Question

A laboratory procedure is used to measure the level of cadmium in soil. It is applied to a specimen that has a controlled level of 1.5 μg/g of cadmium. Three measurements are obtained by splitting the specimen into three parts of equal weight and applying the procedure to each part. Here are the results: 1.64, 1.85, and 1.94. (a) Does a 95% confidence interval give an indication that the laboratory procedure may be biased? b. What are your assumptions that you are making to produce the confidence interval? second part of question: For 95% confidence interval of (1.38, 1.92) is given for the mean of a population based on the normal distribution. Re-express this as a 90% confidence interva.

Answer 1

Answer:

a)

At the point when the example is part into three sections, the CI condition has root n in the denominator which will become n/3 for every one of the split examples. For developing a certainty interim, we should take an adequately huge n more prominent than 30. Expecting an adequately enormous n and afterward representing variety in estimations utilizing standard deviation, we can make a decent certainty interim in order to wipe out any inclination that can emerge out of the strategy.

b)

To give 90% confidence interval

Given,

95% CI = (1.38,1.92)

consider,

95% CI = xbar +/- 1.96*s/sqrt(n)

xbar - 1.96*s/sqrt(n) = 1.38 ------------> (1)

xbar + 1.96*s/sqrt(n) = 1.92 ------------> (2)

Here by adding both, we get

2xbar = 3.3

xbar = 3.3/2 = 1.65

Now substitute xbar = 1.65 in (2)

1.65 + 1.96*s/sqrt(n) = 1.92

s/sqrt(n) = (1.92 - 1.65)/1.96

= 0.1378

Now consider,

90% CI = xbar +/- 1.645*s/sqrt(n)

substitute the known values in above expression

= 1.65 +/- 1.645*0.1378

= 1.65 +/- 0.2267

90% CI = (1.4233 , 1.8767)