Answer :
Answer:
e)The width of I400 will be 1/4 times the width of I100.
Step-by-step explanation:
If the confidence interval is 95% = 0.95, therefore α = 1 - 0.95 = 0.05.
[tex]\frac{\alpha }{2} =\frac{0.05}{2}=0.0025[/tex]
The width of the confidence (W) is given by the equation:
[tex]W=\frac{z_{\frac{\alpha }{2} }*\sigma}{\sqrt{n} }[/tex], where n is the sample size, [tex]\sigma[/tex] is the standard deviation and [tex]z_{\frac{\alpha }{2} }[/tex] is the z score of α/2
From the equation, we can see that the width is inversely proportional to the square root of the sample size.
Since nterval, I400, will be constructed using samples of size 400 from each of R and J, and the other confidence interval, I100, will be constructed using samples of size 100 from each of R and J. and all other things remain the same.
Let [tex]W_{I4[/tex] be the width of I400 and [tex]W_{I1[/tex] be the width of I100. Therefore:
[tex]W_{I4}=\frac{z_{\frac{\alpha }{2} }*\sigma}{\sqrt{400} }=\frac{z_{\frac{\alpha }{2} }*\sigma}{4 }\\And, W_{I1}=\frac{z_{\frac{\alpha }{2} }*\sigma}{\sqrt{100} }=\frac{z_{\frac{\alpha }{2} }*\sigma}{1 }\\Therefore,\frac{W_{I4}}{W_{I1}} =\frac{\frac{z_{\frac{\alpha }{2} }*\sigma}{4 }}{\frac{z_{\frac{\alpha }{2} }*\sigma}{1 }}\\\frac{W_{I4}}{W_{I1}}=\frac{1}{4}\\ W_{I4}=\frac{1}{4}W_{I1}[/tex]
The width of I400 will be 1/4 times the width of I100.
You can use the formula for calculating the width of the confidence interval.
The option that describes the relationship between the two confidence intervals is:
Option d: The width of i400 will be 1/2 time the width of i100
How to calculate width of the confidence interval with normal distribution?
(Assuming that the population was normally distributed)
The limits of confidence interval is calculated by:
[tex]CI = \overline{x} \pm Z_{\alpha/2} \dfrac{\sigma}{\sqrt{n}}[/tex] where the n denotes the sample size and alpha denotes level of significance.
The width is calculated by subtracting upper limit by lower limit of confidence interval which results in:
[tex]W = 2Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex] (two sided width)
For 95% confidence interval, the level of significance = 1-0.95 = 0.05
For first sample, the width will be
[tex]W_{i400} = \dfrac{Z_{\alpha/2} 2\sigma}{\sqrt{400}} = \dfrac{Z_{\alpha/2}\sigma}{10}[/tex]
For second sample, the width will be
[tex]W_{i100} = \dfrac{Z_{\alpha/2} 2\sigma}{\sqrt{100}} = \dfrac{Z_{\alpha/2}\sigma}{5}[/tex]
Thus, we have:
[tex]\dfrac{W_{i400}}{W_{i100}} = \dfrac{\dfrac{Z_{\alpha/2} 2\sigma}{\sqrt{400}}}{\dfrac{Z_{\alpha/2} 2\sigma}{\sqrt{100}}} = \dfrac{1}{2}\\\\W_{i400} = \dfrac{1}{2} \times W_{i100}[/tex]
Thus, the relationship between the two confidence interval is described by:
Option d)The width of i400 will be 1/2 times the width of i100
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