A study of intra-observer variability in the assessment of cervical smears, 3,32
ID: 3337261 • Letter: A
Question
A study of intra-observer variability in the assessment of cervical smears, 3,325 slides were screened for the presence or absence of abnormal squamous cells. Each slide was screened by a particular observer and then rescreened six months later by the same observer. The results are shown below. Test whether time of screening is related to the diagnosis. Was the observer more likely to detect squamous cells at one of the screenings? Use a 5% level of significance. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available. Second Screening First Screening Present Absent Present 1763 489 Absent 403 670
Explanation / Answer
Solution:
Here, we have to use Chi square test for checking whether two categorical variables are related to each other or not. The null and alternative hypothesis for this test is given as below:
1]
Null hypothesis: H0: The time of screening is not related to the diagnosis.
Alternative hypothesis: Ha: The time of screening is related to the diagnosis.
2]
Appropriate test is Chi square test for independence of two categorical variables.
3]
Decision Rule:
We reject the null hypothesis if chi square test statistic is greater than critical value 3.841459149, otherwise we do not reject the null hypothesis.
We reject the null hypothesis if p-value is less than level of significance or alpha value.
4]
The test statistic for this test is given as below:
Chi square = [(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
Expected frequencies = E = row total * column total / Grand total
Calculations for expected values and chi square test statistic are given as below:
Observed Frequencies
Column variable
Row variable
Second Screening
First screening
Total
Present
1763
489
2252
Absent
403
670
1073
Total
2166
1159
3325
Expected Frequencies
Column variable
Row variable
Second Screening
First screening
Total
Present
1467.017143
784.9828571
2252
Absent
698.9828571
374.0171429
1073
Total
2166
1159
3325
(O - E)
295.9829
-295.983
-295.983
295.9829
(O - E)^2/E
59.71699
111.6022
125.3333
234.2295
We are given
Level of significance = = 0.05
Number of rows = r = 2
Number of columns = c = 2
Degrees of freedom = (r – 1)*(c – 1) = (2 – 1)*(2 – 1) = 1*1 = 1
Critical value = 3.841459149 (by using chi square table)
Chi square test statistic = 530.8820835
P-value = 0.00
5]
Here, P-value = 0.00 < = 0.05
So, we reject the null hypothesis that the time of screening is not related to the diagnosis.
There is sufficient evidence to conclude that the time of screening is related to the diagnosis.
Observed Frequencies
Column variable
Row variable
Second Screening
First screening
Total
Present
1763
489
2252
Absent
403
670
1073
Total
2166
1159
3325
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