Question

(a) find the margin of error for the values of c, s, and n, and

(b) construct the confidence interval for μ using the t-distribution. Assume the population is normally distributed.

1. c=0.90, s=25.6, n=16, x= 72.1

2. c=0.95, s=1.1, n=25, x = 3.5

3. c = 0.98, s=0.9, n=12, x= 6.8

4. c = 0.99, s=16.5, n=20, x= 25.2

Answer 1

solution:-

1.a.margin of error
given that
c=0.90, s=25.6, n=16, x= 72.1
df = n - 1 = 16-1 = 15
critical value for t is 1.753
=> margin of error formula
=> t * s/sqrt(n)
=> 1.753 * 25.6/sqrt(16)
=> 11.2192

b.confidence interval formula
=> x +/- ME
=> 72.1 +/- 11.2192
=> (60.8808 , 83.3192)

2.a.margin of error
given that
c=0.95, s=1.1, n=25, x = 3.5
df = n - 1 = 25-1 = 24
critical value for t is 2.064
=> margin of error formula
=> t * s/sqrt(n)
=> 2.064 * 1.1/sqrt(25)
=> 0.4541

b.confidence interval formula
=> x +/- ME
=> 3.5 +/- 0.4541
=> (3.0459 , 3.9541)

3.a.margin of error
given that
c = 0.98, s=0.9, n=12, x= 6.8
df = n - 1 = 12-1 = 11
critical value for t is 2.718
=> margin of error formula
=> t * s/sqrt(n)
=> 2.718 * 0.9/sqrt(12)
=> 0.7062

b.confidence interval formula
=> x +/- ME
=> 6.8 +/- 0.7062
=> (6.0938 , 7.5062)

4.a.margin of error
given that
c = 0.99, s=16.5, n=20, x= 25.2
df = n - 1 = 20-1 = 19
critical value for t is 2.861
=> margin of error formula
=> t * s/sqrt(n)
=> 2.861 * 16.5/sqrt(20)
=> 10.5557

b.confidence interval formula
=> x +/- ME
=> 25.2 +/- 10.5557
=> (14.6443 , 35.7557)

all are rounded with four decimal places