Question

Last year, the mean waiting time for the number 14 bus was 11.4 minutes.  This year, a random sample of 12 waits for the bus yielded the following waiting times, in minutes.  

Use the Wilcoxon signed-rank test to perform the required hypothesis test. Last year, the mean waiting time for the number 14 bus was 11.4 minutes. This year, a random sample of 12 waits for the bus yielded the following waiting times, in minutes. 0.2 14.6 6.0 20.5 2.7 23.6 4.1 8.6 30.6 4.7 12.0 16.9 At the 1% significance level, do the data provide sufficient evidence to conclude that the mean waiting time this year is higher than last year's mean time of 11.4 minutes? Step 1, Ho: μ Step 2, Alpha, α: Step 3. Critical Value Step 4. Test Statistic: W Step 5. Decision: Step 6. Interpretation: At the and H1:μ Ho. significance level, the data sufficient evidence

Answer 1

	Values ( X )	$\Sigma (X_{i} - \bar{X})^{2}$
	10.2	12.3082
	14.6	0.7951
	6	59.4179
	20.5	46.1272
	2.7	121.1827
	23.6	97.8457
	14.1	0.1534
	8.6	26.0947
	30.6	285.3295
	4.7	81.1495
	12	2.9183
	16.9	10.1869
Total	164.5	743.5091

Mean $\bar{X} = \Sigma X_{i} / n$

Y = 164.5/12= 13.7083

Standard deviation $S_{X} = \sqrt{\Sigma (X_{i} - \bar{X})^{2}/n-1}$

sx = V743.509 1/12-1-8.2214

To Test -

H0 :-  

μ = 11.4

H1 :-  

μ>11.4

α 0.01

Test Statistic :-
$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$

t = 0.9726

t = (13.7083-11.4)/(8.2214/v/12)

Test Criteria :-
Reject null hypothesis if

t > ta,n-1

Critical value  

ta,n-l t0.01 12-1 2.718

Result :- Fail to reject null hypothesis

ta,n-1 = 0.9726 < 2.718

Decision based on P value
P - value = P ( t > 0.9726 ) = 0.1758
Reject null hypothesis if P value < level of significance
P - value = 0.1758 > 0.01 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

α 0.01

At 1% significance level, the data does not provide sufficient evidence to support the claim that mean waiting time this year is higher.