Need help, please show work In a sample of n 15 lichen specimens, the researcher

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Need help, please show work

In a sample of n 15 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 0.005 microcurie per illiter, respectively. Suppose the researchers want to increase the sample size in order to estimate the mean 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 b. What is the sampling error desired by the researchers? The sampling error is c. Compute the sample size necessary to obtain the desired estimate. The sample size is(Type a whole number.)

Explanation / Answer

a) Mean = 0.009
t critical = 2.14
sM = ((0.005)^2/15) = 0.0013

= M ± t(sM)
= 0.009 ± 2.14*0.0013
= 0.009 ± 0.00278

M = 0.009, 95% CI [0.00622, 0.01178].

You can be 95% confident that the population mean () falls between 0.00623 and 0.01177.

b) sM = ((0.005)^2/15) = 0.0013

c) Step 1: Find t alpha/2 by dividing the confidence interval by two, and looking that area up in the t-table:

.95/2 = 0.475. The closest t-score for 0.475 and with 15-1=14 d.f is 2.14

Step 2: Multiply step 1 by the standard deviation.
0.005 * 2.14 = 0.0107

Step 3: Divide Step 2 by the margin of error. Our margin of error (from the question), is 0.001.
0.0107/0.001 = 10.7

Step 4: Square Step 3.
10.7*10.7 = 114.49= 114 sample size

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