Question

The table below describes residents of a neighborhood based on their car ownership and public transportation usage. Is there evidence that car ownership is associated with whether or not residents use public transportation? Use α = 0.05. Clearly state your hypotheses and provide the relevant p-value.

	Owns car	Does not own car	Total
Uses public transport	65	63	128
Does not uses public transport	95	48	143
Total	160	111	271

Answer 1

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All points calculated and explained in detail below, please read on.

The hypothesis set is designed as:

Null hypothesis: Ho:  Car ownership is NOT associated with whether or not residents use public transportation

Alternate hypothesis : Car ownership IS associated with whether or not residents use public transportation

The contingency table below provides the following information: the observed cell totals, (the expected cell totals) and [the chi-square statistic for each cell].

The Chi-square of each cell is calculated as : (O-E)^2/E, where

For example for 1st cell:

O = Observed value = 65

E = RowSum*ColumnSum/Total = 128*160/271 = 75.57

The ChiSquare for this cell = (75.57-65)^2/65 = 1.48

Results
	owns cars	does not own caar				Row Totals
uses public	65  (75.57)  [1.48]	63  (52.43)  [2.13]				128
does not use public	95  (84.43)  [1.32]	48  (58.57)  [1.91]				143
Column Totals	160	111				271  (Grand Total)

We add all the Chi-Square values in square brackets to get 6.8427.

Then apply the formula of 1 - CHISQ.DIST(6.8427, 1) = 1-.0011 = 0.0089.

(Above is an Excel formula = CHISQ.DIST(Chi-Square statistic , df), where:

df = (nRow-1)*(nColumn-1) = (2-1)*(2-1) = 1

Answer: p-value = 0.0089

The chi-square statistic is 6.8427. The p-value is .0089. The result is significant at p < .05.

Conclusion: As p-value < alpha, we reject Ho and conclude that : Car ownership IS associated with whether or not residents use public transportation