Complete the following ANOVA table for an experiment that involved five treatments with sample sizes of 12, 12, 14, 15, 14: Degrees of Freedom Sum of Mean F-value p-value Squares Square Treatment ?? 788.23 ?? ?? ?? Residuals ?? ?? ?? Total: ?? 942.87 ?? --
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Complete the following ANOVA table for an experiment that involved five treatments with
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- Consider the following 3 independent sample statistics (from Exercise 11.2.1, pg 454): Sample 1: Mean=43, SD=3.74, n=4, Sample 2: Mean=44, SD=4, n=3, Sample 3: Mean=34, SD=3.92, n=4. Compute the SS(between) and the SS(within). Type the three digit numbers, first representing the SS(between) and second the SS(within), separated by comma with no spaces; e.g., 350,278 SHOW THE STEPS I NEED A NEW ANSWER PLEASEListed below are the lead concentrations (in ug/g) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States. Assume that a simple random sample has been selected. Use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 µg /g. 2.96 6.45 5.99 5.51 20.53 7.45 11.97 20.46 11.52 17.54 D Identify the null and alternative hypotheses. Ho: H1: (Type integers or decimals. Do not round.) Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. V the null hypothesis. There sufficient evidence at the 0.01 significance level to V the claim that the mean lead concentration for all Ayurveda medicines…The effect of graphite coating type (M, A, K, L) on light reading boxes will be investigated. Since these readings may change from day to day, the day effect was taken into the experiment as the block effect. Four coating types were tried in a random order each day. Obtained results are given below. According to this; Create the ANOVA table and interpret the results.
- In the US, 45.1% of all people have type O blood, 40.9% have type A blood, 10.6% have type B blood and 3.4% have type AB blood. A researcher wants to see if the distribution of blood type is different for millionaires. The table below shows the results of a random sample of 3564 millionaires. What can be concluded at the αα = 0.10 significance level? Complete the table by filling in the expected frequencies. Round to the nearest whole number: Outcome Frequency Expected Frequency O 1590 A 1477 B 388 AB 109 What is the correct statistical test to use? What are the null and alternative hypotheses?H0: The distribution of blood type for millionaires is the same as it is for Americans in general. Blood type and income are independent. Blood type and income are dependent. The distribution of blood type for millionaires is not the same as it is for Americans in general. H1: The distribution of blood type for millionaires is not the same as it is for…Consider this only partly filled in a partly labeled ANOVA source table Between: 40 4 10 7.35 Within: 30 22 To the best of your ability to judge, how many subjects do think were in this experiment?Acne is a common skin disease that affects most adolescents and can continue into adulthood. A study compared the effectiveness of three acne treatments and a placebo, all in gel form, applied twice daily for 12 weeks. The study's 517 teenage volunteers were randomly assigned to one of the four treatments. Success was assessed as clear or almost clear skin at the end of the 12 week period. The results of the study can be seen in the table below. Using the appropriate statistical test, determine if there is significant evidence that the four treatments perform differently. If so, how do they compare.
- A physician who specialized in weight control has three different plans he recommends. TTISIT As an experiment, he randomly selected 12 patients and then assigned 4 to each diet. After three weeks, the following weight losses, in pounds, were noted. Plan A Plan B Plan C 15 15 19 16 17 19 Grand Mean=17 14 19 21 15 17 17 Mean 2=17 Mean 3=19 Mean1=15 Use 0.05 significance level, to test if the plans are different. Question: Calculate the following, only: The SStotal, the SSE, the SST, dfT, dfE, VarT, VarE, and the test statistic F. Note: Write the details of your calculations for the Sum of Squares, only, and the final Result for the others. Type your answers on the Response Template. (without the formulas).Listed below are the lead concentrations (in µg/g) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States. Assume that a simple random sample has been selected. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 µg/g. 5.98 5.50 20.54 3.03 6.46 Identify the null and alternative hypotheses. Ho: H 14 H₁: μ 14 (Type integers or decimals. Do not round.) Identify the test statistic. = (Round to two decimal places as needed.) 7.45 12.01 20.47 11.48 17.53 D S Vi I. (1,0) MoreNine sample of each of two types of paint are scored as follows: Paint I.. 85,87,92,80, 84,86, 92,98, 77 Paint II.. 89,89,90, 84,80, 90,85, 92, 96 Analyse this with the Wilicoxon rank sum test.
- The following are the normalized levels of a critical protein in 12 samples of blood: 8.2 , 4.7 , 10.3 , 11.7 , 18.3 , 5.9 , 3.8 , 18.9 , 7.6 , 14.2 , 9.8 , 16.4 Compute the sample mean, the sample median, and the sample standard deviation.ShopEasy Inc. is a specialty e-tailer that operates 87 catalogue websites on the Internet. Larson Brown, Sales Director, feels that the style (colour scheme, graphics, fonts, etc.) of a website may affect its sales. He chooses three levels of design style (neon, old world and sophisticated) and randomly assigns six catalogue websites to each design style. Analysis of Larson's data yielded the following ANOVA table: Source of Variation Between Groups Within Groups Total SS 19.43 2.85 O 3.68 O 19.45 O 3.57 df 68102.33 2 34051.17 Using a = 0.05, the critical F value is MS 29177.67 15 1945.178 97280 17 FIf SSBetween = 225.31 and there are four groups in the study with 10 cases per group, MSBetween = :