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# Chi-Square, Correlation, and Regression (Statistics Project Sample)

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The test requires the calculations of ANOVA, Chi-Square, Correlation and Regression using SPSS. All the outputs are explained in word document. No sources were needed

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Lab 2: ANOVA, Chi-Square, Correlation, and Regression

First Last Name

Name of Institution

Lab 2: ANOVA, Chi-Square, Correlation, and Regression

Question 1 (a)

Categorical

<=30 (n %)

31-40 (n %)

41- 50 (n %)

>50 (n %)

Gender

50

62

51.4

44.7

Exp Fatigue

50

62

51.4

44.7

Exp malaise

50

62

51.4

44.7

Continuous

<=30 (mean & sd)

30-40(mean & sd)

4-50(mean & sd)

>50(mean & sd)

Alk phosphate

Mean = 112.5

Sd = 56.49322

Mean = 88.9032

Sd = 59.98216

Mean= 106.6667

Sd = 41.92374

Mean= 114.5291

Sd = 43.25137

Albumin

Mean = 3.9313

Sd = 0.50637

Mean = 3.971

Sd = 0.40585

Mean =3.6333

Sd = 0.78776

Mean = 3.6412

Sd = 0.55384

Protime

Mean = 66.5625

Sd = 22.65867

Mean = 70.0

Sd = 21.00741

Mean = 54.50

Sd = 20.51132

Mean = 55.4706

Sd = 22.17866

b) ALK Phosphate lab result

Albumin lab result

c) H0: there is no significant departure from normality for each of the groups.

H1: there is significant departure from normality for each of the groups

Results

ALK Phosphate lab result sig. (p â‰¤ 0.001) for all the age groups; .001, .000, .000, .000

Conclusion;

We reject the null hypothesis. (P â‰¤ 0.001). Each of the levels is not normally distributed therefore not meeting the assumption of normality.

Results

Albumin lab result; overall the sig. (p) is greater than Î±; .017, .000, .139, .280

Conclusion

We fail to reject the null hypothesis, (p>Î±), hence the groups are normally distributed, meeting the assumption of normality.

Results

Protime lab result sig (p>Î±) for all age groups .096, .007, .013, .028

Conclusion

We retain the null hypothesis. The groups are normally distributed, meeting the assumption.

d) H0: There is no difference among Alk Phosphate, Albumin and Protime

H1: There is a difference among Alk Phosphate, Albumin and Protime.

Results

The total mean of the three groups, Alk Phosphate, Albumin and Protime was 62.5632

(s.d =21.944). The average for Alk Phosphate was the highest 107.75 (s.d = 50.8475) Albumin had the least mean of 4.1067 (s.d = 0.66482) and Protime had 66.5625 (s.d = 22.65867). The significant results F (3,248) = 1.240, p =0.296, F (3,274) = 7.388, p = 0.0 and F (3, 170) = 6.149, p= 0.001 for Alk Phosphate, Albumin and Protime respectively. We can see that there is a significant difference in all the three groups as the value of F> P in all the cases.

ANOVA

Sum of Squares

df

Mean Square

F

Sig.

ALK Phosphate lab result

Between Groups

9803.508

3

3267.836

1.240

.296

Within Groups

653467.810

248

2634.951

Total

663271.317

251

Albumin lab result

Between Groups

8.767

3

2.922

7.388

.000

Within Groups

108.390

274

.396

Total

117.157

277

Protime lab result

Between Groups

8155.576

3

2718.525

6.149

.001

Within Groups

75155.228

170

442.090

Total

83310.805

173

Conclusion

Since the value of F > P in all the cases we reject null hypothesis as there is a significant mean different across the groups. A post hoc test has to be performed to determine the differences.

ii) Î±= 0.05

H0: there is no difference in the 3 groupâ€™s variance.

H1: there is difference in the 3 groupâ€™s variance.

Result ALK Phosphate lab result; sig. (p) =0.018.

Conclusion We reject the null hypothesis p<Î±, the assumption is not reasonable.

Result: Albumin lab result; sig. (p) =0.054,

Conclusion: we retain the null hypothesis p>Î±. The assumption has been met.

Result: Protime lab result; sig. (p) =0.921,

Conclusion: we retain the null hypothesis p>Î±. The assumption has been met.

Test of Homogeneity of Variances

Levene Statistic

df1

df2

Sig.

ALK Phosphate lab result

3.415

3

248

.018

Albumin lab result

2.579

3

274

.054

Protime lab result

.163

3

170

.921

iii) Turkey HSD= (m1-m2)/â€š (msw/n)

ALK Phosphate lab result; msw=2634.951, n=252, k=4, dfw=248

Q (0.05) =3.66, Q (0.01) =4.45. Judging by 5% confidence,

<=30&31-40 = (107.75-97.7143)/â€š (2634.951/252)=3.1036 not significantly different

<=30&41-50 = (107.75-105.2333)/â€š (2634.951/252)=0.7784 not significantly different

<=30&>50 = (107.75-114.1333)/â€š (2634.951/252)=-1.9744 not significantly different

31-40&41-50= (97.7143-105.2333)/â€š (2634.951/252)=-2.3257 not significantly different

31-40&>50= (97.7143-114.1333)/â€š (2634.951/252)=-5.0785 significantly different

41-50&>50= (105.2333-114.1333)/â€š (2634.951/252)=-2.7528 not significantly different

Albumin lab result; msw=.396, n=278, k=4, dfw=274

Q (0.05) =3.66, Q (0.01) =4.44

<=30&31-40 = (4.1067-3.8565)/ â€š (0.396/278)=6.6292 significantly different

<=30&41-50 = (4.1067-3.6200)/â€š (0.396/278)=12.8954 significantly different

<=30&>50 = (4.1067-3.6788)/â€š (0.396/278)=11.3375 significantly different

31-40&41-50= (3.8565-3.6200)/ â€š (0.396/278)=6.2662 significantly different

31-40&>50= (3.8565-3.6788)/ â€š (0.396/278)=4.7083 significantly different

41-50&>50= (3.6200-3.6788)/ â€š (0.396/278)=-1.5579 not significantly different

Protime lab result; msw=442.090, n=174, k=4, dfw=170

Q (0.05) =3.67, Q (0.01) =4.48. From 5% confidence interval

<=30&31-40= (66.5625-69.5938)/â€š (442.090/174)=-1.9017 not significantly different

<=30&41-50 = (66.5625-53.6000)/â€š (442.090/174)=8.1322 significantly different

<=30&>50 = (66.5625-56.7895)/â€š (442.090/174)=6.1322 significantly different

31-40&41-50= (69.5938-53.6000)/â€š (442.090/174)=10.0339 significantly different

31-40&>50= (69.5938-56.7895)/â€š (442.090/174)=8.0395 significantly different

41-50&>50= (53.6000-56.7895)/â€š (442.090/174)=-2.0010 not significantly different

Question 2

H0: Gender and Age when categorized are independent

H1: Gender and age when categorized are not independent

Results

Chi-Square Tests

Value

df

Asymp. Sig. (2-sided)

Pearson Chi-Square

10.120a

3

.018

Likelihood Ratio

10.415

3

.015

Linear-by-Linear Association

.161

1

.689

N of Valid Cases

310

a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 6.61.

X (1) = (10.120, 3), p= 0.018, thus since p>0.05 we reject the null hypothesis.

Conclusion

We conclude that there is no statistically significance evidence to proof that association between gender and age when categorized are independent.

Question 3

H0: Experiencing Malaise and age when categorized are independent

H1: Experiencing Malaise and age when categorized are not independent

Results

Chi-Square Tests

Value

df

Asymp. Sig. (2-sided)

Pearson Chi-Square

6.621a

3

.085

Likelihood Ratio

6.676

3

.083

Linear-by-Linear Association

2.774

1

.096

N of Valid Cases

308

a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 25.35.

The Pearson chi-Square X (6.621, 3) p = 0.085. Since p > 0.05 we accept the null hypothesis.

Conclusion

There is statistically significant evidence that experiencing Malaise and age when categorized are independent since p (0.085) > 0.05

Question 4

H0: Experiencing fatigue and age when categorized are independent

H1: Experiencing fatigue and age when categorized are not independent

Results

Chi-Square Tests

Value

df

Asymp. Sig. (2-sided)

Pearson Chi-Square

18.731a

3

.000

Likelihood Ratio

19.494

3

.000

Linear-by-Linear Association

15.371

1

.000

N of Valid Cases

308

a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 22.44.

The Pearson Chi-Square X (18.731, 3) p= 0.0. Since P (0.0) < 0.05, we reject the null hypothesis.

Conclusion

There is no statistically significance evidence to proof that experiencing fatigue and age when categorized are independent.

Question 5

H0: Having an abnormal result in the first trimester is related to having an abnormal result in the third trimester.

H1: Having an abnormal result in the first trimester is not related to having an abnormal result in the third trimester.

Chi-Square Tests

Value

df

Asymp. Sig. (2-sided)

Exact Sig. (2-sided)

Exact Sig. (1-sided)

Pearson Chi-Square

7.073a

1

.008

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