The Kyoto Protocol was signed in 1997, and required countries to start reducing their carbonemissions. The protocol became enforceable in February 2005. The table contains a random sampleof CO2 emissions from different countries. Compute a 98% confidence interval to estimate thepopulation mean CO2 emission.Round final answers to three decimal places.CO2 Emissions (metric tons per capita)0.745.216.695.687.330.74.4918.341.546.773.465.651.647.230.31.657.92.468.356.263.611.032.466.97With% confidence, it can be stated that the true populationmean CO2 emission for all participating countries is betweenmetric tons per capita.

Solution-: We find 98% C.I. for the population mean CO2 emission We prepare the following table: xi xi^2 0.74 0.5476 0.7 0.49 6.77 45.8329 0.3 0.09 6.26 39.1876 5.21 27.1441 4.49 20.1601 3.46 11.9716 1.65 2.7225 3.61 13.0321 6.69 44.7561 1 1 5.65 31.9225 7.9 62.41 1.03 1.0609 5.68 32.2624 8.34 69.5556 1.64 2.6896 2.46 6.0516 2.46 6.0516 7.33 53.7289 1.54 2.3716 7.23 52.2729 8.35 69.7225 6.97 48.5809 Total 107.46 645.6156 From this table we get, n=25,∑xi=107.46,∑xi2=645.6156Sample Mean: x¯=∑xin=107.4625=4.2984 and Sample variance: s2=∑(xi2-n*x¯)2n-1=645.6156-25*4.2984225-1=7.6546 ∴s=s2=7.6546=2.7667 Solution-: Given: x¯=4.2984,n=25,s=2.7667,α=0.02 The 98% C.I. is (x¯± t(n-1),α/2*sn) (4.2984± 2.492*2.766725) Where, ttab=t(n-1),α/2=t(25-1),0.02/2=t24,0.01=2.492 By using MS-Excel Function "=TINV(0.02,24)" (4.2984± 1.379) (4.2984- 1.379,4.2984+1.379) (2.919.5.677) The required C.I. is (2.919.5.677) i.e. 2.919<μ<5.677 with 98% confidence, it can be stated that the true population mean CO2 emission for all participating countries is between 2.919<μ<5.677 metric tons per capita