Question: The iris data described in Example 6 are given in the stem-and-leaf diagrams below. The MINITAB output for the analysis of the iris data is
-1.png)
The MINITAB output for the analysis of the iris data is given below.
-2.png)
Individual 95% CIs For Mean Based on Pooled StDev
-3.png)
(a) Identify the SSE and its degrees of freedom. Also locate s.
(b) Check the calculation of F from the given sums of squares and d.f.
(c) Is there one population with highest mean or are two or more alike? Use multiple-r confidence intervals with α= .05.
LEAF UNIT .20 VERSICOLOR a 0 22333 2 4445555s 2 667777 2 9 0000003111 00000000111 ]000 000000000iiii 22222333 2222233 3 4444444445sssss 3 4 3 66677 3 889899 4 03 4 2 2223 3 44 3 6 3 88 3 44 3 6 3 88 ne-vay ANOVA Source DP Ira, 2 11.345 5.672 49.160.000 Error 147 16-962 0 135 Total 149 28.307 S- 0.3397 R-Sq 40.001 80 MD Level Mean stdev 50 50 50 3.4280 2.7700 2.9740 0.3791 0.3 0.3225 130 Pooled StDev 0.3397 C. 2-25 3.00 3.25 3.50
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a From the ERROR line in the table we know that SSE 16962 with df 147 Also under Poo... View full answer
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