Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning." These words begin a report on a statistical study of the effects of logging in Borneo. Here are data on the number of tree species in 12 unlogged forest plots and 9 similar plots logged 8 years earlier: Unlogged 22 18 22 20 15 21 13 13 19 13 19 15 Logged 17 4 18 14 18 15 15 10 12 Use the data to give a 90% confidence interval for the difference in mean number of species between unlogged and logged plots. Compute degrees of freedom using the conservative method.Interval: _____ to ______.

Answer: -8.16 to 15.84Step-by-step explanation: Confidence Interval is an interval in which we are a percentage sure the true mean is in the interval.A confidence interval for a difference between two means and since sample 1 and sample 2 are under 30, will be[tex]x_{1}-x_{2}[/tex] ± [tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex]wherex₁ and x₂ are sample meanst is t-score[tex]S_{p}[/tex] is estimate of standard deviationn₁ and n₂ are the sample numbersThe estimate of standard deviation is calculated as[tex]S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }[/tex]wheres₁ and s₂ are sample standard deviation of each sampleDegrees of freedom is:[tex]df=n_{1}+n_{2}-2[/tex]df = 12 + 9 - 2df = 19Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.Calculating estimate of standard deviaton:[tex]S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }[/tex][tex]S_{p}=\sqrt{\frac{299.07}{19} }[/tex][tex]S_{p}=[/tex] 15.74The difference between means is[tex]x_{1}-x_{2}[/tex] = 17.5 - 13.66 = 3.84Calculating the interval:[tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = [tex]1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }[/tex][tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = [tex]27.21\sqrt{\frac{21}{108} }[/tex][tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = [tex]27.21\sqrt{0.194}[/tex][tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = 12Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:lower limit: 3.84 - 12 = -8.16upper limit: 3.84 + 12 = 15.84The interval is from -8.16 to 15.84.