Respuesta :
Answer:
a) [tex] 78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416[/tex]
[tex] 78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084[/tex]
b) [tex] 78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989[/tex]
[tex] 78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511[/tex]
c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower
d) [tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
Solving for n we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
And replacing the info we have:
[tex]n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55[/tex]
Step-by-step explanation:
Part a
For this case we have the following data given
[tex]\bar X = 78.25[/tex] represent the sample mean for the customer order totals
[tex]\sigma =22.50[/tex] represent the population deviation
[tex] n= 40[/tex] represent the sample size selected
The confidence level is 90% or 0.90 and the significance level would be [tex]\alpha=0.1[/tex] and [tex]\alpha/2 = 0.05[/tex] and the critical value from the normal standard distirbution would be given by:
[tex] z_{\alpha/2}=1.64[/tex]
And the confidence interval is given by:
[tex] \bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
And replacing we got:
[tex] 78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416[/tex]
[tex] 78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084[/tex]
Part b
The sample size is now n = 75, but the same confidence so the new interval would be:
[tex] 78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989[/tex]
[tex] 78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511[/tex]
Part c
For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower
Part d
The margin of error is given by:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
Solving for n we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
And replacing the info we have:
[tex]n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55[/tex]
