Mathematics, 21.04.2020 15:48, iwannasleep

In a sample of nequals16 lichen specimens, the researchers found the mean and standard deviation of the amount of the radioactive element, cesium-137, that was present to be 0.009 and 0.003 microcurie per milliliter, respectively. Suppose the researchers want to increase the sample size in order to estimate the mean mu to within 0.001 microcurie per milliliter of its true value, using a 95% confidence interval. Complete parts a through c. a. What is the confidence level desired by the researchers? The confidence level is 95. b. What is the sampling error desired by the researchers? The sampling error is 0.001. c. Compute the sample size necessary to obtain the desired estimate. The sample size is nothing. (Type a whole number.)

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answered: supersquad907p2m3vo

(a) The confidence level desired by the researchers is 95%.

(b) The sampling error is 0.001 microcurie per millilitre.

Step-by-step explanation:

The mean and standard deviation of the amount of the radioactive element, cesium-137 present in a sample of n = 16 lichen specimen are:

$\bar x=0.009\\s=0.003$

Now it is provided that the researchers want to increase the sample size in order to estimate the mean μ to within 0.001 microcurie per millilitre of its true value, using a 95% confidence interval.

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x \pm z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

(a)

The confidence level is the probability that a particular value of the parameter under study falls within a specific interval of values.

In this case the researches wants to estimate the mean using the 95% confidence interval.

Thus, the confidence level desired by the researchers is 95%.

(b)

In case of statistical analysis, during the computation of a certain statistic, to estimate the value of the parameter under study, certain error occurs which are known as the sampling error.

In case of the estimate of parameter using a confidence interval the sampling error is known as the margin of error.

In this case the margin of error is 0.001 microcurie per millilitre.

(c)

The margin of error is computed using the formula:

$MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the sample size value as follows:

$MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times s}{MOE}]^{2}$

$=[\frac{1.96\times 0.003}{0.001}]^{2}$

$=34.5744\\\approx36$

Thus, the sample size necessary to obtain the desired estimate is 36.

answered: Guest

introduction, methodology, results, discussion

step-by-step explanation:

an observational study is one which consists of actual observation to record data and then analyse the data.

there are a number of factors that should be analyzed before starting an observational study.

the main four factors are:

i) introduction: this should be analysed to check whether the purpose of the study is correctly established and the data recorded are appropriate for the study

methodology: the method of collecting the data, such as whether randomly selected, and whether sample size matches with the population to be studied, whether questions asked are easy to answer by common men, etc.

results: then results such as summary of data with comments should be studied thoroughly

discussion: the last is the discussion, this can be either by management people or persons who are all interested , etc.

answered: Guest

answer: .75 km

step-by-step explanation: