Respuesta :
The critical value for a 95% C.I. is about [tex]z_{0.95}=1.96[/tex], so the 95% C.I. for this sample is
[tex]110\pm1.96\dfrac{10}{\sqrt{150}}=(108.4,111.6)[/tex]
[tex]110\pm1.96\dfrac{10}{\sqrt{150}}=(108.4,111.6)[/tex]
Answer:
(106.805, 113.195)
Step-by-step explanation:
Given that A random sample of 150 students is chosen from a population of 5,000 students.
Mean of sample is 110 and std dev =10
s=10
Std error of sample = 20/sqrt 150
=20/12.25
=1.63
Mean = 110
Since confidence level is 95% z = 1.96
Margin of error = 1.96 (std error) =3.195
Confidence interval lower bound = mean-3.195 = 106.805
Upper bound = mean+3.195 = 113.195