Respuesta :
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 [tex]\sigma[/tex] be the standard deviation, then
[tex]\text{Range}\approx4\times\sigma\\\\\Rightarrow\ \sigma\approx\dfrac{\text{Range}}{4}=\dfrac{650}{4}=162.5[/tex]
Thus, [tex]\sigma\approx162.5[/tex]
For confidence interval of 95% , the critical z value = [tex]z_{\alpha/2}=1.96[/tex]
Formula to find the sample size : [tex](\dfrac{z_{\alpha/2}\ \sigma}{E})^2[/tex]
Now, for the margin of error of E=[tex]\pm 10[/tex], we have
[tex](\dfrac{1.96\times162.5}{10})^2=1014.4225\approx1015[/tex]
Hence, the required sample size = 1015
Using the uniform and the z-distribution, it is found that 1353 specimens should be tested.
For an uniform distribution of bounds a and b, the standard deviation is given by:
[tex]\sigma = \sqrt{\frac{(b - a)^2}{12}}[/tex]
In this problem, [tex]a = 50, b = 700[/tex], thus, the estimate is:
[tex]\sigma = \sqrt{\frac{(700 - 50)^2}{12}} = 187.64[/tex]
The margin of error of a z-confidence interval is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- z is the critical value.
- [tex]\sigma[/tex] is the population standard deviation.
- n is the sample size.
The first step is finding the critical value, which is z with a p-value of [tex]\frac{1 + \alpha}{2}[/tex], in which [tex]\alpha[/tex] is the confidence level.
In this problem, [tex]\alpha = 0.95[/tex], thus, z with a p-value of [tex]\frac{1 + 0.95}{2} = 0.975[/tex], which means that it is z = 1.96.
Estimate of the standard deviation is of 187.64, thus, [tex]\sigma = 187.64[/tex].
We want the sample for a margin of error of 10, thus, we have to solve for n when [tex]M = 10[/tex]. Then:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]10 = 1.96\frac{187.64}{\sqrt{n}}[/tex]
[tex]10\sqrt{n} = 1.96(187.64)[/tex]
Simplifying by 10:
[tex]\sqrt{n} = 0.196(187.64)[/tex]
[tex](\sqrt{n})^2 = [0.196(187.64)]^2[/tex]
[tex]n = 1352.6[/tex]
Rounding up:
1353 specimens should be tested.
A similar problem is given at https://brainly.com/question/25177638
