Two different analytical methods were used to determine residual chlorine in sewage effluents. Both methods were used on the same samples, but each sample came from various locations with differing amounts of contact time with the effluent. Two methods were used to determine the concentration of Cl in mg/L, and the results are shown in the following table: Sample Method A Method B 0.36 1.35 0.39 2 0.84 3 1.76 2.56 3.92 3.35 4.69 4 5.35 8.33 10.70 5 6. 7.70 7 10.52 8 10.92 10.91 a. What type of t test should be used to compare the two methods and why? b. Do the two methods give different results? State and test the appropriate hypotheses. c. Does the conclusion depend on whether the 90%, 95%, or 99% confidence levels are used?
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- Penicillin is produced by the Penicillin fungus, which is grown in a broth whose sugar content must be carefully controlled. Several samples of broth were taken on three successive days, and the amount of dissolved sugars, in milligrams per milliliter, was measured on each sample. The results were as follows. Day 1 : 4.9 5.4 5.3 4.9 5.2 5.1 5.4 4.9 5.1 5.1 4.9 5.4 Day 2 : 5.5 5.2 5.1 5.0 5.3 5.4 5.3 5.2 5.4 5.3 5.4 5.1 Day 3 : 5.8 5.0 5.4 5.5 5.5 5.5 4.8 5.5 5.2 4.9 5.5 5.0 Construct an ANOVA table. Round your answers to four decimal places as needed. One-way ANOVA: Sugar Concentration Source DF SS MS F P Days Error Total Is there enough evidence to conclude that the mean sugar concentration…Two different analytical methods were used to determine residual chlorine in sewage effluents. Both methods were used on the same samples, but each sample came from various locations with differing amounts of contact time with the effluent. Two methods were used to determine the concentration of Cl in mg/L, and the results are shown in the following table: Sample Method A Method B 1 0.39 0.36 2 0.84 1.35 3 1.76 2.56 4 3.35 3.92 5 4.69 5.35 6 7.70 8.33 7 10.52 10.70 8 10.92 10.91 a) What type of t-test should be used to compare the two methods and why? b) Do the two methods give different results? State and test the appropriate hypotheses. c) Does the conclusion depend on whether the 90%, 95%, or 99% confidence levels are used?Two different analytical methods were used to determine residual chlorine in sewage effluents. Both methods were used on the same samples, but each sample came from various locations with differing amounts of contact time with the effluent. Two methods were used to determine the concentration of Cl in mg/L, and the results are shown in the following table: Answer the following questions as per the table: - 1.What type of t test should be used to compare the two methods and why? 2.Do the two methods give different results? State and test the appropriate hypotheses. 3.Does the conclusion depend on whether the 90%, 95%, or 99% confidence levels are used?
- Life-saving drug: Penicillin is produced by the Penicillin fungus, which is grown in a broth whose sugar content must be carefully controlled. Several samples of broth were taken on three successive days, and the amount of dissolved sugars, in milligrams per milliliter, was measured on each sample. The results were as follows.The owner of a new car conducts a series of six gas mileage tests and obtains the following results, expressed in miles per gallon: 3., 22.7, 21.4, 20.6, and 21.4. 20.9. Find the mode for these data.A sample is counted for 10 minutes and results in 530 gross counts. A 30-minute background count results in 1,500 counts. The detector efficiency is 0.10 cpm/dpm. What is the critical level (net count rate) that corresponds to a risk of making a Type I error, where α is 0.05? a. 3.0 cpm b. 4.2 cpm c. 50 cpm d. 53 cpm
- Periodically, the county Water Department tests the drinking water of homeowners for contaminants such as lead and copper. The lead and copper levels in water specimens collected in 1998 for a sample of 10 residents of a subdevelopement of the county are shown below. lead (μμg/L) copper (mg/L) 4.4 0.643 2.4 0.57 1.5 0.46 2.6 0.895 5.9 0.2 3.4 0.54 3.8 0.245 1.6 0.583 5.7 0.769 1.7 0.215 (a) Construct a 9999% confidence interval for the mean lead level in water specimans of the subdevelopment. ≤μ≤Analysis of several plant-food preparations for potassium ion yielded the following data:Periodically, the county Water Department tests the drinking water of homeowners for contaminants such as lead and copper. The lead and copper levels in water specimens collected in 1998 for a sample of 10 residents of a subdevelopement of the county are shown below. lead (μμg/L) copper (mg/L) 4.44.4 0.6430.643 2.42.4 0.570.57 1.51.5 0.460.46 2.62.6 0.8950.895 5.95.9 0.20.2 3.43.4 0.540.54 3.83.8 0.2450.245 1.61.6 0.5830.583 5.75.7 0.7690.769 1.71.7 0.2150.215 (a) Construct a 9999% confidence interval for the mean lead level in water specimans of the subdevelopment. ≤μ≤≤μ≤ (b) Construct a 9999% confidence interval for the mean copper level in water specimans of the subdevelopment. ≤μ≤≤μ≤
- The vapor pressure of 1-chlorotetradecane at several temperatures are tabulated below. T (ºC) P∗ mm Hg 98.5 1 131.8 5 148.2 10 166.2 20 199.8 60 215.5 100 What error is there in using two-point linear interpolation to find the value of vapor pressure at 185ºC?Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800.Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.