Respuesta :
Answer:
a) [tex]654.16-2.01\frac{165.4}{\sqrt{52}}=608.06[/tex]
[tex]654.16+2.01\frac{165.4}{\sqrt{52}}=700.26[/tex]
b) [tex]n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189[/tex]
So the answer for this case would be n=189 rounded up to the nearest integer
Step-by-step explanation:
Part a
[tex]\bar X=654.16[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=165.4 represent the sample standard deviation
n =52represent 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] (1)
The degrees of freedom aregiven by:
[tex]df=n-1=52-1=51[/tex]
Since the Confidence is 0.95 or 95%, the significance [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value would be [tex]t_{\alpha/2}=2.01[/tex]
Now we have everything in order to replace into formula (1):
[tex]654.16-2.01\frac{165.4}{\sqrt{52}}=608.06[/tex]
[tex]654.16+2.01\frac{165.4}{\sqrt{52}}=700.26[/tex]
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 =25 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} s}{ME})^2[/tex] (b)
The critical value for this case wuld be [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189[/tex]
So the answer for this case would be n=189 rounded up to the nearest integer
