In: Statistics and Probability
Percent Wound Healing |
||||
Treatment |
0-25% |
26-50% |
51-75% |
76-100% |
New Compound (n=125) |
15 |
37 |
32 |
41 |
Placebo (n=125) |
36 |
45 |
34 |
10 |
Is there a difference in the extent of wound healing by treatment? (Hint: Are treatment and the percent wound healing independent?) Run the appropriate test at a 5% level of significance.
Percent Wound Healing |
|||||
Treatment |
0-25% |
26-50% |
51-75% |
76-100% |
Total |
Number of Patients |
51 |
82 |
66 |
51 |
250 |
Chi-square test of independence
The Chi-square test of independence is performed to test the association between the extent of wound healing and the type of treatment
Hypothesis:
H0: The two variables are independent
Ha: The two variables are dependent
Chi-Square statistic:
The observed values are,
Percent Wound Healing | |||||
Treatment | 0-25% | 26-50% | 51-75% | 76-100% | Total |
New Compound (n=125) | 15 | 37 | 32 | 41 | 125 |
Placebo (n=125) | 36 | 45 | 34 | 10 | 125 |
Total | 51 | 82 | 66 | 51 | 250 |
The expected values are obtained using the following formula,
Percent Wound Healing | |||||
Treatment | 0-25% | 26-50% | 51-75% | 76-100% | Total |
New Compound (n=125) | 125*51/250=25.5 | 125*82/250=41 | 125*66/250=33 | 125*51/250=25.5 | 125 |
Placebo (n=125) | 125*51/250=25.5 | 125*82/250=41 | 125*66/250=33 | 125*51/250=25.5 | 125 |
Total | 51 | 82 | 66 | 51 | 250 |
The Chi-Square value is obtained using the formula
Observed, Oi | Expected, Ei | |
15 | 25.5 | 4.32 |
37 | 41 | 0.39 |
32 | 33 | 0.03 |
41 | 25.5 | 9.42 |
36 | 25.5 | 4.32 |
45 | 41 | 0.39 |
34 | 33 | 0.03 |
10 | 25.5 | 9.42 |
Sum | 28.33 |
Critical value:
The critical value for the chi-square statistic is obtained from the chi-square distribution table for,
degree of freedom = (r-1)(c-1) = (2-1)(4-1) = 3
Where r is the number of rows and c is the number of columns in the table.
For significance level = 0.05 and degree of freedom = 3, the critical value is,
Conclusion:
Since the chi-square value is greater than the critical value, It can be concluded that the null hypothesis is rejected at a 5% significance level.
Hence the extent of wound healing and the type of treatment are dependent.
Chi-Square Goodness of fit test
The Chi-Square Goodness of fit test is used here to test whether the samples are from normal distribution within a given proportion.
Hypothesis:
Null hypothesis: There is no significant difference between the observed frequency and the expected frequency.
Chi-Square statistic:
The observed values are,
Treatment | Observed |
0-25% | 51 |
26-50% | 82 |
51-75% | 66 |
76-100% | 51 |
Total | 250 |
The expected values are obtained by multiplying the total population by the given proportion,
Treatment | Expected |
0-25% | 250*0.30=75 |
26-50% | 250*0.40=100 |
51-75% | 250*0.20=50 |
76-100% | 250*0.10=25 |
The Chi-Square statistic is obtained using the formula,
Treatment | Observed | Expected | |
0-25% | 51 | 75 | 7.68 |
26-50% | 82 | 100 | 3.24 |
51-75% | 66 | 50 | 5.12 |
76-100% | 51 | 25 | 27.04 |
Total | 43.08 |
Critical value:
The critical value for the chi-square statistic is obtained from the chi-square distribution table for,
degree of freedom = 4-1 = 3 and significance level = 0.05
Conclusion:
Since the chi-square statistic is greater than the critical value at a 5% significance level. it can be concluded that the null hypothesis is rejected. Hence we can conclude that the distribution of percent wound healing doesn't follow a normal distribution.