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Mathematics, 27.06.2019 09:10 bdMack932

For a group experiment, your science class measured the fine-particulate
concentrations in the air at random places around campus, and estimated a
sample average of 12 ug/m3 (micrograms per cubic meter). if 196 readings
were taken, and the standard deviation of the sample measurements was
3.5 pg/m3, you are 99.7% confident that actual concentration of fine
particulates at the school is

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Answer from: cbless972

The 99.7% confidence interval for the mean here is

$\rm (11.2, 12.8)\; \mu g\cdot m^{-3}$ .

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?

$\displaystyle \left(\bar{x} - t\cdot \frac{s_{n-1}}{\sqrt{n}}, \; \bar{x} + t\cdot \frac{s_{n-1}}{\sqrt{n}}\right)$ ,

where

$\bar{x}$ is the unbiased estimate of the population mean (a.k.a sample mean.) For this question, $\bar{x} = 12$ . $s_{n-1}$ is the unbiased estimate of the standard deviation (the square root of variance) of the population. For this question, $s_{n-1} = 3.5$ .

is the sample size. For this question, n = 196

n = 196

.

What is not given is

, 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 of . The degree of freedom for a one-variable estimate is usually equal to n -1 (that is: sample size minus one.) For this question, n = 196 so df = 196 - 1 = 195 .

If represent a random variable that follows the $t_{195}$ -distribution, the value of t shall ensure that

$P(-t \le T \le t) = \text{Confidence Interval} = 0.997$ .

The t-distribution is symmetric. As a result,

$\displaystyle P(T t) = 0.997 + \frac{1}{2}\times (1 - 0.997) = 0.9985$ .

Find the value of either with a t-distribution table or with technology. Keep in mind that for this question, df = n - 1 = 195 .

$t \approx 3.00549$ .

Apply the formula for the confidence interval:

$\displaystyle \left(\bar{x} - t\cdot \frac{s_{n-1}}{\sqrt{n}}, \; \bar{x} + t\cdot \frac{s_{n-1}}{\sqrt{n}}\right)$ .

Lower bound:

$\displaystyle \bar{x} - t\cdot \frac{s_{n-1}}{\sqrt{n}} &= 12 - 3.00549\times \frac{3.5}{\sqrt{196}}\approx 11.2$ .

Upper bound:

$\displaystyle \bar{x} + t\cdot \frac{s_{n-1}}{\sqrt{n}} &= 12 + 3.00549\times \frac{3.5}{\sqrt{196}}\approx 12.8$ .

In other words, the 99.7% confidence interval for the actual concentration is $\rm (11.2, 12.8)\; \mu g\cdot m^{-3}$ .

ansver

Answer from: Quest

answer: ouch. thats difficult

step-by-step explanation:

ansver

Answer from: Quest

your answer is d 2,500

step-by-step explanation:

The circle graph represents a family’s monthly budget if the total monthly budget is 5,000 how much

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