Question

Viewing the return rate of lost letters as a measure of social responsibility in neighborhoods, a social psychologist intentionally ‘loses’ self-addressed, stamped envelopes near mailboxes. Furthermore, to determine whether social responsibility, as inferred from the mailed return rates, varies with the type of neighborhood, lost letters are scattered throughout three different neighborhoods: downtown, suburbia, and a college campus.

Letters are ‘lost’ in each of the three types of neighborhoods according to procedures that control for possible contaminating factors, such as the density of pedestrian traffic and mailbox accessibility. (Ordinarily, the social psychologist would probably scatter equal numbers of letters among the three neighborhoods, but to maximize the generality of the current example, we will assume that a total of 200 letters were scattered as follows: 60 downtown, 70 in suburbia, and 70 on campus.) Each letter is cross classified on the basis of the type of neighborhood where it was lost and whether or not it was returned. For instance, of the 60 letters lost downtown, 39 were returned, while of the 70 letters lost in suburbia, 40 were not returned. When observations are cross classified according to two qualitative variables, as with the lost letter study, the test is a two-variable test. Answer the following.

A) Defined null and alternative hypothesis?

B) Obtain all expected frequencies from table of observed frequencies?

C) Find test statistic value?

D) Identify degrees of freedom?

E) Find the chi-squared value?

F) What is the conclusion?

Answer 1

a)

Ho: given two variable are independent
H1: Given two variables are not independent

b)

	Observed Frequencies
	0
0	downtown	suburbia					Total
returned	39	30					69
not returned	21	40					61

	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	downtown	suburbia					Total
returned	60*69/130=31.846	70*69/130=37.154					69
not returned	60*61/130=28.154	70*61/130=32.846					61

	(fo-fe)^2/fe
returned	1.607	1.377
not returned	1.818	1.558

c)

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   6.360

d)

Level of Significance =   0.05
Number of Rows =   2
Number of Columns =   2
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 2- 1 ) =   1

e)

Critical Value =   3.841   [ Excel function: =CHISQ.INV.RT(α,DF) ]

f)

Decision: test stat,X² > critical value , So, Reject the null hypothesis      

Please revert in case of any doubt.

Please upvote. Thanks in advance