Question

1.) The Maine Department of Transportation mea sures compaction of hot mix asphalt paving material. Compaction refers to the process of using rollers to compress the particles of the mixture closer together to remove air voids. Effort required for compaction can vary based on many factors, including particle size or shape, asphalt binder type &content, thickness of the asphalt mix layer, and environmental conditions at the jobsite. The main objective for compaction is to reduce the volume of air in the mix to the desired level (between 3 and 5% air) so that the pavement will be less susceptible to damage by exposure to water (fewer ruts and pot-holes & lon ger lasting pavement-less SS). The DOT suspected that changes in the asphalt paving material were causing problems in pavement quality. They compared compaction data measured from the summer of 2014 to older data from 2007 to see if any changes could be detected statistically

Answer 1

mean of difference = -0.567

standard deviation of difference = 1.598

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ha: u> 3 This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used (2) Rejection Region Based on the information provided, the significance level is 0.05, and the critical value for a right-tailed test is te- 1.699 The rejection region for this right-tailed test is R (3)_Test Statistics The t-statistic is computed as follows t:t 1.699 0.567 -3 s/Vn 1.598/V30 (4) Decision about the null h othesi Since it is observed that t --12.226 < tc 1.699, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p-1, and since p = 1 > 0.05, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is greater than 3, at the 0.05 significance level

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ha: μ < 5 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used (2) Rejection Region Based on the information provided, the significance level is a- 0.05, and the critical value for a left- 1.699 tailed test is tc-- The rejection region for this left-tailed test is R- t:t<-1.699 (3)_Test Statistics The t-statistic is computed as follows 0.567 -5 s/Vn 1.598/V30 (4) Decision about the null hypothesis Since it is observed that t -19.081 < tc rejected. Using the P-value approach: The p-value is p-0, and since p = 0 〈 0.05, it is concluded that the null hypothesis is rejected (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 5, at the 0.05 significance level 1.699, it is then concluded that the null hypothesis is