Which statistical analysis to use?

Question

I am comparing the number of positive screens and the number of negative screens to see if there is a significant difference. If there is a significant number of positive screens, then the anxiety screening will be supported.

Answer 1

Example data set up:

Screening Outcome	Observed Count	Expected Count
Positive	62	50
Negative	38	50
Total	100	100

The expected distribution was 50% positive and 50% negative screens (100x0.5=50)

The chi-square test had 1 degree of freedom (df = 1)

Chi-square critical (table) value at α = 0.05 with df=1, χ²₀.₀₅,₁ = 3.84

Kind of data you have: Categorical data

Counts, number of positive screens vs negative screens

Question being asked: Is the number of positive screens significantly different from what we would expect by chance or from a reference value?

You are testing whether one category occurs more than expected.

Statistical test: Chi-squre goodness-of-fit test

Why?:

One sample

Counts in categories

Comparing observed counts to expected counts

Assumptions:

Observations are independent

Categories are mutually exclusive

Expected counts are at least 5.

Hypotheses:

Null: The observed screening outcomes follow the expected distribution.

Alternative: The observed screening outcomes differ from the expected distribution.

Using the standard chi-square goodness-of-fit formula:

X² = Sum ((O-E)²) / E)

(where O = observed count, E = expected count)  

Solve:

Positive: (62-50)^2/50 = (12)^2/50 = 144/50 = 2.88

Negative: (38-50)^2/50 = (-12)^2/50 = 144/50 = 2.88

Sum these:

X² = 2.88 + 2.88 = 5.76

Chi-square test statistic is χ²(1) = 5.76

Now compare 5.76 to the chi-square distribution with df = 1 (two categories minus one), 3.84, to get a p-value.

Your test statistic: χ² = 5.76

Critical value: 3.84

Because the calculated chi-square value (5.76) exceeds the critical value (3.84), the null hypothesis is rejected. 5.76 > 3.84

Results:

p < 0.05, Reject H₀.

Conclusion:

A chi-square goodness-of-fit test showed that observed screening outcomes differed significantly from the expected distribution, χ²(1) = 5.76, p < 0.05.