In a study of exhaust emissions from school buses, the pollution intake by passe
ID: 3154612 • Letter: I
Question
In a study of exhaust emissions from school buses, the pollution intake by passengers was determined for a sample of nine school buses used in the Southern California Air Basin. The pollution intake is the amount of exhaust emissions, in grams per person, that would be inhaled while traveling on the bus during its usual 18-mile trip on congested freeways from South Central LA to a magnet school in West LA. (As a reference, the average intake of motor emissions of carbon monoxide in the LA area is estimated to be about 0.000046 grams per person.) Here are the amounts for the nine buses when driven with the windows open. (a) Make a stemplot. Are there outliers or strong skewness that would preclude use of the f procedures? The distribution is symmetric so use of t procedures is appropriate. The distribution has a slight right skew, but no potential outliers, so the use of t procedures is still appropriate. The sample is small and the stemplot is left skewed with possible outliers, so use of t procedures is not appropriate. The sample is small and the stemplot is right skewed with possible outliers, so use of f procedures is not appropriate. (b) A good way to judge the effect of outliers is to do your analysis twice, once with the outliers and a second time without them. Give a 90% confidence Interval, with all the data, for the mean pollution intake among all school buses used in the Southern California Air Basin that travel the route investigated in the study. (Round your answers to three decimal places.) Give a 90% confidence interval with the outliers removed. (Round your answers to three decimal places.) c) Compare the two intervals in part (b). What is the most important effect of removing the outliers? Without the outliers, there is a smaller margin of error. Without the outliers, there is larger margin of error.Explanation / Answer
b. WITHOUT OUTLIER DATA
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=0.5156
Standard deviation( sd )=0.4342
Sample Size(n)=9
Confidence Interval = [ 0.5156 ± t a/2 ( 0.4342/ Sqrt ( 9) ) ]
= [ 0.5156 - 1.86 * (0.145) , 0.5156 + 1.86 * (0.145) ]
= [ 0.246,0.785 ]
b. WITH OUTLIER DATA
Confidence Interval = [ 0.3057 ± t a/2 ( 0.1323/ Sqrt ( 7) ) ]
= [ 0.3057 - 1.943 * (0.05) , 0.3057 + 1.943 * (0.05) ]
= [ 0.209,0.403 ]