Question

Amy has two ways to travel from her home in Norco to her office in Los Angeles. One is to go via the 10 Freeway, and the other is to go via 60 Freeway. In order to determine which way she should travel on a daily basis, Amy has recorded the travel times for samples of ten trips via the 10 Freeway and ten trips via the 60 Freeway. The following table gives the travel times (in minutes) for the twenty trips: Travel times in minutes 10 Freeway 70,73.78, 74, 72, 74, 72, 70, 71, 70 60 Freeway 68,80, 72, 72, 77, 74, 74, 75, 72,75 E Assume that the two populations of travel times are normally distributed and that the population variances are equal. Can we conclude, at the 0.1 level of significance, that the mean travel times of the two routes are different? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table.

Answer 1

Let denote the mean travel times for 10 Frequncy and 60 Frequency respectively.

111, 112

Hypotheses : To test null hypothesis Hopi - Ml2 = 0 against alternative hypothesis HV 41-412 70 This is a two tailed test as the alternative hypothesis contains '#' sign. Here, Ti = 72.44, s = 7.0278, n1 = 9,72 = 73.90, sź = 10.5444, n2 = 10 The test statistic can be written as (21-02) t = Sp V + 1 112 which under Ho follows a t distribution with 17 df. where pooled variance (n1 – 1)s + (n2 – 1) s ni + n2 - 2 = 8.889542 and standard error for the difference between means 1 1 1 1 SE= + = 8.889542 * + = 4.084463 11 n2 9 10 Decision rule / Rejection region : We reject Ho at 0.1 level of significance if P-value < 0.1 or if |tstat| > t0.05,71 +112–2 = to.05,17 = 1.739607 0.4- 0.3- C 0.2- 0.1- 0.0 - t Now, Value of the test statistic : The value of the test statistic is tstat = (72.44 - 73.90) 8.889542 * 1 + 1 Vg -1.062511 -1.063 10 associated degrees of freedom = n1 + n2 - 2 = 9 + 10 - 2 = 17 The critical value = tcritical = $t0,05,17 == 1.739607 1.740 P-value = P(|tni +12-2| > tstat) = P(|t17> tstat) = 2 * P(t17 < -1.062511) = 2 * 0.151432 = 0.302865 -0.3029 Conclusion : Since p-value > 0.1 and tstat| tcritical = 1.740, so we fail to reject Hat 0.1 level of significance Hence, we can conclude that there is no significant difference between two population means