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When would a t-test be used instead of a z-test to assess the Ho? O when the is unknown, and the sample size is large.

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When would a t-test be used instead of a z-test to assess the Ho? O when the is unknown, and the sample size is large. O when the is known, and the sample size is large. when the is unknown, and the sample size is small. O when the is known, and the sample size is small.For the following graph below, choose the best answer to represent the significant tests. * LovelaoffacliorA 0 All nonsignificant. 0 Significant A main effect; B main effect and interaction nonsignificant. 0 Significant A and B main effects; no interaction. 0 A and B main effects nonsignificant; significant interaction. Imagine that you conducted an independent groups t-test with 10 participants in each group. For a one-tailed test, the tcv at a = 0.05 would be [Note that the tcv is a critical value] * O 1.729 O 2.101 O 1.734 O 2.093What is the type of hypothesis specified in the following statement? * Statement: Vitamin BB has no effect on the emotional on the emotional status of schizophrenic patients. 0 Alternative hypothesis. 0 Null hypothesis. 0 Experimental hypothesis. 0 Alternate hypothesis. H1: H1 # 12 is the hypothesis for a tailed test. O null; two O alternative; two O null; one O alternative, oneIn a randomized ANOVA, if there are four groups with 15 participants in each group, then the df for F-ratio is equal to * O 60 O 59 O 3, 56 O 3,57If you had rejected the H0 in the vitamin supplement experiment when in fact, the Vitamin supplement did not influence intelligence, you would have made a ' O Typelerror. O Typell error. 0 Type I\" error 0 None of the above. For an F-ratio with df = (3, 20), then the Fcv for a = 0.05 would be O 3.10 O 4.94 O 8.66 O 5.53Given the following values of t and n. Which of the following pairs is incorrect if the Ho be rejected using a two-tailed test and a = 0.05?* I. t = 2.086, n =20 Il. t = 1.990, n =121 Ill. t = 2.201, n =8 O l and II I and Ill O Il and Ill 1, ll and Ill Option 5Let tcv = 2.15 and tcal = -2.20. Based on these results, we 0 reject Ho 0 do not reject Ho O reject H1 0 do not reject H1

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