# Jane’s study

Write a non-directional null hypothesis for Jane’s study. (2)

There is no significant difference between people with and without schizophrenia in terms of whether they have a romantic partner or not and any difference that does arise is due to chance alone.

Explain which statistical test Jane would use for her study (2)

Jane should use a chi squared as she has an independent measures deign, i.e. she is looking at two separate groups of people, those with and without diagnosis of schizophrenia and she has nominal data, that is she has divided them into categories those who do have a romantic partner and those who don’t.

Calculate the chi-squared for Jane’s data. (4)

Workings out the expected frequencies:

A. 17 x 27=459/40 =11.475

B. 17 x 13=221/40 =5.525

C. 23 x 27=621/40 =15.525

D. 23 x 13=299/40 =7.475

Working out chi squared:

A. 10 – 11.475 = -1.475 x -1.475 = 2.175625 / 11.475 =0.18959695

B. 7 – 5.525 = 1.475 x 1.475 = 2.175625 / 5.525 = 0.39377828

C. 17 – 15.525 = 1.475 x 1.475 = 2.175625 / 15.525 = 0.14013688

D. 6 – 7.475 = -1.475 x -1.475 = 2.175625 / 7.475 = 0.29105351

chi squared = 0.18959695 + 0.39377828 + 0.14013688 + 0.29105351 = 1.01456562

or 1.02 to 2 dp.

Explain whether Jane should reject her null hypothesis or not (3)

Jane should accept her null hypothesis as her observed value of chi squared is 1.02, which is less than the critical value of 3.84 for a two tailed test at 5% level of significance with one degree of freedom, meaning that having schizophrenia does not affect whether you have a romantic partner or not.