ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS
ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS
4th Edition
ISBN: 9781319509286
Author: BALDI
Publisher: MAC HIGHER
Question
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Chapter 11, Problem 11.51E
To determine

To use a software to create Normal quantile plots for each of the three groups of the forest plots and explain are the three distributions roughly Normal and what are the most prominent deviations from Normality that you see.

Expert Solution & Answer
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Answer to Problem 11.51E

Yes, the three groups are roughly Normal.

Explanation of Solution

In the question, the data is given that described of the effects of logging on counts in the Borneo rainforest as:

Thus, we will use the excel to construct the Normal quantile plot by using z-scores. And the excel function used is as:

  =NORMSINV(prob)

So, for group one, we have the following calculation as:

    Group 1RankRank proportionRank based z scoreGroup 1
    161=(AA81-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB81)16
    192=(AA82-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB82)19
    193=(AA83-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB83)19
    204=(AA84-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB84)20
    215=(AA85-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB85)21
    226=(AA86-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB86)22
    247=(AA87-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB87)24
    278=(AA88-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB88)27
    279=(AA89-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB89)27
    2810=(AA90-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB90)28
    2911=(AA91-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB91)29
    3312=(AA92-0.5)/COUNT($Z$81:$Z$92)=NORMSINV(AB92)33

The result is as:

    Group 1RankRank proportionRank based z scoreGroup 1
    1610.041667-1.7316616
    1920.125-1.1503519
    1930.208333-0.8122219
    2040.291667-0.5485220
    2150.375-0.3186421
    2260.458333-0.1046322
    2470.5416670.10463324
    2780.6250.31863927
    2790.7083330.54852227
    28100.7916670.81221828
    29110.8751.15034929
    33120.9583331.73166433

Thus, we will select the last two column and then go to the insert tab and select the chart option and then select the plot, so the normal quantile plot is as:

  ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS, Chapter 11, Problem 11.51E , additional homework tip  1

So, for the group two we have the calculation as:

    Group 2RankRank proportionRank based z scoreGroup 2
    21=(AG81-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH81)2
    92=(AG82-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH82)9
    123=(AG83-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH83)12
    124=(AG84-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH84)12
    145=(AG85-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH85)14
    146=(AG86-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH86)14
    157=(AG87-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH87)15
    178=(AG88-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH88)17
    179=(AG89-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH89)17
    1810=(AG90-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH90)18
    1911=(AG91-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH91)19
    2012=(AG92-0.5)/COUNT($AF$81:$AF$92)=NORMSINV(AH92)20

Thus, the result is as:

    Group 2RankRank proportionRank based z scoreGroup 2
    210.0416667-1.7316643962
    920.125-1.150349389
    1230.2083333-0.81221780112
    1240.2916667-0.54852228312
    1450.375-0.31863936414
    1460.4583333-0.10463345614
    1570.54166670.10463345615
    1780.6250.31863936417
    1790.70833330.54852228317
    18100.79166670.81221780118
    19110.8751.1503493819
    20120.95833331.73166439620

Thus, we will select the last two column and then go to the insert tab and select the chart option and then select the plot, so the normal quantile plot is as:

  ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS, Chapter 11, Problem 11.51E , additional homework tip  2

So, for the group three we have the calculation as:

    Group3RankRank proportionRank based z scoreGroup3
    41=(T81-0.5)/COUNT($S$81:$S$89)=NORMSINV(U81)4
    122=(T82-0.5)/COUNT($S$81:$S$89)=NORMSINV(U82)12
    123=(T83-0.5)/COUNT($S$81:$S$89)=NORMSINV(U83)12
    154=(T84-0.5)/COUNT($S$81:$S$89)=NORMSINV(U84)15
    185=(T85-0.5)/COUNT($S$81:$S$89)=NORMSINV(U85)18
    186=(T86-0.5)/COUNT($S$81:$S$89)=NORMSINV(U86)18
    197=(T87-0.5)/COUNT($S$81:$S$89)=NORMSINV(U87)19
    228=(T88-0.5)/COUNT($S$81:$S$89)=NORMSINV(U88)22
    229=(T89-0.5)/COUNT($S$81:$S$89)=NORMSINV(U89)22

The result is as:

    Group3RankRank proportionRank based z scoreGroup3
    410.055555556-1.5932188184
    1220.166666667-0.96742156612
    1230.277777778-0.58945579812
    1540.388888889-0.28221614715
    1850.5018
    1860.6111111110.28221614718
    1970.7222222220.58945579819
    2280.8333333330.96742156622
    2290.9444444441.59321881822

Thus, we will select the last two column and then go to the insert tab and select the chart option and then select the plot, so the normal quantile plot is as:

  ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS, Chapter 11, Problem 11.51E , additional homework tip  3

Thus, we can see that in the three groups the points are approximately in a straight line and increasing in nature. Thus, we can say that the three groups are approximately follows Normal distribution and they shows Normality in nature.

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