI = (57.524, 59.076) (d) n = 100, x̄ = 58.3, 82% CI z-score = 1.15 for 82% CI E = 1.15 * (3.0/√100) = 0.331 82% CI = (57.969, 58.631) Q3 Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation .75. a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85. b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average porosity of 4.56 (a) Given:

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Standard deviation (σ) = 0.75 Sample size (n) = 20 Sample mean (x̄ ) = 4.85 Confidence level = 95% To compute 95% CI: * z-score for 95% CI is 1.96 (from z-table) * Margin of error (E) E = z * (σ/√n) E = 1.96 * (0.75/√20) E = 1.96 * (0.75/4.47) = 0.32 * 95% CI = (x̄ - E, x̄ + E) = (4.85 - 0.32, 4.85 + 0.32) = (4.53, 5.17) (b) Given: n = 16 x̄ = 4.56 σ = 0.75 Confidence level = 98% * z-score for 98% CI is 2.33 * Margin of error E = 2.33 * (0.75/√16) E = 2.33 * (0.75/4) E = 0.433 * 98% CI = (x̄ - E, x̄ + E) = (4.56 - 0.433, 4.56 + 0.433) = (4.127, 4.993) Therefore, the 98% CI for true average porosity is (4.127, 4.993) LECTURE 6-II Q1 A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of 0.81 seconds and a sample standard deviation of 0.34 seconds. ("Lighting Strikes to an Airplane in a Thunderstorm," Journal of Aircraft, 1984). Calculate a 99% (two-

sided) confidence interval for the true average echo duration μ, and interpret the resulting interval. Given: Sample size (n) = 110 Sample mean (x̄ ) = 0.81 seconds Sample standard deviation (s) = 0.34 seconds 99% confidence level (two-sided) Steps: 1) The z-score corresponding to 99% confidence level is z = 2.58 (from z-table) 2) The margin of error (E) is: E = z * (s / √n) = 2.58 * (0.34 / √110) = 2.58 * (0.34/10.49) = 0.086 seconds 3) The 99% confidence interval is: (x̄ - E, x̄ + E) = (0.81 - 0.086, 0.81 + 0.086) = (0.724, 0.896) seconds Interpretation: We are 99% confident that the true average radar echo duration for lightning flashes in this region lies between 0.724 seconds and 0.896 seconds. So there is a 99% probability that the 99% confidence interval of (0.724, 0.896) seconds captures the true unknown population average echo duration. Q2 (i). Determine the values of the following quantities: a. t .1,15 b. t .05,15 c. t .05,25 d. t .05,40 e. t .005,40 (ii). Determine the t critical value that will capture the desired t curve area in each of the following cases: a. CI = .95, df = 10 b. CI = .95, df = 15

c. CI= .99, df = 15 d. CI = .99, n = 5 e. CI = .98, df = 24 f. CI= .99, n = 38 (i) a) t.1,15 = 1.341 (Obtain from t-distribution table with df = 15, alpha/2 = 0.05 which gives t value for 90% CI) b) t.05,15 = 1.753 (Obtain from t-distribution table with df = 15, alpha/2 = 0.025 which gives t value for 95% CI) c) t.05,25 = 1.708 (Obtain from t-distribution table with df = 25, alpha/2 = 0.025 which gives t value for 95% CI) d) t.05,40 = 1.684 (Obtain from t-distribution table with df = 40, alpha/2 = 0.025 which gives t value for 95% CI) e) t.005,40 = 2.423 (Obtain from t-distribution table with df = 40, alpha/2 = 0.0025 which gives t value for 99% CI) (ii) a. CI = 0.95, df = 10: The t critical value for a 95% confidence interval with 10 degrees of freedom is approximately 2.228. b. CI = 0.95, df = 15: The t critical value for a 95% confidence interval with 15 degrees of freedom is approximately 2.131. c. CI = 0.99, df = 15: The t critical value for a 99% confidence interval with 15 degrees of freedom is approximately 2.947. d. CI = 0.99, n = 5: ??? e. CI = 0.98, df = 24: The t critical value for a 98% confidence interval with 24 degrees of freedom is approximately 2.492. f. CI = 0.99, n = 38: The z-score for a 99% confidence interval is approximately 2.576. ??? Q3 A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of 8.48 MPa and a sample standard deviation of 0.79 MPa. (a) Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress?

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Okay, let me re-work this with the updated information: (a) Given information: n (sample size) = 14 Sample mean (x̄ ) = 8.48 MPa Sample standard deviation (s) = 0.79 MPa Assuming the proportional limit stress follows a normal distribution in the population. Steps: 1) Degrees of freedom = n - 1 = 14 - 1 = 13 2) For 95% lower confidence bound, from t-table with df = 13, t = 1.771 3) Margin of error, E = t * (s/√n) = 1.771 * (0.79/√14) = 1.771 * (0.79/3.74) = 0.389 MPa 4) Lower 95% confidence bound = x̄ - E = 8.48 - 0.389 = 8.091 MPa Interpretation: We are 95% confident that the true average proportional limit stress in the population is at least 8.091 MPa. This is based on the assumption that the population data follows a normal distribution. In summary, the lower 95% confidence bound for the true mean proportional limit stress is 8.091 MPa, assuming normality. Q4 An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 418, 421, 421, 422, 425, 427, 431, 434, 437, 439, 446, 447, 448, 453, 454, 463, 465. Calculate a two-sided 95% confidence interval for true average degree of polymerization Okay, let's go through this step-by-step: Given: Sample size (n) = 17

Sample degree of polymerization values are provided 95% confidence interval (two-sided) Steps: 1) Calculate sample mean (x̄ ): Sum of all values = 7,543 x̄ = Sum/n = 7,543/17 = 443.47 2) Calculate sample standard deviation (s): * Find deviation of each value from the mean * Square each deviation * Find the sum of squared deviations = 172.47 * Calculate s: s = √(Sum of squared deviations/n-1) = √(172.47/16) = 3.95 3) Calculate margin of error (E): * Degrees of freedom = n-1 = 16 * t-value from t-distribution table = 2.120 (for 95% CI) * E = t * (s/√n) = 2.120 * (3.95/√17) = 2.120 * (3.95/4.12) = 2.120 * 0.959 = 2.03 4) Calculate 95% confidence interval: (x̄ - E, x̄ + E) = (443.47 - 2.03, 443.47 + 2.03) = (441.44, 445.50) Therefore, the 95% two-sided confidence interval for the true average degree of polymerization is (441.44, 445.50). Q5 The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi sqrt(in), given in increasing order) 69.5, 71.9, 72.6, 73.1, 73.3, 73.8, 75.8, 76.1, 76.2 76.2, 77.0, 79.7 79.9, 80.1, 82.2, 83.7 Calculate a 99% CI for the standard deviation of the feature-toughness distribution. Is this interval valid whatever the nature of the distribution? Explain

Given: Sample size (n) = 16 Fracture toughness values are provided Steps: 1) Calculate the sample standard deviation (s): * Find the deviations and sum of squared deviations * s = √(sum of squared deviations/(n-1)) * s = 2.901 2) For 99% CI, the t statistic = 3.355 (from t-table with df = 15) 3) The margin of error E = t * (s/√n) = 3.355 * (2.901/√16) = 3.355 * (2.901/4) = 2.326 4) 99% Confidence Interval for true standard deviation σ: (s - E, s + E) = (2.901 - 2.326, 2.901 + 2.326) = (0.575, 5.227) Regarding validity: This interval is only valid if we assume the underlying population is normally distributed. The t- interval relies on normality of the data. If normality does not hold, for example for highly skewed distributions, then this interval may not have 99% coverage for σ. We would have to use alternative methods that do not assume normality. So in summary, the 99% CI is (0.575, 5.227), but it is strictly valid only if normality holds. Q6 For each of 18 preserved cores from oil-wet carbonate reservoirs, the amount of residual gas saturation after a solvent injection was measured at water flood-out. Observations, in percentage of pore volume, were: 23.5, 31.5, 34.0, 46.7, 45.6, 32.5, 41.4, 37.2, 42.5, 46.9, 51.5, 36.4, 44.5, 35.7, 33.5, 39.3, 22.0, 51.2 Is it plausible that the sample was selected from a normal population distribution? Calculate a 98% CI for the true average amount of residual gas saturation.

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1) Check for normality assumption: Looking at the data, there does not seem to be any obvious skewness. The sample size is decently large as well (n=18). So it is plausible that the sample comes from an underlying normal distribution. However, to formally test normality, we would have to perform a normality test like Shapiro-Wilk. But based on visual inspection, normality seems like a reasonable assumption. 2) Calculate 98% CI: Sample size (n) = 18 Sample mean (x̄ ) = 39.4% Sample standard deviation (s) = 8.95 Degrees of freedom = n-1 = 17 t-value (for 98% CI) = 2.11 Margin of error E = t * (s/√n) = 2.11 * (8.95/√18) = 2.11 * (8.95/4.24) = 4.56 98% CI = (x̄ - E, x̄ + E) = (39.4 - 4.56, 39.4 + 4.56) = (34.84, 43.96) In summary, assuming normality, the 98% confidence interval for true average residual gas saturation is (34.84%, 43.96%). Q7 A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°F. A sample of n=9 systems, when tested, yields a sample average activation temperature of 131.08°F. If the distribution of activation temperature is normal with standard deviation 1.5°F, you would like to know whether the test result contradict the manufacturer’s claim at a significance level α = 0.01 Null hypothesis H0: True average activation temperature = 130°F (Manufacturer's claim) Alternate hypothesis H1: True average activation temperature ≠ 130°F Sample size (n) = 9 Sample mean (x̄ ) = 131.08°F Population standard deviation (σ) = 1.5°F (given)

Test statistic: z = (x̄ - μ) / (σ/√n) Here μ is the value under null hypothesis which is 130°F Plugging in values: z = (131.08 - 130) / (1.5 / √9) z = 1.08 / 0.5 z = 2.16 Using significance level α = 0.01, the two-tailed critical value from z table is ±2.575 Since z = 2.16 lies within -2.575 and +2.575, we FAIL TO REJECT the null hypothesis. Conclusion: At 1% significance level, there is not enough evidence to contradict the manufacturer's claim that the true average activation temperature is 130°F. The sample mean obtained is reasonably close to the claimed value. Q8 The melting point of each of 16 samples of a certain brand of vegetable oil was determined, resulting in xbar =94.32. Assume that the distribution of the melting point is normal with = 1.2. 𝜎 a) Test whether the true melting point of vegetable oil differs from 95 using a significance level of = 0.1 𝛼 Null hypothesis (H0): The true mean melting point = 95°F Alternate hypothesis (H1): The true mean melting point ≠ 95°F Given: Sample mean (x̄ ) = 94.32°F Sample size (n) = 16 Standard deviation (σ) = 1.2°F (population standard deviation) Significance level (α) = 0.1 Test statistic: Z = (x̄ - μ) / (σ/√n) Here μ is the hypothesized value under null hypothesis which is 95°F Calculate the test statistic: Z = (94.32 - 95) / (1.2/√16) Z = -0.68/0.3

Z = -2.267 Using α = 0.1, the two tailed critical value from standard normal table is ±1.645. Since the calculated Z value = -2.267 falls in the rejection region (less than -1.645), we REJECT the null hypothesis at the 10% significance level. Conclusion: There is sufficient statistical evidence to indicate the true mean melting point differs from 95°F at the 10% significance level. Q9 The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. The article “Nutrient Intakes and Dietary Patterns of Older Americans: A National Study” (J. Gerontology, 1992: M145–150) reports the following summary data on intake for a sample of males age 65–74 years: n=115, _x_ˉ=11.3, and s=6.43. Does this data indicate that average daily zinc intake in the population of all males age 65–74 falls below the recommended allowance? Hypotheses: H0: True average daily zinc intake ≥ 15 mg/day H1: True average daily zinc intake < 15 mg/day Given information: Sample mean (x̄ ) = 11.3 mg/day Sample size (n) = 115 Sample standard deviation (s) = 6.43 Test statistic: Z = (x̄ - μ) / (s / √n) (Here μ is 15 mg/day from the null hypothesis) Calculate the Z statistic: Z = (11.3 - 15)/ (6.43/√115) Z = -3.7/0.6 Z = -6.167 Using α = 0.05, critical value for one tailed test is -1.645 (from Z table). Since calculated Z = -6.167 < critical value = -1.645, we reject the null hypothesis. Conclusion:

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There is enough evidence that the true average daily zinc intake for ages 65-74 years is less than 15 mg/day. So the sample data indicates intake below recommended allowance. Q10 A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCi/L of radon: The resulting readings were as follows: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6. 107.7, 103.3, 92.4 Does this data suggest that the population mean reading under these conditions differs from 100. State and test the appropriate hypotheses using alpha=0.05 Okay, let's go through this step-by-step: Hypotheses: H0: True mean reading for the population = 100 pCi/L H1: True mean reading ≠ 100 pCi/L Sample data: n = 12 Sample readings are given (Calculate sample mean x̄ and standard deviation s) Test statistic: Z-score = (x̄ - μ) / (s / √n) Calculating statistics from data: x̄ = 98.375 pCi/L s = 6.10 Calculate Z-score: Z = (98.375 - 100) / (6.1 / √12) Z = -0.921 Using α = 0.05, critical Z value is ±1.96 Since calculated Z score lies within -1.96 to +1.96, we FAIL TO REJECT the null hypothesis. Conclusion: At 5% significance level, there is not enough evidence to suggest the true mean differs from 100 pCi/L. The sample data does not indicate the population mean statistically differs from 100. Q11 Q12 Q13

Q14 Q15 LECTURE 7 Q16 Q17 Q18 Q19 Q20 Q21 Q22 In a study of copper deficiency in cattle, the copper values (ug Cu/100mL blood) were determined both for cattle grazing in an area known to have well-defined molybdenum anomalies (metal values in excess of the normal range of regional variation) and for cattle grazing in a nonanomalous area, resulting in s1 = 21.5(m = 48) for the anomalous condition and s2 = 19.45 (n = 45) for the nonanomalous condition. Test for the equality versus inequality of population variances at significance level .10 by using the P-value approach. Okay, let's go through this step-by-step: Hypotheses: H0: σ12 = σ22 (Variances are equal) H1: σ12 ≠ σ22 (Variances are unequal) Sample 1 (anomalous): n1 = 48 s1 = 21.5 Sample 2 (nonanomalous): n2 = 45 s2 = 19.45 Test Statistic: F = s1/s2 (Take larger variance in numerator) Calculate F statistic: F = (21.5)^2/(19.45)^2 = 1.226

Degrees of freedom: df1 = n1 - 1 = 48 - 1 = 47 df2 = n2 - 1 = 45 - 1 = 44 Using significance level (α) = 0.10 The critical value from F distribution tables, with df1 = 47, df2 = 44 is 1.59 (or 1.64?) Ho σ1 = σ2 will be rejected in favor of H1 f>=F( .025,47,44): ~.56 or f<= F(.975,47,44) ~ 1.8. As f = 1.22, Ho is not rejected. The data does not suggest a difference in the two variances. Conclusion: At 10% significance level, there is not enough evidence to suggest the population variances are unequal. The data indicates the variances are statistically equal. Q23 Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of toxaphene exposure on animals, groups of rats were given toxaphene in their diet. The article “Reproduction Study of Toxaphene in the Rat” (J. of Environ. Sci. Health, 1988: 101-126) reports weight gains (in grams) for rats given a low dose (4 ppm) and for control rats whose diet did not include the insecticide. The sample standard deviation for 23 female control rats was 32 g and for 20 female low-dose rats was 54 g. Does this data suggest that there is more variability in low-dose weight gains than in control weight gains? Assuming normality, carry out a test of hypotheses at significance level .05 Okay, let's go through this step-by-step: Hypotheses: H0: σ12 = σ22 (Equal variances in the two groups) H1: σ12 > σ22 (More variability in low-dose group) Control group: n1 = 23 s1 = 32 g Low-dose group: n2 = 20 s2 = 54 g Test statistic: F = s22/s12 (Larger variance in numerator since we are testing if s22 > s12)

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Calculate test statistic: F = (54)2/(32)2 = 2.84 Degrees of freedom: df1 = n2 - 1 = 20 - 1 = 19 df2 = n1 - 1 = 23 - 1 = 22 Using α = 0.05, critical F value (from F table) = 2.07 Since calculated F = 2.84 > Critical F = 2.07, We reject the null hypothesis. Conclusion: At 5% significance level, there is evidence to suggest there is more variability in weight gains for the low-dose group compared to the control group, assuming normality. LECTURE 8

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