Respuesta :
Answer:
a) Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]t_{\alpha/2}=1.96[/tex]
And the margin of error is given by:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
And replacing we got:
[tex] ME=1.96\frac{125}{\sqrt{16}}=61.25[/tex]
b) [tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =45 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
[tex]n=(\frac{1.960(125)}{45})^2 =29.64 \approx 30[/tex]
So the answer for this case would be n=30 rounded up to the nearest integer
b. 30
c) And for this case the conditions in order to use the confidence interval are satisfied since:
a. Yes. The distribution of sample means is normal because the data are normal.
Step-by-step explanation:
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 for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma =125[/tex] represent the population standard deviation
n=16 represent the sample size
Part a
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]t_{\alpha/2}=1.96[/tex]
And the margin of error is given by:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
And replacing we got:
[tex] ME=1.96\frac{125}{\sqrt{16}}=61.25[/tex]
c. 61.25 kg
Part b
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =45 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
[tex]n=(\frac{1.960(125)}{45})^2 =29.64 \approx 30[/tex]
So the answer for this case would be n=30 rounded up to the nearest integer
b. 30
Part c
And for this case the conditions in order to use the confidence interval are satisfied since:
a. Yes. The distribution of sample means is normal because the data are normal.
