Question

(Hypothetical.) Suppose five measurements of the volume of a liquid yield the following data (in milliliters).

Answer 1

Answer:

(48.106 ; 53.494)

Step-by-step explanation:

Given the data:

X : 52 48 49 52 53

Sample mean = ΣX / n

n = 5

Sample mean, xbar = 254 / 5 = 50.8

Standard deviation, s = 2.17 (using calculator)

The standard error (SE) : s/√n =2.17/√5 = 0.970

The degree of freedom, df = n-1

df = 5 - 1 = 4

Tscore(0.05, 4) = 2.776

Confidence interval :

Xbar ± Tscore*standard error

50.8 ± (2.776 * 0.970)

50.8 ± 2.694

Lower boundary = 50.8 - 2.694 = 48.106

Upper boundary = 50.8 + 2.694 = 53.494

(48.106 ; 53.494)