Question

I. Conduct a paired t-test for testing μD--μι-μ2: H0 : μD = 0 H1 : μD (two-sided) 0 vs. Group 1 Group 2 Pair 1 Pair 2 Pair 3 air 4 Pair 5 Pair 6 Pair 7 Pair 8 Pair 9 84 73 69 71 76 64 72 74 52 56 58 54 56 39 45 50 2. Provide a complete ANOVA table from the following information for one-way design of 3 groups df2) function in R when computing with 5( = r) within-group repetitions. You may use pf (t , df1, p-value Group 1 5 40 10 Group 2 5 50 10 Group 3 5 60 10 3. Notice for the data given in Problem 1 that the pair can be considered as a block. Using R, conduct the two-way design analysis (including multiple testing)

please solve at least 1, 2. It would be grateful if you solve 3

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

Group 1 Group 2 84 73 69 71 76 64 72 74 52 Pair 1 Pair 2 Pair 3 Pair 4 Pair 5 Pair 6 Pair 7 Pair 8 Pair 9 Sum Mean St.Dev 12 28.44444444 56 58 54 56 39 527 45 50 587 738 82 60 65.2222222258.55555556 6.6666667 13.22 13.51 15.89

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 9.6047

SE = s / sqrt(n)

S.E = 3.202

DF = n - 1 = 9 -1

D.F = 8

t = [ (x₁ - x₂) - D ] / SE

t = 2.08

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 8 degrees of freedom is more extreme than 2.08; that is, less than - 2.08 or greater than 2.08.

Thus, the P-value = 0.071

Interpret results. Since the P-value (0.071) is greater than the significance level (0.05), we failed to reject the null hypothesis.