Yield of potato in kilogram per plot with different fertilizer treatments from RCBD experiment with four replications.
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- you randomly select 15 apple tree locations and 15 citrus tree locations from different forests. Each location is 1000 m² in size. Now you randomly create 10 zones within each forest (300 total) and measure plant biomass in each for three years (900 being the final tally) determine the population of interest, the dependent variable, the independent variable, and if the independent variable is being observed or controlled/manipulated, and number of replicates pseudoreplicates.It is known that the average yield per plant is 180 g and the variance is 250 g for a type of strawberry. 10 of this plant seed was grown in soils cultivated with a special method and average yield was observed to be 200 gr. Soil cultivation Test whether the method raises the yield average. (a = 0.05)Q4
- A 2x2 experiment looks at how obesity and hunger affect the amount of food people eat. The results are shown below. n=10 participants per condition. Additional info if needed: N=40, G=120, ∑?2=640∑x2=640 please conduct a twoway ANOVA and draw conclusions about the main effects and interaction of obesity and hunger on food intake. Obesity (Factor B) B1 (Obese) B2 (normal weight) Hunger A1 (not hungry) M=4 SS=50 M=1 SS=30 Factor A A2 (hungry) M=5 SS=60 M=2 SS=40The Paleo diet allows only for foods that humans typically consumed over the last 2.5 million years, excluding those agriculture-type foods that arose during the last 10,000 years or so. Researchers randomly divided 500 volunteers into two equal-sized groups. One group spent 6 months on the Paleo diet. The other group received a pamphlet about controlling portion sizes. A randomized treatment assignment was performed, and at the beginning of the study, the average difference in weights between the two groups was about 0 pounds. After the study, the Paleo group had lost an average of 7.1 pounds with a standard deviation of 22.75 pounds while the control group had lost an average of 4.5 pounds with a standard deviation of 11.5 pounds. After failing to reject the null hypothesis support your answer with a p-value (a) P-value = _______A two-way ANOVA experiment with no interaction is conducted. Factor A has four levels (columns) and Factor B has eight levels (rows). Assume normality in the underlying populations. The results include the following sum of squares terms: SST=367.5 SSA = 207.0 SSE = 52.4 a. Construct an ANOVA table. (Round "SS" and "MS" to 2 decimal places and "F" to 3 decimal places.) ANOVA Source of Variation Rows Columns Error Total SS X Answer is complete but not entirely correct. 207.00 X 108.10 X 52.40 367.50 df 7 3 21 31 MS 29.57 X 36.03 X 2.50 F 11.850 X 14.412 X p- value 0.001 0.000
- Determine your independent and dependent variables, which type of ANOVA would you choose, and how much replication would be needed. Farmer Joe has eight different plots of land. He applies two different nitrogen levels randomly to each of the 8 plots, low and high. In addition, four different strawberry varieties are randomized within the plots. He is measuring mass of fruit for his outcome. I would like to see if there is a difference in birth weight among cat breeds. I will measure 5 different cat breeds.Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on arganisms, an expert claimed that the amount of gamma radiation needed to sterilize a colany of tardigrades no longer has a mean of 1200 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy.) A study was conducted on a sample of 21 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean af 1225 Gy, with a sample standard deviation of 65 Gy. Assume that the population of amounts of gamma radiation needed to sterilize a colony of tardigrades is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.10 level af significance, to support the claim that H, the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is nat equal to 1200 Gy. (a) State the null hypothesis and the…Data originally from DASL http://lib.stat.cmu.edu/DASL/Datafiles/CancerSurvival.html (Links to an external site.). Some bronchial cancer patients had the following survival times (in days) after receiving a certain treatment: 20, 37, 63, 64, 72, 81, 138, 151, 155, 166, 166, 223, 245, 246, 450, 459. (The second instance of “166” is not a mistake.) Do these numbers suggest that the average survival time, for all bronchial cancer patients who receive that treatment, is significantly different from 200 days? CANCER SURVIVAL Report an appropriate one-sample hypothesis test and use a 10% significance level. For Step 4 of the six-step process that was taught in this class, choose the BEST option to complete the following sentence: We assume that the sample of subjects is a SRS and we assume that the population of observations has no outliers ... and is Normally distributed. and is without a doubt bell-shaped. and is Uniformly distributed. (nothing needs to go here because the sample…
- A two-way ANOVA experiment with interaction was conducted. Factor A had three levels (columns), factor B had five levels (rows), and six observations were obtained for each combination. Assume normality in the underlying populations. The results include the following sum of squares terms: SST=1542 SSA= 1042 SSB = 358 SSAB = 39 a. Construct an ANOVA table. (Round "MS" to 4 decimal places and "F" to 3 decimal places.) ANOVA Source of Variation Rows Columns Interaction Error Total SS 0 df Answer is not complete. 0 MS F p- value 0.000 0.000 0.002Data originally from DASL http://lib.stat.cmu.edu/DASL/Datafiles/CancerSurvival.html (Links to an external site.). Some bronchial cancer patients had the following survival times (in days) after receiving a certain treatment: 20, 37, 63, 64, 72, 81, 138, 151, 155, 166, 166, 223, 245, 246, 450, 459. (The second instance of “166” is not a mistake.) Do these numbers suggest that the average survival time, for all bronchial cancer patients who receive that treatment, is significantly different from 200 days? CANCER SURVIVAL Report an appropriate one-sample hypothesis test and use a 10% significance level. For Step 5 of the six-step process that was taught in this class, which would be the appropriate procedure and why? Choose the correct option. Group of answer choices We would do a t-procedure because we have the standard deviation. We would do a t-procedure because the sample size is large. We would do a t-procedure because we do not have the population standard deviation and…The basic principles of experimental designs are randomization, replication and error or local control. These principles make a valid test of significance possible. Discuss briefly what makes them indispensable requisite to make field experiments valid?