Question

5. Dave rides the different routes on bike A and bike B. He wants to figure out if bike A is faster than bike B. Perform a hypothesis test for this using the data below. (5 points) Route Bike A (minutes) Bike B (minutes) 152 121 135 143 29 95 67 78 45 47 25 161 122 129 145 31 94 69 80 43 48 25 Mountains Flat A Canada A Canada B Michigan Work Cali A Cali B Cali C Work 1 Work 2 Daycare ride The standard deviation of differences is s-3.62. Hint: This is a one-tail test.

Answer 1

• EXPALANTION
• Here since the same samples of routes has been use to test the speed we will use paired samples test the hypothesis.
• Here the random variables follow normal distribution , since the sample size is small and population standard deviation is unknown we will use t distribution to test the statistics
• Here Let XD be the random variable denoting the speed of the bikes A and B respectively in distance per minute .
• The differences is given by

• bike A	bike B	A-B
  152	161	-9
  121	122	-1
  135	129	6
  143	145	-2
  29	31	-2
  95	94	1
  67	69	-2
  78	80	-2
  45	43	2
  47	48	-1
  25	25	0

• The mean of the difference is given by
• $\bar{X}=\frac{\sum X}{n}$ =-0.90909
• The standard deviation is given by
• =3.62 as in question.
• TO TEST THE HYPOTHESIS.
• The null hypothesis is given by the that there is no difference in the speed covered by bikes A and B
• Let the alternate hypothesis be that bike A is faster that bike B.
• Ho: $\mu d=0$ vs H1: $\mu d<0$
• TEST STATISTICS IS GIVEN BY
• $\frac{\bar{xd}-\mu d }{\frac{s}{\sqrt{n}}}\sim t_{n-1,\alpha }$
• $\frac{-0.90909-0}{3.62/\sqrt{11}}$ =-0.8329
• CONCLUSION
• The upper and lower critical value at 5% significant level for t distribution with n-1=11-1=10 degrees of freedom $\pm$ 1.812
• The test statistics obatined lies between the upper and lower critical value.
• Hence we do not have enough evidence to reject the null hypothesis.
• Hence we conclude that there is no difference in the speed covered by bikes A and B