Respuesta :
Answer:
(a) n≥4 (b) n≥7
Step-by-step explanation:
The confidence interval is defined as
[tex]\bar{X}\pm z*\frac{s}{\sqrt{n} }[/tex]
So the difference between upper limit (UL) and lower limit (LL) must be
[tex]UL-LL=(\bar{X}+ z*\frac{s}{\sqrt{n} })-(\bar{X}- z*\frac{s}{\sqrt{n} })\\\\UL-LL=2*z*\frac{s}{\sqrt{n} }[/tex]
We can clear the number of observations rearranging the last equation
[tex]UL-LL=2*z*\frac{s}{\sqrt{n} }\\\\\sqrt{n}=\frac{2*z*s}{(UL-LL)}\\ \\n=(\frac{2*z*s}{(UL-LL)})^{2}[/tex]
(a) UL-LL must be less than 30. For a 95% CI, z=1.96.
[tex]n=(\frac{2*z*s}{(UL-LL)})^{2}= (\frac{2*1.96*15}{30} )^{2}=(\frac{58.8}{30})^{2} =1.96^{2}=3.8416[/tex]
The sample needs to be at least 4 observations.
(b) UL-LL must be less than 30. For a 99% CI, z=2.576.
[tex]n=(\frac{2*z*s}{(UL-LL)})^{2}= (\frac{2*2.576*15}{30} )^{2}=(\frac{77.28}{30})^{2} =2.576^{2}=6.6358[/tex]
The sample needs to be at least 7 observations.
The confidence interval is the range of value, of a sample that represents the population.
- The sample size must not be greater than 4, if the 95% CI is not greater than 30
- The sample size must not be greater than 7, if the 99% CI is not greater than 30
The given parameter is:
[tex]\mathbf{\sigma = 15}[/tex]
ME is calculated as:
[tex]\mathbf{ME = z_{critical} \times \frac{\sigma}{\sqrt n}}[/tex]
Takes square of both sides
[tex]\mathbf{ME^2 = z_{critical}^2 \times \frac{\sigma^2}{n}}[/tex]
Make n the subject
[tex]\mathbf{n = z_{critical}^2 \times \frac{\sigma^2}{ME^2}}[/tex]
(a) The sample size, if 95% CI is not greater than 30
We have:
[tex]\mathbf{\sigma = 15}[/tex]
For a CI of 30, ME = 15
The z critical at 95% CI is 1.96
So, we have:
[tex]\mathbf{n = z_{critical}^2 \times \frac{\sigma^2}{ME^2}}[/tex]
[tex]\mathbf{n = 1.96^2 \times \frac{15^2}{15^2}}[/tex]
[tex]\mathbf{n = 3.8416\times \frac{1}{1}}[/tex]
[tex]\mathbf{n = 3.8416}[/tex]
Approximate
[tex]\mathbf{n = 4}[/tex]
The sample size must not be greater than 4, if the 95% CI is not greater than 30
(b) The sample size, if 99% CI is not greater than 30
We have:
[tex]\mathbf{\sigma = 15}[/tex]
For a CI of 30, ME = 15
The z critical at 99% CI is2.576
So, we have:
[tex]\mathbf{n = z_{critical}^2 \times \frac{\sigma^2}{ME^2}}[/tex]
[tex]\mathbf{n = 2.576^2 \times \frac{15^2}{15^2}}[/tex]
[tex]\mathbf{n = 6.635776 \times \frac{1}{1}}[/tex]
[tex]\mathbf{n = 6.635776 }[/tex]
Approximate
[tex]\mathbf{n = 7}[/tex]
The sample size must not be greater than 7, if the 99% CI is not greater than 30
Read more about confidence intervals at:
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