For a group experiment, your science class measured the fine-particulateconcentrations in the air at random places around campus, and estimated asample average of 12 ug/m3 (micrograms per cubic meter). If 196 readingswere taken, and the standard deviation of the sample measurements was3.5 pg/m3, you are 99.7% confident that actual concentration of fineparticulates at the school is

Answer:The 99.7% confidence interval for the mean here is [tex]\rm (11.2, 12.8)\; \mu g\cdot m^{-3}[/tex].Step-by-step explanation:The data in this question comes from random samples. In other words, the true population data are not known. Assume that both the sample average and the sample stdev are unbiased estimates for the population mean and stdev.  The sample size 196 is sufficient large, such that the central limit theorem will apply. By the central limit theorem, the distribution of the mean (as well as the sum) of a sufficiently large samples resembles normal distributions. However, since stdev is only an estimate, the confidence interval can only be found using the Student's t-distribution.What is the confidence interval for the mean of a random variable that follows the t-distribution?[tex]\displaystyle \left(\bar{x} - t\cdot \frac{s_{n-1}}{\sqrt{n}}, \; \bar{x} + t\cdot \frac{s_{n-1}}{\sqrt{n}}\right)[/tex],where[tex]\bar{x}[/tex] is the unbiased estimate of the population mean (a.k.a sample mean.) For this question, [tex]\bar{x} = 12[/tex]. [tex]s_{n-1}[/tex] is the unbiased estimate of the standard deviation (the square root of variance) of the population. For this question, [tex]s_{n-1} = 3.5[/tex].[tex]n[/tex] is the sample size. For this question, [tex]n = 196[/tex].What is not given is[tex]t[/tex], the test statistics of the t-distribution. This value depends on the confidence level of the estimate (99.7% in this case.)Start by determining the degree of freedom [tex]df[/tex] of [tex]t[/tex]. The degree of freedom for a one-variable estimate is usually equal to [tex]n -1[/tex] (that is: sample size minus one.) For this question, [tex]n = 196[/tex] so [tex]df = 196 - 1 = 195[/tex].If [tex]T[/tex] represent a random variable that follows the [tex]t_{195}[/tex]-distribution, the value of t shall ensure that[tex]P(-t \le T \le t) = \text{Confidence Interval} = 0.997[/tex].The t-distribution is symmetric. As a result,[tex]\displaystyle P(T > t) = 0.997 + \frac{1}{2}\times (1 - 0.997) = 0.9985[/tex].Find the value of [tex]t[/tex] either with a t-distribution table or with technology. Keep in mind that for this question, [tex]df = n - 1 = 195[/tex].[tex]t \approx 3.00549[/tex].Apply the formula for the confidence interval:[tex]\displaystyle \left(\bar{x} - t\cdot \frac{s_{n-1}}{\sqrt{n}}, \; \bar{x} + t\cdot \frac{s_{n-1}}{\sqrt{n}}\right)[/tex].Lower bound:[tex]\displaystyle \bar{x} - t\cdot \frac{s_{n-1}}{\sqrt{n}} &= 12 - 3.00549\times \frac{3.5}{\sqrt{196}}\approx 11.2[/tex].Upper bound:[tex]\displaystyle \bar{x} + t\cdot \frac{s_{n-1}}{\sqrt{n}} &= 12 + 3.00549\times \frac{3.5}{\sqrt{196}}\approx 12.8[/tex].In other words, the 99.7% confidence interval for the actual concentration is [tex]\rm (11.2, 12.8)\; \mu g\cdot m^{-3}[/tex].