A study of intra-observer variability in the assessment of cervical smears, 3,32

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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