Question

1. Find the margin of error for the given values of​ c, sigma​, and c= 0.90​, sigma =2.3​, n= 49

2.

Use the confidence interval to find the margin of error and the sample mean.

(0.512,0.690)

3.

Find the minimum sample size n needed to estimate μ for the given values of​ c, σ​, and E. c=0.98​, σ=5.8​, and E =2

Assume that a preliminary sample has at least 30 members.

Answer 1

Solution :

Given that,


Population standard deviation =    = 2.3

Sample size = n =49

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2 * ( /n)

= 1.645 * (2.3 /  49 )

E= 0.5405

(b)

Solution :

  = (Lower confidence interval + Upper confidence interval ) / 2

Sample mean = = (0.512+0.690) / 2=0.601

Sample mean = =0.601

Margin of error = E = Upper confidence interval -   = 0.690-0.601=0.089

Margin of error = E = 0.089

(c)

Solution :

Given that,

standard deviation =   =5.8

Margin of error = E = 2

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

sample size = n = [Z/2* / E] 2

n = ( 2.326* 5.8 / 2 )2

n =45.50

Sample size = n =46