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An environmental agency is examining the relationship between the ozone level (i

ID: 3319085 • Letter: A

Question

An environmental agency is examining the relationship between the ozone level (in parts per million) and the population (in millions) of a number cf cities. Part of the regression analysis is shown to the right. Complete partsa and b. Dependent variable is: Ozone R squared-70.9% s= 1.731 with 16-2= 14 df Variable Coefficient SE(Coeff) Intercept 17.982 Pop 1.853 1.052 6.149 a) Give a 90% confidence interval for the approximate increase in ozone level associated with each additional milion city inhabitants. Each additional million residents corresponds to an increase in average ozone level of between Round to two decimal places as needed. Use ascending order.) andpp %confidence. with b) For the cities studied, the mean population was 1.7 million people. The population of a particular city is approximately 0.8 million people. Predict the mean ozone level for cities of that size with an interval in which you have 90% confidence. ppm, with 90% confidence. The mean ozone level for cities with 0.8 milion people is between (Round to two decimal places as needed. Use ascending order.) and

Explanation / Answer

As we know the formula of LCL/UCL = X' = mu + z * s/sqrt(n)

a) here we have z value at 90% conf level is +/- 1.64

s=1.731 n= 1 (1millinon population already given) , mu is the mean ozone level which can be found by the below equation provided in the analysis of regression

Ozone = 17.982 +6.149*Pop = 17.982 +6.149*1 = 24.131

so mu =24.131

So., X' = 24.131 -/+ 1.64*1.731 = 21.29216 to 26.96984 ppm

b)

same as the above example here

here we have z value at 90% conf level is +/- 1.64

s=1.731 n= 0.8 (0.8millinon population already given) , mu is the mean ozone level which can be found by the below equation provided in the analysis of regression

Ozone = 17.982 +6.149*Pop = 17.982 +6.149*1.7 = 28.4353

so mu =28.4353

So., X' =28.4353 -/+ 1.64*1.731/sqrt(0.8) = 25.26138 to 31.60922 ppm

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