1) pH in Rain An environmentalist wanted to determine if the mean acidity of rai
ID: 3183604 • Letter: 1
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1) pH in Rain An environmentalist wanted to determine if the mean acidity of rainwater differed among Alaska. Florida, and Texas. He randomly selects five rain date records from each of the three states, with no specification of dates or locations within state, and obtains the data given below. Use this data to test whether the mean pH of rainwater differs between these three states. at level a 0.05. Alaska Florida Texas 5.39 5.18 5.89 4.90 4,52 5.57 5.14 5.12 5.15 Grand 4.80 4.89 5.45 15 ir, 5.068 4.916 5,504 5,163 si l 0.2295 0.2602 0.2662 s 0.3181 la) Complete the ANOVA table below. Write only your final calculations in the given table, and use the available space below the table to show all of your work. Source of P-value SS MS Variation 0,9316 Treatment 0.7651 1.6967 Total lb) State both your statistical and proper conclusions for the ANOVA F-test.Explanation / Answer
Alaska= X1=5.068
Florida= X2=4.916
Texas= X3=5.504
X=5.16, n=15, c=3
SSA=5(5.068-5.16)2+5(4.916-5.16)2+5(5.504-5.16)2=0.93
SSW= (5.11-5.068)2…..+ (5.45-5.504)2=0.76512
MSA=0.93/ (3-1) =0.465787
MSW=0.76512/ (15-3) =0.06376
FSTAT=0.465787/0.06376=7.31
FSTAT (7.31) is greater than FCRIT (3.885294) and hence reject the null hypothesis
xTX-xFL=5.504-4.916=0.588
xTX-xAK=5.504-5.068=0.436
xAK-xFL=5.068-4.916=0.152
Q value from the table c=3 and (n-c)=(15-3)=12 degrees of freedom is 3.77
Critical range is
Q*MSW/2(1/ni+1/nj)
3.77*0.06376/2*(1/5+1/5)
3.77*0.1129
0.4257
All except xAK-xFL are greater than the critical range. Therefore there is a significant difference between each pair of means at 5% level of significance except for xAK-xFL.
Anova: Single Factor SUMMARY Groups Count Sum Average Variance Alaska 5 25.34 5.068 0.05267 Florida 5 24.58 4.916 0.06773 Texas 5 27.52 5.504 0.07088 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 0.931573 2 0.465787 7.305312 0.008409 3.885294 Within Groups 0.76512 12 0.06376 Total 1.696693 14xTX-xFL=5.504-4.916=0.588
xTX-xAK=5.504-5.068=0.436
xAK-xFL=5.068-4.916=0.152
Q value from the table c=3 and (n-c)=(15-3)=12 degrees of freedom is 3.77
Critical range is
Q*MSW/2(1/ni+1/nj)
3.77*0.06376/2*(1/5+1/5)
3.77*0.1129
0.4257
All except xAK-xFL are greater than the critical range. Therefore there is a significant difference between each pair of means at 5% level of significance except for xAK-xFL.
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