# Calculation Chi Square Static

1. Table 7-52 displays summary statistics on the participants involved in the study described in Problem 1. Are any of the characteristics significantly different between groups? Justify briefly.
 New Compound Placebo p-value Mean age 47.2 46.1 0.7564 Men (%) 44 59 0.0215 Mean education 13.1 14.2 0.6898 Mean income 36,560 37,470 0.3546 Mean BMI 24.7 25.1 0.0851

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Solution

1. Table 7-52 displays summary statistics on the participants involved in the study described in Problem 1. Are any of the characteristics significantly different between groups? Justify briefly.
 New Compound Placebo p-value Mean age 47.2 46.1 0.7564 Men (%) 44 59 0.0215 Mean education 13.1 14.2 0.6898 Mean income 36,560 37,470 0.3546 Mean BMI 24.7 25.1 0.0851

The p-value corresponding to percentage of men is lower than that of 0.05 which means there are enough statistical evidence to reject the null hypothesis. hence it can be inferred that percentage of men is significantly different in two groups

4.

Null value for total cholesterol is 191. Sample size is 40 which reveals the mean cholesterol level is 175 and standard deviation of 19.5 . T –test would be conducted for testing following null and alternate hypothesis:

H0: Mean cholesterol level of the sample = 191

Ha: Mean cholesterol level of the sample ≠ 191

t static is

t static is = (175 – 191) / (19.5/ sqrt(40))

t static is = -5.189378724

Degree of freedom is 40-1 = 39

p value corresponding to t static is 0.0000 which is lower than 0.05 which means there are enough statistical evidence to reject the null hypothesis hence mean cholesterol level of the sample ≠ 191.

Since t-static is negative which means, hence there are sufficient statistical evidence of reduction of cholesterol level.

13.

Recent recommendation suggests 60 minutes of physical activity. Sample size is 50 which reveals the mean physical activity of 38 minutes and standard deviation of 19 minutes. T –test would be conducted for testing following null and alternate hypothesis:

H0: Mean physical activity of 60 minutes is statistically similar to that of mean physical activity of 38 minutes

Ha: Mean physical activity of 60 minutes is not statistically similar to that of mean physical activity of 38 minutes

t static is

t static is = (38 – 60) / (19/ sqrt(50))

t static is = -8.18755

Degree of freedom is 50-1 = 49

p value corresponding to t static is 0.0000 which is lower than 0.05 which means there are enough statistical evidence to reject the null hypothesis hence mean physical activity of 60 minutes is not statistically similar to that of mean physical activity of 38 minutes

20.

The mean lifetime for cardiac stents is 8.9 years. Sample size is 40 which reveals the mean lifetime of 9.7 years and standard deviation of 3.4 years. T –test would be conducted for testing following null and alternate hypothesis :

H0: Mean lifetime of new equipment is equal to that of generic cardiac stents

Ha: Mean lifetime of new equipment is different than that of generic cardiac stents

t static is

t static is = (9.7 – 8.9) / (3.4/ sqrt(40))

t static is = 1.488

Degree of freedom is 40-1 = 39

p value corresponding to t static is 0.072379172 which is higher than 0.05 which means there are not enough statistical evidence to reject the null hypothesis hence mean lifetime of new equipment is equal to that of generic cardiac stents.

1.

Chi –square test would be conducted to test the following hypothesis:

H0: There is no difference in the extent of wound healing by treatment

H1: There is difference in the extent of wound healing by treatment

Formulae for calculating Chi –square static is as follows

While formulae for calculating degree of freedom is

Degree of freedom = (number of rows- 1)- (number of columns-1)

As a first step observed frequencies are determined

 Observed Frequencies Percent Would Healing Treatment 0-25 26-50 51-75 76-100 Total New Compound 15 37 32 41 125 Placebo 36 45 34 10 125 Total 51 82 66 51 250

In second step, expected step frequencies are as follows:

 Expected Frequencies Percent Would Healing Treatment 0-25 26-50 51-75 76-100 Total New Compound 25.5 41 33 25.5 125 Placebo 25.5 41 33 25.5 125 Total 51 82 66 51 250

The expected frequencies are calculated as per the following formula

 (Observed – Expected)^2 /Expected New Compound 4.32 0.39 0.03 9.42 Placebo 4.32 0.39 0.03 9.42
 Data Level of Significance 0.05 Number of Rows 2 Number of Columns 4 Degrees of Freedom 3 Results Critical Value 7.8147 Chi-Square Test Statistic 28.33 p-Value 0.00

Since p –value of the test is 0 which is less than 0.05, hence there are sufficient statistical evidence to reject the null hypothesis. Therefore there is difference in the extent of wound healing by treatment

1. The data in Problem 1 is Table 7-51

Chi –square test would be conducted to test the following hypothesis:

Formulae for calculating Chi –square static is as follows

While formulae for calculating degree of freedom is

Degree of freedom = (number of rows- 1)- (number of columns-1)

The expected frequencies are calculated as per the following formula

 Observed Frequencies Percent Would Healing Treatment 0-25 26-50 51-75 76-100 Total New Compund 15 37 32 41 125 Placebo 36 45 34 10 125 Total 51 82 66 51 250 Observed Percentage 20.4% 32.8% 26.4% 20.4% Given distribution 30% 40% 20% 10% Expected values 75 100 50 25 Observed – Expected -24 -18 16 26 (Observed – Expected)^2 /Expected 7.68 3.24 5.12 27.04
 Data Level of Significance 0.05 Number of Rows 2 Number of Columns 4 Degrees of Freedom 3 Results Critical Value 7.8147 Chi-Square Test Statistic 43.08 p-Value 0.00

Since p –value of the test is 0 which is less than 0.05, hence there are sufficient statistical evidence to reject the null hypothesis. Therefore the proportion of the percent wound healing is not normal

1. The data in Problem 25 is Table 7-64

The analysis has been done in two step. In first stem, it has been examined if there exist any association between gender and treatment

H0           There is no association between gender and treatment

H1           There is association between gender and treatment

Formulae for calculating Chi –square static is as follows

While formulae for calculating degree of freedom is

Degree of freedom = (number of rows- 1)- (number of columns-1)

As a first step observed frequencies are determined

 Observed Frequencies Percent Would Healing Treatment Placebo Standard Drug New drug Total Male 12 13 9 34 Female 28 27 31 86 Total 40 40 40 120

In second step, expected step frequencies are as follows:

 Expected Frequencies Percent Would Healing Treatment Placebo Standard Drug New drug Total Male 11.33 11.33 11.33 34 Female 28.67 28.67 28.67 86 Total 40 40 40 120

The expected frequencies are calculated as per the following formula

 (Observed – Expected)^2 /Expected Male 0.04 0.25 0.48 Female 0.02 0.10 0.19
 Data Level of Significance 0.05 Number of Rows 2 Number of Columns 3 Degrees of Freedom 2 Results Critical Value 5.9915 Chi-Square Test Statistic 1.07 p-Value 0.59

Since p –value of the test is 0.59 which is higher than 0.05, hence there are not sufficient statistical evidence to reject the null hypothesis. Therefore there is no association between gender and treatment. Hence gender and treatment are independent. So there is no significant difference in proportion of men assigned to each treatment.

Further equality of proportions need to be tested.

Formulas are as follows:

 Proportion Sample 0.30 40 p1 0.33 40 p2 0.23 40 p3 1st hypothesis 2nd  hypothesis 3rd  hypothesis H0 p1 = P2 H0 p1 = P3 H0 p2 = P3 H1 p1 ≠ P2 H1 p1 ≠ P3 H1 p2 ≠ P3 p 0.3125 p 0.2625 p 0.275 SE 0.0733 SE 0.0696 SE 0.0706 Z -0.34112 Z 1.07807 Z 1.41643 p 0.36651 p 0.85950 p 0.92167 Could not reject null hypothesis Could not reject null hypothesis Could not reject null hypothesis p1= p2 p1=p3 p2=p3

So there is no significant difference in proportion of men assigned to each treatment.

29.

Chi –square test would be conducted to test the following hypothesis:

H0: There is no association between group (married/not married) and alcohol addiction                                                                      H1: There is association between group (married/not married) and alcohol addiction                                                                          Formulae for calculating Chi –square static is as follows

While formulae for calculating degree of freedom is

Degree of freedom = (number of rows- 1)- (number of columns-1)

As a first step observed frequencies are determined

 Observed Relation Social isolation and Alcohol Group Diagnosed Alcoholic Undiagnosed Alcoholic Not Alcoholic Total Married 21 37 58 116 Not married 59 63 42 164 Total 80 100 100 280

In second step, expected step frequencies are as follows:

 Expected Relation Social isolation and Alcohol Group Diagnosed Alcoholic Undiagnosed Alcoholic Not Alcoholic Total Married 33.14 41.43 41.43 116 Not married 46.86 58.57 58.57 164 Total 80 100 100 280

The expected frequencies are calculated as per the following formula

 (Observed – Expected)^2 /Expected Male 4.45 0.47 6.63 Female 3.15 0.33 4.69
 Data Level of Significance 0.05 Number of Rows 2 Number of Columns 3 Degrees of Freedom 2 Results Critical Value 5.9915 Chi-Square Test Statistic 19.72 p-Value 0.00

Since p –value of the test is 0 which is less than 0.05, hence there are sufficient statistical evidence to reject the null hypothesis. Hence, there is association between group (married/not married) and alcohol addiction