Answer:
Confidence interval [tex]159.366[/tex] to [tex]180.633[/tex]
Margin of error is 10.633
Step-by-step explanation:
Confidence interval formula : [tex]\bar{x}\pm z \times \frac{\sigma}{\sqrt{n}}[/tex]
Mean = [tex]\bar{x}=170[/tex]
Standard deviation = [tex]\sigma= 33[/tex]
Sample size = n = 37
Confidence interval = 95%
z value fro 95% confidence level is 1.96
Substitute the values in the formula :
Confidence interval [tex]\bar{x}\pm z \times \frac{\sigma}{\sqrt{n}}[/tex]
Confidence interval [tex]170 \pm 1.96 \times \frac{33}{\sqrt{37}}[/tex]
Confidence interval [tex]170 - 1.96 \times \frac{33}{\sqrt{37}}[/tex] to [tex]170 + 1.96 \times \frac{33}{\sqrt{37}}[/tex]
Confidence interval [tex]159.366[/tex] to [tex]180.633[/tex]
Formula of margin of error : [tex]z \times \frac{\sigma}{\sqrt{n}}[/tex]
Margin of error : [tex]1.96 \times \frac{33}{\sqrt{37}}[/tex]
= [tex]10.633[/tex]
Hence Confidence interval [tex]159.366[/tex] to [tex]180.633[/tex] and Margin of error is 10.633
