Question

A courier service advertises that its average delivery time is less than 6 hours for local deliveries. A sample of 16 local deliveries was recorded and yielded the statistics     x =   5.83 hours and s = 1.59 hours. At the 5% significance level, conduct a hypothesis test to determine whether there is sufficient evidence to support the courier’s advertisement.  

Answer 1

Let be the true mean delivery time for local deliveries. The appropriate hypothesis are,

H0: = 6 hours

H1: < 6 hours

Sample mean = 5.83 hours

Sample standard deviations, s = 1.59 hours

Standard error of mean = s / = 1.59 / = 0.3975

Since we do not know the population standard deviation, the test statistic will follow t distribution with degree of freedom = n -1 = 16 - 1 = 15

Test statistic, t = ( - ) / Std error = (5.83 - 6) / 0.3975 = -0.4277

P-value = P(t < -0.4277, df = 15) = 0.3375

Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that true mean delivery time for local deliveries is less than 6 hours. Thus, there is no sufficient evidence to support the courier’s advertisement.