Exhibit 13-12 SSTR = 736 SSE = 900 nT = 18 Ho: M1 = μ₂=M3 Ha: At least one mean is different
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A: Bottom 0.43 0.266 0.567 0.531 0.707 0.716 0.651 0.589 0.469 0.723 Surface 0.415 0.238 0.39 0.41…
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Q: n = 64 X= 64.8 a=24 Ho: μ = 60 H₂:μ*60 Refer to Exhibit 9-31. The test statistic equals
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Given that,
SSTR=736
SSE=900
nT=18
Step by step
Solved in 3 steps
- Consider the following table: SS DF MS FL Among Treatments 4468.45 Error Total ? 10 706.87 15 Step 1 of 8: Calculate the sum of squares of experimental error. Please round your answer to two decimal places.(c) Assume that the resist ance of a cert ain type of resistors is normally distributed with a mean of 250 ohms and a st and ard deviat ion of 15 ohms. What percent age of the resistors of this type will have a resistance above 270 ohms?Refer to the following morbidity and mortality statistics for Community X for 2008. Total One-Year Population: 250,000 Population of Women 15-44: 75,000 Number of live births: 6,000 Total Deaths: 2,260 Number of infant deaths: 120 Deaths from heart disease: 222 Deaths from Cancer: 115 Deaths from Stroke: 90 Number of persons diagnosed with cancer 10,500 Calculate the following: Infant Mortality Rate _____________________________________ Prevalence of cancer for 2008 _____________________________ Crude Death Rate ______________________________________ Cause-specific mortality rate for stroke…
- 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 (g/L) 2.9 0.2 5.1 4.2 5.5 1.2 0.133 0.774 0.214 0.671 0.444 0.234 0.357 0.761 0.176 0.888 (a) Construct a 99% confidence interval for the mean lead level in water specimans of the subdevelopment. OSMOSO copper (mg/L) 0.3 1.3 4.9 1.7 (b) Construct a 99% confidence interval for the mean copper level in water specimans of the subdevelopment. OSHSO8.6 Cherry Trees: Timber yield is approximately equal to the volume of a tree, however, this value is difficult to measure without first cutting the tree down. Instead, other variables, such as height and diameter, may be used to predict a tree's volume and yield. Researchers wanting to understand the relationship between these variables for black cherry trees collected data from 31 such trees in the Allegheny National Forest, Pennsylvania. Height is measured in feet, diameter in inches (at 54 inches above ground), and volume in cubic feet. (Hand, 1994) Estimate Std. Error t value P(>|t|) (Intercept) -57.99 8.64 -6.71 0.00 height 0.34 0.13 2.61 0.01 diameter 4.71 0.26 17.82 0.00The measurements of suspended solid material concentration in water samples is illustrated in the figure below. Determine approximately the IQR value of the data.
- NonePls show complete solutionTrace metals in drinking water affect the flavor and an unusually high concentration can pose a health hazard. Ten pairs of data were taken measuring zinc concentration in bottom water and surface water. Bottom 0.430 0.266 0.567 0.531 0.707 0.716 0.651 0.589 0.469 0.723 Surface 0.415 0.238 0.390 0.410 0.605 0.609 0.632 0.523 0.411 0.612 Does the data suggest that the true average concentration in the bottom water is different than that of surface water? Note: You can assume that the conditions for inference are met.
- Listed below are pulse rates (beats per minute) from samples of adult males and females. Does there appear to be adifference? Find the coefficient of variation for each of the two samples; then compare the variation. A. The coefficient of variation for the male pulse rates iS (Type an integer or decimal rounded to one decimal place as needed.) B. The coefficient of variation for the female pulse rates is (Type an integer or decimal rounded to one decimal place as needed.)Consider the following table: SS DF MS ? 10 10 7 706.87 15 Among Treatments 4468.45 Error Total Step 3 of 8: Calculate the mean square among treatments. Please round your answer to two decimal places. LL FGiven the population: {0, 1, 2, 3, 4}, calculate the mean. Ομ3.0 Ομ2.5 I = 2.0 Ομ2.0