Question

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

52 48 49 52 5.3
Assume that the different measurements are due only to measurement error, and that measurement error is Normally distributed. The goal is to set up an approximate 95% confidence interval for the correct measurement. In each part, answer correct to three decimal places. First find the average and SD of the data. Calculator welcome and recommended. Average of the data: SD of the data (remember, it's a small sample): Estimate the correct measurement. Attach a give-or-take value to the estimate (that is, estimate the standard error). There are 5 measurements. If measurements follow a Normal distribution (e.g., if they vary only due to chance measurement error), then the standardized average of 5 measurements follows a Student t-distribution with So, for 95% confidence, we include our estimate. degrees of freedom. standard error(s) on each side of An approximate 95% confidence interval for the correct measurement is ______.

Answer 1

(48.106 ; 53.494)

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)