Respuesta :
Answer:
a) The 99% confidence interval would be given by (24.409;24.979) Â
b) n=464
Step-by-step explanation:
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". Â
The margin of error is the range of values below and above the sample statistic in a confidence interval. Â
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". Â
Part a
The confidence interval for the mean is given by the following formula: Â
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1) Â
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by: Â
[tex]df=n-1=47-1=46[/tex] Â
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 table to find the critical value. The excel command would be: "=-T.INV(0.025,46)".And we see that [tex]t_{\alpha/2}=2.01[/tex] Â
Now we have everything in order to replace into formula (1): Â
[tex]525-2.01\frac{75}{\sqrt{47}}=503.01[/tex] Â
[tex]525+2.01\frac{75}{\sqrt{47}}=546.99[/tex] Â
So on this case the 95% confidence interval would be given by (503.01;546.99)
Part b
The margin of error is given by this formula: Â
[tex] ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1) Â
And on this case we have that ME =7 msec, we are interested in order to find the value of n, if we solve n from equation (1) we got: Â
[tex]n=(\frac{t_{\alpha/2} s}{ME})^2[/tex] (2) Â
The critical value for 95% of confidence interval is provided, [tex]t_{\alpha/2}=2.01[/tex] from part a, replacing into formula (2) we got: Â
[tex]n=(\frac{2.01(75)}{7})^2 =463.79 \approx 464[/tex] Â
So the answer for this case would be n=464 rounded up to the nearest integer Â