Question

The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:

	Male	Female	SUM
Local	50	90	140
Non local	80	60	140
SUM	130	150	280

1= She will use this information to perform a chi-square hypothesis test using a level of significance of 0.01. The critical value is : ( )

2= She will use this information to perform a chi-square hypothesis test using a level of significance of 0.01. Which of the following is most correct?

		Reject the null hypothesis and conclude that the proportion of males and females has no difference
		Reject the null hypothesis and conclude that the proportion of males and females has a difference
		Do not reject the null hypothesis and conclude that the proportion of males and females has no difference
		Do not reject the null hypothesis and conclude that the proportion of males and females has a difference

3= What is the value of the test statistic to use in evaluating the alternative hypothesis that there is a difference in the two population proportions using α = 0.01?

4= The expected cell frequency in the Male and Local cell is ( )

5= The expected cell frequency in the Female and Non Local cell is ( )

6 =

which represents the relevant hypotheses?

		H0: π1 − π2 ≤ 0 versus H1: π1 − π2 > 0
		H0: π1 − π2 ≠ 0 versus H1: π1 − π2 = 0
		H0: π1 − π2 ≥ 0 versus H1: π1 − π2 < 0
		H0: π1 − π2 = 0 versus H1: π1 − π2 ≠ 0

answers only !!

Answer 1

1).

Level of Significance =   0.01
Number of Rows =   2
Number of Columns =   2
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 2- 1 ) =   1
  

Critical Value =   6.635

2)

Reject the null hypothesis and conclude that the proportion of males and females has a difference

3)

Chi-Square Test of independence
	Observed Frequencies
	male	female					Total
local	50	90					140
non local	80	60					140
Total	130	150					280
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	male	female					Total
local	130*140/280=65	150*140/280=75					140
non local	130*140/280=65	150*140/280=75					140
Total	130	150					280
(fo-fe)^2/fe	male	female					total
local	3.462	3.000					6.462
non local	3.462	3.000					6.462

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   12.923

4) Expected frequency of a cell = sum of row*sum of column / total sum = 130*140/280=65

5) Expected frequency of a cell = sum of row*sum of column / total sum = 150*140/280=75

6)  H0: π1 − π2 = 0 versus H1: π1 − π2 ≠ 0