Question

Question 10 of 10 (4 points) | Attempt 1 of 1 View question in a popup 02h 47m Remaining 11.3 Section Exercise 12 (p-value, calc) ol. Truck pollution: In an experiment to determine the effect of ambient temperature on the emissons of oxides of nitrogen (NOR) of diesel trucks, 10 trucks were run at temperatures of 40 °F and 80 °F. The emissions, in parts per billion, are presented in the following table. 0 Truck 40 °F 80 °F 1 909.8 888.6 2 828.5 849.4 3 924.2 918.4 4 982.3 860.9 5 895.4 863.7 6 957.8 917.8 7 971.0 893.3 8 829.1 770.9 9 830.3 865.6 10 916.5 866.2 Send data to Excel Let u, represent the mean emission at 40 °F and u d =ļi - 42: Can you conclude that the mean emissions are lower at 40 °F? Use the a=0.1 level of significance and the TI-84 Plus calculator to answer the following. Part: 0 / 4 Next Part Submit Assignment

Answer 1

Solution: The data provided in the question statement can be tabulated as below. Calculating a few additional columns: "! Truck 40°F 80°F (d - d) X1 X2 Difference d = X1-X2 21.2 1 909.8 888.6 2 -20.9 828.5 924.2 190.716 3125.928 853.224 849.4 918.4 860.9 3 5.8 4 982.3 121.4 7463.232 5 895.4 863.7 31.7 10.956 6 957.8 917.8 40 24.900 7 971 893.3 77.7 1822.436 537.776 8 829.1 770.9 58.2 9 830.3 865.6 -35.3 4943.496 10 916.5 866.2 50.3 d =350.1 2 (d 233.784 d) 2 = 19206.449 d=d / n = 350.1/10 = 35.01 sa = £ (d – d) 2/(n − 1) = 19206.449/(10 – 1) = 46.196 Level of significance = a = 0.1 Null Hypothesis: Ho: Up = 0 Alternate Hypothesis: Hi: Up < 0 t = Test Statistic: (ā - Ma) / (sa/ Vn) t = (35.01 - 0)/(46.196/V10) t = 2.397

Degrees of freedom = df = n - 1 = 10 – 1 = 9 From t-distribution table; the area in the left tail corresponding to (t = 2.397) and for (df = 9), is given by = 0.9799 Hence, p-value = 0.9799 The p-value is greater than a, hence, we fail to reject the null hypothesis. There is not sufficient evidence to conclude that emissions are lower at 40°F