Question 2 College of Arts and Sciences Department of...

Question 2 College of Arts and Sciences Department of Biological and Environmental Sciences In preparation for an environmental determination of lead (Pb) in soil, a standard reference material (SRM) was analyzed using the recommended laboratory method. The SRM is certified to contain 13.9 µg dL$^{-1}$ (micrograms per deciliter) of Pb. An environmental analyst obtained the following values for Pb in the SRM: 15.7, 16.2, 13.2, 14.8, and 15.1 µg dL$^{-1}$ [Mean = 15.0 and Standard deviation = 1.1]. Determine whether or not the analyst obtained a Pb concentration in the SRM that was significantly different from the certified value at the 95% confidence level.

Transcript

00:01 So to begin, for finding the f value, that is going to be the ratio of the two sample variances.

00:08 So, looking at our data, we would have s1, or s -s in this case, is 0 .199, or 199, so we have 0 .199 to the power of 2 as our numerator.

00:21 We divide by 0 .110 squared to get an f statistic of 3 .27, roughly.

00:30 Now, i don't have access to the particular table or a table value that you're looking for there.

00:39 But we can find, or what i'll do is i'll use my software here for finding the inverse cdf for...

00:49 One moment here.

00:50 Yeah, so i want to find the inverse cdf for an f -ratio distribution, where the ratios here will be the...

01:03 Or pardon me, the numerator and denominator degrees of freedom will be equal to the sample size minus 1 for the two different samples.

01:11 So f ratio distribution 5 -5.

01:13 We have sample size 6 for both s -1 and s -2, and then we want to find the probability to, this will give me the probability to the left of my result.

01:28 Oh, pardon me, no, i want to do, not inverse cdf, excuse me, that should just be cdf.

01:33 So we have a probability to the left of 0 .89, which then means that our probability to the right is 0 .197.

01:43 So because of that, we have such a large probability to the right, that means that we would be, in a hypothesis test for difference of, standard, or difference of variances, we would fail to reject the null hypothesis.

02:02 So, we would then conclude that they do not significantly differ.

02:12 So, using that then, we're asked to find the appropriate t value.

02:18 Well, since we know that our variances do not differ significantly, the way that we'll calculate our t score is that that's going to be equal to x bar 1 minus x bar 2, divided by the pooled standard deviation times the square root of 1 over n1, 1 plus 1 over n2, where the pooled standard deviation, we calculate by taking, let's see here, it's the square root of sample size 1 minus 1 times variance 1, 0 .192 squared, plus sample size 2 minus 1, so that's 5 times variance 2, 0 .11 to the power of 2, divided by n1 plus n2 minus 2.

03:06 So we have our pooled standard deviation is equal to roughly 0 .1608...

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