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1) pH in Rain An environmentalist wanted to determine if the mean acidity of rai

ID: 3183604 • Letter: 1

Question

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 14

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.