t-test for avg. walking time of men and women

Variable 1 = Walking time for men

Variable 2 = Walking time for women

t-Test: Two-Sample Assuming Unequal Variances
	Variable 1	Variable 2
Mean	32.60606061	35.85714286
Variance	162.8138821	218.4005468
Observations	297	210
Hypothesized Mean Difference	0
df	407
t Stat	-2.579738612
P(T<=t) one-tail	0.005118641
t Critical one-tail	1.64860612
P(T<=t) two-tail	0.010237281
t Critical two-tail	1.965809661

As t stat > 1.96 hence the difference between avg. walking time of men and women is significant. (Rejecting null hypothesis at 95% confidence level).

Problem for Chi-Square goodness of fit test

In this example the headway data collected on a road near Guwahati has been fitted to the negative exponential distribution and theoretical headway frequencies are obtained from the negative exponential distribution.

Table 8.2: Calculations involved in testing the closeness of the observed data with the assumed probability distribution

Class	Headway, t	Observed frequnecy	Obs.  Freq. h>t	Relative freq. (h>t)	Theor. Rel. freq.	Theor. Freq. (h>t)	Theor. Freq.
0 to 2	0.00	47	522.00	100.00	100.00	522.00	159	77.98
2 to 4	2.00	125	475.18	91.03	69.59	363.28	110	2.05
4 to 6	4.00	128	398.18	76.28	48.43	252.82	77	35.58
6 to 8	6.00	67	270.18	51.76	33.71	175.94	53	3.70
8 to 10	8.00	49	203.18	38.92	23.46	122.45	37	3.89
10 to 12	10.00	32	154.18	29.54	16.32	85.21	26	1.96
12 to 14	12.00	11	122.18	23.41	11.36	59.30	18	2.72
14 to 16	14.00	11	111.18	21.30	7.91	41.27	13	0.08
16 to 18	16.00	13	100.18	19.19	5.50	28.72	9	3.13
18 to 20	18.00	7	87.18	16.70	3.83	19.99	6	0.17
20 to 22	20.00	9	80.18	15.36	2.66	13.91	14	1.23
								132.48

The statistic resulting from the last column of Table 8.2 follows chi-square distribution. Hypotheses formulation for this problem is given below;

H₀: Headway data follows negative exponential distribution

H₁: Headway data doesn't follow the negative exponential distribution

Chi-square calculated is 132.48 and the critical Chi-square value, for 11-1-1 = 9 degrees of freedom (1 for the mean headway and the other one for cumulative frequency) at 5 % significance level, is 3.33. Since the critical value is less than the calculated value the decision is “reject the null hypothesis that the headway data follows negative exponential distribution”.