Question

Below is the output from an ANOVA test Note that the degrees of freedom between groups is 2,

treatment Residuals Df Sum Sq Mean Sq F value Pr(>F) 2 639.48 319.74 3.33 0.0461 393740.43 95.91 Spooled = 9.793 on df = 39

The degrees of freedom within is given to be 39, the F statistic is given to be 3.33. and the p-value of 0.0461. (a) How many groups are there in this test? (b) What was the sample size? (c) What is the conclusion of the test at the 5% significance level?

An independent random sample is selected from an approxi- mately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and T test statistic. Also determine if the null hypothesis would be rejected at \alpha α = 0.05.

(a) H_A: \mu>\mu_0 H A : μ > μ 0 , n = 14, T = 1.19

(b) H_A:\mu<\mu_0, H A : μ < μ 0 , n =15, T = -3.04

(c) H_A:\mu\ne\mu_0, H A : μ ≠ μ 0 , n = 17, T = 0.77

Answer 1

Number o groupS = 2+ = 3 Sample size b 2+39 O As the p-value is les than the significance level Co-05) We veject the nul hypo thenis and conclude that three aroup means Sigmikeanely 4- \3 @ T = 19, dof right-tailed test O 12766 p-value Do not As P-value is greater than d=0 05;reject the nulk hypothesis = \5-\ = {4- T = -3.04, dof left tailed test Pvalue O 004 4 As p-Value is \ess than dA =0-05; Reject the null hpothais C T- 077 91=1-ヒ」 = fop tuo- tailed test D-value= O.45 252 As p-Yalue is greater than dzo05; Do not reject tho null upothanis