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postnew [5]
3 years ago
8

The Bureau of Alcohol, Tobacco, and Firearms (BATF) has been concerned about lead levels in California wines. In a previous test

ing of wine specimens, lead levels ranging from 50 to 700 parts per billion were recorded. How many wine specimens should be tested if the BATF wishes to estimate the true mean lead level for California wines to within 10 parts per billion with 95% confidence?
Mathematics
1 answer:
scoundrel [369]3 years ago
4 0

Answer:  1015

Step-by-step explanation:

Given :In a previous testing of wine specimens, lead levels ranging from 50 to 700 parts per billion were recorded.

Here , Range = 700-50=650    [∵ Range = Max-Min]

According to the thumb rule , the range is approximately 4 times the standard deviation.

Let \sigma be the standard deviation, then

\text{Range}\approx4\times\sigma\\\\\Rightarrow\ \sigma\approx\dfrac{\text{Range}}{4}=\dfrac{650}{4}=162.5

Thus, \sigma\approx162.5

For confidence interval of 95% , the critical z value = z_{\alpha/2}=1.96

Formula to find the sample size : (\dfrac{z_{\alpha/2}\ \sigma}{E})^2

Now, for the margin of error of E=\pm 10, we have

(\dfrac{1.96\times162.5}{10})^2=1014.4225\approx1015

Hence, the required sample size = 1015

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Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

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Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

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P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.842

And if we solve for a we got

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