Regression Analysis

Question 1

In a study of intra-observer variability in the assessment of cervical smears, 3341 slides were screened for the presence or absence of abnormal squamous cells. Each slide was screened by a particular observer and then re-screened six months later by the same observer. The results of this study are shown in Table 1.1 below.

Table 1.1: Cross tabulation of the results of the first and second screening

Second screening

Present

Absent

First

screening

Present

1764

498

Absent

412

667

Do these data support the null hypothesis that there is no association between time of screening and diagnosis? Briefly explain by carrying out an appropriate test.

To check the association among the screen timings and the time of diagnosis we will carry out a Chi-square test. On the basis of chi-square test we will conclude our test results for the accepting or rejecting our null hypothesis.

Test Hypothesis

: There is no association between timing of screening and diagnosis.

: There is association between timing of screening and diagnosis.

To obtain the chi-square statistics for the given table first we will generate the following table:

Second Screening

Present

Absent

First Screening

Present

1764

498

2264

Absent

412

667

1079

2176

1165

3341

The chi-square statistics used to conclude the test result were obtained by using the following formula:

The chi-square statistics obtained through the above formula is:

The p-value of the test, i.e. the significance level of the test appears to be 0.0001. Therefore, we can conclude that there is a strong association among the timings of screening and the diagnosis. Hence on the basis of our p-value we can finally accept our null hypothesis.

Question 2

A nested case-control study was conducted in the Hunter Region of NSW to examine whether patients who were current smokers of cigarettes had more complications after surgery than patients who were non-smokers. Tables 2.1 and 2.2 show the smoking status for cases (those with complications at the 6 week follow-up visit) and controls (those with no complications), separately by hospital.

Table 2.1: Count of the number of complications among smokers and non-smokers at Hospital 1

Hospital 1

Complication

Present

Absent

Smoking status

Smoker

42

22

Non-smoker

85

62

Table 2.2: Count of the number of complications among smokers and non-smokers at Hospital 2

Hospital 2

Complication

Present

Absent

Smoking status

Smoker

254

348

Non-smoker

96

172

(a)Estimate the odds of being a smoker among patients who had a complication at 6 weeks and who had their surgery performed at Hospital 1.

Through the results of SPSS, the odds ratio of being a smoker who had a complication at six weeks and also had a surgery among the patients of hospital 1 can be given as:

Risk Estimate

Value

95% Confidence Interval

Lower

Upper

Odds Ratio for Smokingstatus (non-smoker / Smoker)

1.393

.756

2.565

For cohort Complication = Absent

1.227

.833

1.808

For cohort Complication = Present

.881

.704

1.103

N of Valid Cases

211

Therefore, form the above table we can conclude that the odds ratio for being a smoker among those who had complication at six weeks and also had a surgery in hospital 1 is .881

(b) Estimate the odds of being a smoker among patients who DID NOT have a complication at 6 weeks and who had their surgery performed at Hospital 1.

From the above table, the odds ratio of being a smoker who did not had a complication at six weeks and had a surgery among the patients ...