Question

Descriptive Statisticsa
Sex		N	Minimum	Maximum	Mean	Std. Deviation	Variance
Female	YearsSick	44	2	30	16.61	7.254	52.615
	Sex	44	1	1	1.00	.000	.000
	Valid N (listwise)	44
Male	YearsSick	48	4	35	17.44	8.008	64.124
	Sex	48	2	2	2.00	.000	.000
	Valid N (listwise)	48
Other	YearsSick	2	3	15	9.00	8.485	72.000
	Sex	2	3	3	3.00	.000	.000
	Valid N (listwise)	2
a. No statistics are computed for one or more split files because there are no valid cases. Based on the above, how can I conduct the appropriate t-test to test the hypothesis ‘females with schizophrenia have a longer duration than males” Interpret findings –is there a significant difference? What does this mean? Do we assume equal variance? Why? Why Not?

Answer 1

We need to test the following hypothesis:

Ha: Females with schizophrenia have a longer duration than males i.e mu1 > mu2

From the given table, v1 - 52.615, v2 = 64.124
As one of the variance is not twice as large as the other, we will assume equal variances and conduct the two sample t-test

Let's compute the t-statistic using the given formula:

Here s is the pooled variance
s² = (52.615*43 + 64.124*47)/90 = 58.625

Inputting the values into the formula:

t = (16.61 - 17.44)/1.598 = -0.52

p-value at df = n1 + n2 - 2 = 90 is: 0.3021

As the p-value > 0.05, we cannot reject the null hypothesis. We do not have enough evidence to say that Females with schizophrenia have a longer duration than males.