Respuesta :
Answer:
II and III
See explanation above
Step-by-step explanation:
When all other rings remain the same, which of the following conditions would have resulted in a wider interval than the one constructed?
I. A sample size of 20 with 95 percent confidence
II. A sample size of 15 with 99 percent of confidence
III. A sample size of 12 with 95 percent confidence
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
[tex]\bar X[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
The margin of error is given by:
[tex]ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
I. A sample size of 20 with 95 percent confidence
The original sample size was 15, the degrees of freedom for the original interval would be n-1=14 , the value for [tex]\alpha=0.05[/tex] and [tex]\alpha/2=0.025[/tex] and then the critical value is[tex]t_{critc}=2.14[/tex]
And then the Margin of error would be:
[tex]ME=2.14 \frac{s}{\sqrt{15}}=0.553 s[/tex]
For the new sample size n=20 the degrees of freedom for the new interval interval would be n-1=19 , the value for [tex]\alpha=0.05[/tex] and [tex]\alpha/2=0.025[/tex] and then the critical value is[tex]t_{critc}=2.09[/tex]
And then the Margin error would be:
[tex]ME=2.09 \frac{s}{\sqrt{20}}=0.467 s[/tex]
So then we have a lower margin of error so then we will have a shorter interval
II. A sample size of 15 with 99 percent of confidence
The original sample size was 15, the degrees of freedom for the original interval would be n-1=14 , the value for [tex]\alpha=0.05[/tex] and [tex]\alpha/2=0.025[/tex] and then the critical value is[tex]t_{critc}=2.14[/tex]
And then the Margin of error would be:
[tex]ME=2.14 \frac{s}{\sqrt{15}}=0.553 s[/tex]
For the new interval the sample size n=15 the degrees of freedom for the new interval interval would be n-1=14 , the value for [tex]\alpha=0.01[/tex] and [tex]\alpha/2=0.005[/tex] and then the critical value is[tex]t_{critc}=2.98[/tex]
And then the Margin error would be:
[tex]ME=2.98 \frac{s}{\sqrt{15}}=0.769 s[/tex]
So then we have a greater margin of error so then we will have a wider interval. And this makes sense since with more confidence the interval needs to be wider.
III. A sample size of 12 with 95 percent confidence
The original sample size was 15, the degrees of freedom for the original interval would be n-1=14 , the value for [tex]\alpha=0.05[/tex] and [tex]\alpha/2=0.025[/tex] and then the critical value is[tex]t_{critc}=2.14[/tex]
And then the Margin of error would be:
[tex]ME=2.14 \frac{s}{\sqrt{15}}=0.553 s[/tex]
For the new interval the sample size n=12 the degrees of freedom for the new interval interval would be n-1=11 , the value for [tex]\alpha=0.05[/tex] and [tex]\alpha/2=0.025[/tex] and then the critical value is[tex]t_{critc}=2.20[/tex]
And then the Margin error would be:
[tex]ME=2.20 \frac{s}{\sqrt{12}}=0.635 s[/tex]
So then we have a greater margin of error so then we will have a wider interval.
The conditions that have resulted in a wider interval are II and III and this can be determined by using the formula of the margin of error.
Given :
Last week, a random sample of 15 rings was measured, and the mean and standard deviation of the sample was used to construct a 95 percent confidence interval for the population mean width of the rings.
Check all the conditions in order to determine the correct one:
I. A sample size of 20 with 95 percent confidence
The formula of the margin of error is given by:
[tex]\rm ME = t_{\alpha /2}\dfrac{s}{\sqrt{n} }[/tex]
Now, substitute the value of known terms in the above formula.
[tex]\rm ME = 2.09*\times \dfrac{s}{\sqrt{20} }=0.467s[/tex]
When t = 2.14 and n = 15, ME becomes:
[tex]\rm ME = 2.14*\times \dfrac{s}{\sqrt{15} }=0.553s[/tex]
In this condition, there is a lower margin of error so the interval is shorter.
II. A sample size of 15 with 99 percent confidence
At t = 2.14 and n = 15, the ME becomes:
[tex]\rm ME = 2.14*\times \dfrac{s}{\sqrt{15} }=0.553s[/tex]
At t = 2.98 and n = 15, the ME becomes:
[tex]\rm ME = 2.98\times \dfrac{s}{\sqrt{15} }=0.769s[/tex]
In this condition, there is a greater margin of error so the interval is wider.
III. A sample size of 12 with 95 percent confidence
At t = 2.14 and n = 15, the ME becomes:
[tex]\rm ME = 2.14*\times \dfrac{s}{\sqrt{15} }=0.553s[/tex]
At t = 2.20 and n = 12, the ME becomes:
[tex]\rm ME = 2.20\times \dfrac{s}{\sqrt{12} }=0.635s[/tex]
In this condition, there is a greater margin of error so the interval is wider.
For more information, refer to the link given below:
https://brainly.com/question/21586810
