(a) c = 4, therefore df1 = c - 1 = 3
N = 15 * 4 = 60, therefore df2 = N - c = 60 - 4 = 56
SS within = SST - SSTR = 78.95 - 18.16 = 60.79
MSTR = SSTR / df1 = 18.16 / 3 = 6.0533
MS Within = SS within / df2 = 60.79 / 56 = 1.0855
F = MSTR / MS within = 6.0533 / 1.0855 = 5.577
The ANOVA table is below
Source | SS | DF | Mean Square | F | p - value |
Between | 18.16 | 3 | 6.0533 | 5.577 | 0.002 |
Within/Error | 60.79 | 56 | 1.0855 | ||
Total | 78.95 | 59 |
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(b) Since p value is < (0.10), we reject H0
Option 1: reject H0. We can conclude that some means differ.
a. Given the following information obtained from four normally distributed populations, construct an ANOVA table. (Round...
a. Given the following information obtained from three normally distributed populations, construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS" to 2 decimal places, "MS" to 4 decimal places, and "P' to 3 decimal places.) SSTR = 220.7; SSE = 2,252.2; c = 3; ni = n2 = n3 = 8 ANOVA Source of Variation SS df MS F p-value Between Groups 0.375 Within Groups 0.00 0 Total b. At the 1% significance level,...
CH13 Q2 a. Given the following information obtained from four normally distributed populations, construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS" to 2 decimal places, "MS" to 4 decimal places, and "F' to 3 decimal places.) ANOVA Source of Variation Between Groups Within Groups Total df MS p-value 0.018 0.00 0
A one-way analysis of varlance experlment produced the following ANOVA table. (You may find it useful to reference the g table). SUMMARY Count Groups Column 1 Column 2 olumn 3 Source of Variation Between Groups Within Groups Total Average 8.89 1.31 2.35 SS 8.65 df 15 17 MS 4.33 0.26 16.65 8.6882 12.48 a. Conduct an ANOVA test at the 1% significance level to determine if some population means differ. o Reject Ho, we can conclude that some population means...
CH13Q4 4 The following statistics are calculated by sampling from four normal populations whose variances are equal: (You may find it useful to reference the t table and the g table.) r1 = 137, n1 = 4; = 144, n2 = 4; X3 = 136, n3 = 4; 되 = 124, n4 4; MSE = 57.4 a. Use Fisher's LSD method to determine which population means differ at α=0.01. (Negative values should be indicated by a minus 10 points sign....
CH13 Q4 The following statistics are calculated by sampling from four normal populations whose variances are equal: (You may find it useful to reference the t table and the g table.) r1 = 137, n1 = 4; Tz = 144, n2 = 4; = 136, n3 = 4; T4 = 124, n4 = 4; MSE = 57.4 a. Use Fisher's LSD method to determine which population means differ at a- 0.01. (Negative values should be indicated by a minus sign....
An analysis of variance experiment produced a portion of the accompanying ANOVA table. (You may find it useful to reference the F table.) a. Specify the competing hypotheses in order to determine whether some differences exist between the population means. H0: μA = μB = μC = μD; HA: Not all population means are equal. H0: μA ≥ μB ≥ μC ≥ μD; HA: Not all population means are equal. H0: μA ≤ μB ≤ μC ≤ μD; HA: Not...
Consider the following data drawn independently from normally distributed populations: (You may find it useful to reference the appropriate table: z table or t table) X1 = 27.1 012 = 89.5 n1 = 25 X2 = 30.3 022 = 92.3 n2 = 31 a. Construct the 90% confidence interval for the difference between the population means. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 4 decimal places and final answers to 2...
The data shown to the right are from independent simple random samples from three populations. Use these data to complete parts (a) through (d). Sample 1 Sample 2 Sample 3 Click the icon to view a table of values of Fa Calculate SST, SSTR, and SSE using the computing formulas. SST = SSTR= SSE (Type an integer or a decimal. Do not round.) (Type an integer or a decimal. Do not round.) (Type an integer or a decimal. Do not...
A random sample of five observations from three normally distributed populations produced the following data: (You may find it useful to reference the F table.) Treatments A C 20 1 9 25 25 22 27 21 24 24 26 2.1 22 23 19 XR - 23 SR6.5 S 4.5 S 4.5 Click here for the Excel Data File f. At the 5% significance level, what is the conclusion to the test? Reject Ho since the p-value is less than significance...
The following statistics are calculated by sampling from four normal populations whose variances are equal: (You may find it useful to reference the t table and the gtable.) X1 163, ni = 5; 2 = 171, n2 = 5; J3 = 166, n3 = 5; X4 = 158, n4 = 5; MSE = 41.2 a. Use Fisher's LSD method to determine which population means differ at a = 0.05. (Negative values should be indicated by a minus sign. Round intermediate...