Use the spinner below. 12 1 11 10 9. 4 8 7 5 6. P(5) = 3. 2.
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- Q3: oleasae answFind answer using ANOVA table and by handwriting it out Got the answer reject H0 at α = 0.05 beacuse F = 40.127 and Fcritical = 3.682. Is this right?When you open a bag of M&M's you expect to see 30% brown, 20% yellow, 20% red, 10% orange, 10% green and 10% blue. You look in your bag of 150 M&M;s and find the following results. Is there evidence to say your bag has differing results at α=.05α=.05? Brown 49 Yellow 40 Red 40 Orange 10 Green 5 Blue 6 The Pvalue is P = . (4decimals) There is to say the results vary.
- 2. Can you please help me understand theese grapsh by answering theese question. THank you in advance.3. What length ribbon will you need to stretch from the top of a 25-foot pole to a spot on the ground that is 10 feet from the bottom of the pole? Explain. Beckmann, Sybilla. Mathematics for Elementary Teachers with Activities, (p. 576). Pearson Education. Kindle Edition.question is attached inss below thanks for the hlep 2lph5hl2 hl25p hl5p h2 ph5 plhl5 hpl 5hp 5p h5 p h5 h5 p
- Express the null and alternative hypotheses in symbolic form for this claim: the mean weight of male dancers in a local modern dance company is less than 174 lbs. Note that the mean weight in the sample taken is 165 lbs.H0:μH1:μUse the following codes to enter the following symbols: ≥ enter >= ≤ enter <= ≠ enter !=am. 503.Might I ask where did u get the p value as 0.140951 in the table?
- he personality trait of "Conscientiousness" (someone who is organized, responsible, and can control their impulses) has μ = 120 and σ =9. Test whether ARC students (n = 9, M = 126) differ on Conscientiousness. α = .05. What is the correct result based on the data?A medical researcher says that less than 84% of adults in a certain country think that healthy children should be required to be vaccinated. In a random sample of 400 adults in that country, 81% think that healthy children should be required to be vaccinated. At α=0.01, is there enough evidence to support the researcher's claim? Complete parts (a) through (e) below.30. 35% of women with long hair have split ends. A hair stylist believes this proportion is smaller for her clients using her salon. A sample of 178 women with long hair using her salon, showed 33.96% have split ends. Is there enough evidence to support her claim at α = .03