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Writing Prompts Use your bivariate data sets from WH08, and use the computations for the regression lines as in RL08. For each data set, check the conditions required to perform Inference for linear...

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Writing Prompts

Use your bivariate data sets from WH08, and use the computations for the regression lines as in RL08.


For each data set, check the conditions required to perform Inference for linear regression.


Choose whichever one more convincingly satisfies these criteria for the following questions.

  • Conduct a hypothesis test for slope of your data set. Explain and interpret each step of your test.

  • Conduct a hypothesis test for correlation of your data set. Explain and interpret each step of your test.

  • Choose a level of confidence and create a confidence interval for the slope of your regression line.

Reflection

Answer the following two questions in 3-4 detailed sentences each.

  1. After attempting this assignment, what ideas do you feel like you understand deeply?

  2. After attempting this assignment, what ideas do you feel like you are still struggling to understand?

  3. Give at least one specific question you have about this material.

greynodessolutiondataset-1-pv30hmhr-czynmxwu.xlsx greynodessolutiondataset-2-321or2n3-zj02w3y1.xlsx
Answered Same Day Jun 13, 2021

Solution

Biswajit answered on Jun 15 2021
127 Votes
Assignment Solution
Conditions for inference in linear regression
1. The relationship is linea
2. Data points are independent
3. Residuals & hence dependent variable is normally distributed
4. Zero mean value of e
or
5. No autoco
elation of residuals
6. E
ors are homoscedastic
7. No co
elation between e
or & independent variable
8. No multicollinearity
Dataset 1 (Income vs Hours per week)
For this dataset, assumption of linear relation, independence of datapoints are established, zero mean of residuals, homoscedastic e
ors etc. are established.
Find attached below the Excel outputs.
    Observation
    Predicted Yearly Income ('000's) (Y)
    Residuals
    Standard Residuals
    1
    49.20227453
    -5.402274528
    -0.875978531
    2
    45.78071353
    -1.28071353
    -0.207667632
    3
    49.20227453
    -4.402274528
    -0.713828584
    4
    54.33461603
    -8.334616025
    -1.351457551
    5
    44.75424523
    -3.554245231
    -0.576320678
    6
    49.20227453
    -5.902274528
    -0.957053505
    7
    47.83365013
    -4.233650129
    -0.686486146
    8
    55.36108432
    -9.161084325
    -1.485469343
    9
    55.70324042
    -8.903240425
    -1.443659968
    10
    57.75617702
    -9.556177023
    -1.549533603
    11
    53.65030383
    -4.350303826
    -0.705401537
    12
    59.12480142
    -5.324801423
    -0.863416271
    13
    53.99245993
    -0.092459926
    -0.014992372
    14
    52.96599163
    1.234008374
    0.200094393
    15
    57.75617702
    -7.256177023
    -1.176588724
    16
    55.70324042
    -4.503240425
    -0.730200199
    17
    56.04539652
    -4.545396524
    -0.737035808
    18
    54.67677213
    -2.076772125
    -0.336748491
    19
    54.33461603
    -1.534616025
    -0.248837908
    20
    55.01892823
    -2.118928225
    -0.343584101
    21
    59.46695752
    -9.966957522
    -1.61614164
    22
    47.14933793
    2.650662071
    0.429804716
    23
    55.70324042
    -5.403240425
    -0.876135151
    24
    61.51989412
    -7.219894121
    -1.170705453
    25
    55.36108432
    -0.261084325
    -0.04233481
    26
    60.49342582
    -5.193425822
    -0.842113725
    27
    63.57283072
    -1.87283072
    -0.303679403
    28
    62.20420632
    0.095793679
    0.01553294
    29
    58.44048922
    4.959510777
    0.804184412
    30
    62.20420632
    1.495793679
    0.242542867
    31
    53.65030383
    1.849696174
    0.299928138
    32
    56.04539652
    -0.445396524
    -0.072221023
    33
    49.88658673
    5.813413272
    0.942644657
    34
    63.23067462
    -5.03067462
    -0.815723626
    35
    62.54636242
    -4.246362421
    -0.688547444
    36
    49.54443063
    8.855569372
    1.435930109
    37
    58.78264532
    0.417354677
    0.067674039
    38
    60.83558192
    -1.535581922
    -0.248994528
    39
    61.51989412
    -2.119894121
    -0.343740721
    40
    63.57283072
    -3.07283072
    -0.49825934
    41
    53.65030383
    3.049696174
    0.494508075
    42
    54.67677213
    3.123227875
    0.506431236
    43
    62.88851852
    0.911481479
    0.147796674
    44
    67.33654782
    -3.136547818
    -0.508591064
    45
    60.49342582
    -4.693425822
    -0.761038751
    46
    60.49342582
    -4.293425822
    -0.696178772
    47
    64.25714292
    0.04285708
    0.006949273
    48
    65.28361122
    -0.783611219
    -0.127062518
    49
    60.49342582
    4.206574178
    0.682095782
    50
    65.83106098
    0.268939021
    0.043608448
    51
    60.49342582
    11.80657418
    1.914435384
    52
    56.04539652
    17.15460348
    2.781618054
    53
    62.88851852
    11.31148148
    1.834156129
    54
    58.44048922
    10.05951078
    1.631149145
    55
    61.51989412
    8.180105879
    1.326403739
    56
    62.54636242
    8.653637579
    1.40318688
    57
    51.93952333
    14.36047667
    2.328550539
    58
    59.80911362
    6.690886378
    1.084926875
    59
    62.20420632
    4.495793679
    0.728992709
    60
    66.65223562
    8.147764382
    1.321159567
    61
    50.57089893
    11.42910107
    1.85322814
    62
    60.83558192
    -3.535581922
    -0.573294423
    63
    53.30814773
    1.991852274
    0.322978742
    64
    50.57089893
    5.529101072
    0.896543449
    65
    64.94145512
    -3.441455119
    -0.558031767
Sum of residuals is zero as shown in table.
    Regression Statistics
     
     
     
     
     
     
     
     
    Multiple R
    0.665956616
     
     
     
     
     
     
     
    R Square
    0.443498215
     
     
     
     
     
     
     
    Adjusted R Square
    0.434664853
     
     
     
     
     
     
     
    Standard E
o
    6.21588398
     
     
     
     
     
     
     
    Observations
    65
     
     
     
     
     
     
     
    ANOVA
     
     
     
     
     
     
     
     
     
    df
    SS
    MS
    F
    Significance F
     
     
     
    Regression
    1
    1939.865694
    1939.865694
    50.20718
    1.41668E-09
     
     
     
    Residual
    63
    2434.14446
    38.63721365
     
     
     
     
     
    Total
    64
    4374.010154
     
     
     
     
     
     
     
    Coefficients
    Standard E
o
    t Stat
    P-value
    Lower 95%
    Upper 95%
    Lower 95.0%
    Upper 95.0%
    Intercept
    36.88465494
    2.986349408
    12.35108485
    1.76E-18
    30.91691184
    42.85239803
    30.91691
    42.8524
    Hours Per Week (X)
    0.6843122
    0.096576475
    7.085702786
    1.42E-09
    0.491319514
    0.877304886
    0.49132
    0.877305
Hypothesis test for slope
1. Define the Hypothesis
H0 : Slope is equal to zero
Ha :Slope is not equal to zero or greater than zero
2. Assume level of significance (.05)
3. Decide test statistic (either t or p value)
4. Calculate the test statistic
T statistic =coefficient/standard e
or =0.68431/.09657 =7.085
P value =1.42E-09 (two sided in excel output) & for one sided, half of it.
5. Decision rule: If t statistic > t critical or p value is less than level of significance then reject null hypothesis
6. Conclusion: Here p value < .05, so we reject null hypothesis.
Also t critical is 1.95 at degree of freedom=63, t.inv(.975,63),so t calculated 7.085 > 1.95.Hence reject null hypothesis that slope is zero. In a way, slope is significant.
Hypothesis test for co
elation of dataset.
1. Define the Hypothesis
H0 : coefficient of co
elation is equal to zero
Ha :coefficient of co
elation is not equal to zero
2. Assume level of significance (.05)
3. Decide test statistic ( p value or t statistic)
4. Calculate the test statistic
T statistic =r *sqrt (n-2)/sqrt(1-r^2) = 0.6659 *sqrt (65-2)/sqrt(1-.6659^2) =7.0857
P value =1.416E-09
5. Decision rule :If p value is less than level of significance then reject null hypothesis
6. Conclusion :Here p value < .05,so we reject null hypothesis.
Also t critical is 1.998 at degree of freedom=63, t.inv(.975,63),so t calculated 7.085 > 1.95.Hence reject null hypothesis that slope is zero.In a way,slope is significant.
    Co
elation coefficient 
    Hours Per Week (X)
    Yearly Income ('000's) (Y)
    Hours Per Week (X)
    1
     
    Yearly Income ('000's) (Y)
    0.665956616
    1
Confidence interval
1. Level of confidence is 95%
2. Slope coefficient of Hours per week is 0.6843,standard e
or is 0.0965
3. So confidence interval is (.6843+/- t.inv (.025,63=degree of freedom)*.0965 i.e (0.4913,0.8773)
Dataset 2 (Price vs quantity)
For this dataset,assumption of linear relationship in parameter but approximate linear relationship of variables ,independence of datapoints are established,zero mean of residuals,homoscedastic e
ors etc are established.
Find attached below the Excel outputs.
    Regression Statistics
     
     
     
     
     
     
     
     
    Multiple R
    0.31349225
     
     
     
     
     
     
     
    R Square
    0.09827739
     
     
     
     
     
     
     
    Adjusted R Square
    0.08273045
     
     
     
     
     
     
     
    Standard E
o
    126.528626
     
     
     
     
     
     
     
    Observations
    60
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
    ANOVA
     
     
     
     
     
     
     
     
     
    df
    SS
    MS
    F
    Significance...
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