Question

Consider a normal population distribution with the value of σ known.

(a) What is the confidence level for the interval x ± 2.88σ/sqrt(n) ? (Round your answer to one decimal place.)
___________%

(b) What is the confidence level for the interval x ± 1.42σ/sqrt(n) ? (Round your answer to one decimal place.)
___________%

(c) What value of z_α/2 in the CI formula below results in a confidence level of 99.7%? (Round your answer to two decimal places.)

( x − z_α/2 ·σ/sqrt(n), x + z_α/2 ·σ/sqrt(n) )

z_α/2 = ____________

(d) Answer the question posed in part (c) for a confidence level of 78%. (Round your answer to two decimal places.)
z_α/2 = ________

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution :

(a) given confidence interval is x +/- 2.88σ/sqrt(n)

=> P(-2.88 < z < 2.88) = 0.996 = 99.6%

=> the confidence level is 99.6%

(b) given confidence interval is x +/- 1.42σ/sqrt(n)

=> P(-1.42 < z < 1.42) = 0.8444

= 84.44%

= 84.4% (rounded)

(c)

=> for 99.7% confidence interval , Z = 3.00

=> zα/2 = 3.00

(d)
=> for 78% confidence interval , Z = 1.23

=> zα/2 = 1.23