Question

The accompanying data on degree of spirituality for a sample of natural scientists and a sample of social scientists working at research universities appeared in a paper. Assume that it is reasonable to regard these two samples as representative of natural and social scientists at research universities. Is there evidence that the spirituality category proportions are not the same for natural and social scientists? Test the relevant hypotheses using a significance level of 0.01.

	Degree of Spirituality
	Very	Moderate	Slightly	Not at All
Natural Scientists	54	162	194	216
Social Scientists	56	221	238	239

Calculate the test statistic. (Round your answer to two decimal places.)
χ² = ___

What is the P-value for the test? (Round your answer to four decimal places.)
P-value = ___

Answer 1

Based on the given data, we are asked to test the claim - spirituality category proportions are not the same for natural and social scientists.

The appropriate statistical test to test the above claim would be a Chi-square test of Goodness of fit, where we test:

  (Spirituality category proportions are the same for natural and social scientists). Vs   (Spirituality category proportions are not the same for natural and social scientists).

where, O_i denote the observed cell frequencies and E_i is the expected frequencies when we assume that spirituality category proportions are the same for natural and social scientists.

From the given data,

	Degree of Spirituality
	Very	Moderate	Slightly	Not at all	Total
Natural Scientists	54	162	194	216	626
Social Scientists	56	221	238	239	754
Total	110	383	432	455	1380

The expected frequencies are obtained by equally distributing the total column frequency between Natural and Social scientists, as stated in the null hypothesis:

	Degree of Spirituality
	Very	Moderate	Slightly	Not at all
Natural Scientists	55	191.5	216	227.5
Social Scientists	55	191.5	216	227.5

The test statistic is given by:

Substituting the values,

Oi	Ei	(Oi-Ei)2	(Oi-Ei)2 /Ei
54	55	1.00	0.018
56	55	1.00	0.018
162	192	870.25	4.544
221	192	870.25	4.544
194	216	484.00	2.241
238	216	484.00	2.241
216	228	132.25	0.581
239	228	132.25	0.581
SUM	14.77

We get = 14.77

The p-value of the test can be obtained for 8 - 1 = 7 df using the excel function:

We get p-value = 0.0391