Wilcoxon test statistic The following data are nitrogen content analyses perform

Question

Wilcoxon test statistic

The following data are nitrogen content analyses performed by two different analysts on six water samples. Use these data for Questions 1-4: Define d_i = (Analyst 1 - Analyst 2). Calculate d and S_d. Consider the test of hypothesis Ho:^= 0 Ha:^t 0 a = 0.05 Calculate the test statistic t'. Find the P-value and make a decision about Ho. Construct a 95% confidence interval for mu_d. The following data are blood pressure readings of 16 runners before and after the Pepsi Challenge 10K. Use these data for Questions 5-8: Define d_1 = (BP after - BP before). Calculate d and S_d. Consider the test of hypothesis Ho: mu_d = 4 Ha: mu_d > 4 a = 0.05 Calculate the test statistic t'. Find the P-value and make a decision about Ho. Construct a 95% confidence interval for mu_d.

Explanation / Answer

Answer1-4:

Answer1-4:

Analyst1 (X-M) Sq. Diff (X-M)^2 31.4 -8.4 70.3 37 -2.8 7.7 44 4.2 17.8 28.8 -11.0 120.6 59.9 20.1 404.7 37.6 -2.2 4.8 Mean (M) 39.8 SS 625.89 Analyst2 (X-M) Sq. Diff (X-M)^2 28.7 -9.7 93.1 37.5 -0.9 0.7 40.1 1.8 3.1 27 -11.4 128.8 58.1 19.8 390.1 38.7 0.4 0.1 Mean (M) 38.4 0.0 615.92 Difference Scores Calculations Treatment 1 N1: 6 df1 = N - 1 = 6 - 1 = 5 M1: 39.78 SS1: 625.89 s21 = SS1/(N - 1) = 625.89/(6-1) = 125.18 Treatment 2 N2: 6 df2 = N - 1 = 6 - 1 = 5 M2: 38.35 SS2: 615.92 s22 = SS2/(N - 1) = 615.92/(6-1) = 123.18 T-value Calculation s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((5/10) * 125.18) + ((5/10) * 123.18) = 124.18 s2M1 = s2p/N1 = 124.18/6 = 20.7 s2M2 = s2p/N2 = 124.18/6 = 20.7 t = (M1 - M2)/(s2M1 + s2M2) = 1.43/41.39 = 0.22 The t-value is 0.22278. The p-value is .828189. The result is not significant at p < .05. MSE = (S1^2 + S2^2)/2 = (125.18+123.18)/2 = 124.18 S(M1-M2) = sqrt (2MSE/n) = sqrt(2*124.18/6) = 6.433 Confidence Interval Lower Limit = 1-(0.22)*6.433 = -0.41526 Upper Limit = 1+(0.22)*6.433 = 2.41526

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