Respuesta :
Answer:
a) [tex]4.85-2.33\frac{0.75}{\sqrt{20}}=4.46[/tex]
[tex]4.85+2.33\frac{0.75}{\sqrt{20}}=5.24[/tex]
b) [tex]4.56-2.33\frac{0.75}{\sqrt{16}}=4.12[/tex]
[tex]4.56+2.33\frac{0.75}{\sqrt{16}}=4.99[/tex]
c) [tex]n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55[/tex]
Step-by-step explanation:
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that [tex]z_{\alpha/2}=2.33[/tex]
Now we have everything in order to replace into formula (1):
[tex]4.85-2.33\frac{0.75}{\sqrt{20}}=4.46[/tex]
[tex]4.85+2.33\frac{0.75}{\sqrt{20}}=5.24[/tex]
Part b
[tex]4.56-2.33\frac{0.75}{\sqrt{16}}=4.12[/tex]
[tex]4.56+2.33\frac{0.75}{\sqrt{16}}=4.99[/tex]
Part c
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =0.4/2 =0.2 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)
The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55[/tex]
