Statistics Assignment 11 - Solutions

Transcription

1 Statistics 44.3 Assignment 11 - Solutions 1. Samples were taken of individuals with each blood type to see if the average white blood cell count differed among types. Eleven individuals in each group were sampled. The results are given in the table below: Average White Blood Cell count by Blood Type Σx Σx a. Construct an ANOVA Table for testing the equality of average white blood cell counts among blood types. T x = , = , G = 31330, N = i i= 1 j= 1 i= 1 ni The ANOVA table Source S.S. d.f. M.S. F Between Within Total Page 1

2 b. State conclusions and plot appropriate graphs to illustrate the results. Conclusions: Since F = < F 0.05 =,84 (df 1 = 3, df = 40), we conclude that there is no significant difference in average white cell count among the four blood types (A, B, AB, O). Table: mean white cell count for blood types (A, B, AB, O) Figure: mean white cell count for blood types (A, B, AB, O) An alternative way of illustrating the results is through box- plots White Blood Cell Count COUNT TYPE Non-Outlier Max Non-Outlier Min Median; 75% 5% Outliers

3 . Researchers studied the association between birth mother s smoking habits and the birth weights of their babies. A sample of size n = 11 subjects was selected from each of the four groups. Group 1 nonsmokers Group smokers who smoked less that 1 pack per day Group 3 smokers who smoke more than 1 Pack but less than packs per day Group 4 smokers who smoke more than packs per day. The data is tabulated below: Table: Birth weights (in grams) of infants of mothers (n = 11) in four smoking groups 1 pack to nonsmokers < 1pack packs > packs T i, Σx Σx x i Use the above data to construct an ANOVA table to determine if there is a significant difference in the average birth weight amongst the four groups. Illustrate your findings graphically T x = , = , G = 15016, N = i i= 1 j= 1 i= 1 ni The ANOVA table Source S.S. d.f. M.S. F Between Within Total Conclusions: Since F = > F 0.05 =,84 (df 1 = 3, df = 40), we conclude that there is a significant difference in average birth weight among the four Smoking groups.

4 nonsmokers < 1pack 1 to packs > packs This graph indicates that average birth weight decreases as the level of smoking increases. 3. In the following study the investigator was interested in determining if the Presence of Heart Disease was related to Systolic Blood pressure. The study consisted of four groups of subjects with differing levels of Systolic Blood pressure (<17, , , 167+). The data is tabulated below: Coronary Systolic Blood pressure (mm Hg) Heart Disease < Total Present Absent Total Determine if there is a relationship between the Presence of Heart Disease and Systolic Blood pressure. Expected frequencies Coronary Systolic Blood pressure (mm Hg) Heart Disease < Total Present Absent Total

5 Standardized residuals r x = E E Coronary Systolic Blood pressure (mm Hg) Heart Disease < Present Absent ( x E ) r χ = = = i j i j E Since χ > χ 0.05 = for ( 3)( 1) = 3 df. The Null hypothesis of independence is rejected. Examining the standardized residuals, we see there is higher incidence of heart disease when BP is 167+ then one would expect if the two variables were independent. This is illustrated with the following graph. Presence of Heart Disease 18.00% 16.00% 14.00% 1.00% 10.00% % 8.00% 6.00% 4.00%.00% 0.00% < to to Systolic BP 4. A study of reading errors made by second grade pupils was carried out to help decide whether the use of different sorts of drills for pupils of different reading abilities was warranted. Errors were categorized as follows. DK: Did not know the word at all C: Substitution of a word of similar configuration (e.g. "bad" for "had") T: Substitution of a synonym suggested by the context. OS: Other substitution.

6 The students had been clustered into three relatively homogeneous reading groups on the basis of (1) their reading achievement scores at the end of first grade and () their verbal IQ's. Group A consisted of the least able readers, Group C consisted of the most able readers while Group B made up the middle group. Five children were in Group A, while nine children were in Group B and eleven children were in Group C. The number of errors of each type made by each child were added to obtain the group totals given below: Analyze this data. Group A T 5 C 10 OS 15 DK 53 Total 83 Group B Group C Total Expected frequencies T C OS DK Total Group A Group B Group C Total Standardized residuals r x = E E T C OS DK Group A Group B Group C ( x E ) r χ = = = i j i j E Since χ < χ 0.05 =.59 for ( 3)( ) = 6 df. The Null hypothesis of independence is accepted. This test is not whether one group makes more errors on the average than other groups. The chi-square test is testing if the proportion of times a student makes one type of error (T, C, OS or DK) differs amongst the three groups (A-least able, B- middle, C-most able). The test concludes that there is no significant difference.