Respuesta :
Answer:
[tex] ME = \frac{25.09-21.01}{2}= 1.69[/tex]
The general formula for the margin of error is given by:
[tex] ME= t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
And for this case the width is:
[tex] Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
And if we decrease the confidence level from 95% to 90% then the critical value [tex]t_{\alpha/2}[/tex] would decrease and in effect the width for this new confidence interval decreases.
As confidence level decreases, the interval width decreases
Step-by-step explanation:
For this cae we know that the sample size selected is n =41
And we have a confidence interva for the true mean of foot length for students at a college selected.
The confidence interval is given by this formula:
[tex]\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
And for this case the 95% confidence interval is given by: (21.71,25.09)
A point of etimate for the true mean is given by:
[tex]\bar X = \frac{21.71+25.09}{2}= 23.4[/tex]
And the margin of error would be:
[tex] ME = \frac{25.09-21.01}{2}= 1.69[/tex]
The general formula for the margin of error is given by:
[tex] ME= t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
And for this case the width is:
[tex] Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
And if we decrease the confidence level from 95% to 90% then the critical value [tex]t_{\alpha/2}[/tex] would decrease and in effect the width for this new confidence interval decreases.
As confidence level decreases, the interval width decreases
