Calculation Chi Square Static
- 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 |
4.
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20.
Solution
- 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
- 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
- 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