Question

Measurements of ten samples are made for the thickness of specific mechanical components, which gives 21.628, 22.962, 20.439, 21.589, 20.357, 20.341, 22.091, 20.824, 19.857, and 22.171 mm. Determine (a) the mean value and standard deviation of the ten measurement, and (b) the range of population mean for a 95% confidence level.

Answer 1

(a) The mean value is 21.226 mm. The standard deviation is 0.956 mm

(b) 95% confidence is between a lower limit of 20.542 mm and an upper limit of 21.910 mm

Step-by-step explanation:

(a) Mean = summation of thicknesses ÷ number of measurements (n) = (21.628+22.962+20.439+21.589+20.357+20.341+22.091+20.824+19.857+22.171) ÷ 10 = 212.259 ÷ 10 = 21.226 mm (to 3 decimal places)

Standard deviation = sqrt [(summation (thickness - mean)^2 ÷ number of measurements] = sqrt [((21.628-21.226)^2 + (22.962-21.226)^2 + (20.439-21.226)^2 + (21.589-21.226)^2 + (20.357-21.226)^2 + (20.341-21.226)^2 + (22.091-21.226)^2 + (20.824-21.226)^2 + (19.857-21.226)^2 + (22.171-21.226)^2) ÷ 10] = 0.956 mm

b) Confidence Interval = mean + or - Error margin (E)

mean = 21.226 mm

sd = 0.956 mm

n = 10

degree of freedom = n - 1 = 10 - 1 = 9

confidence level = 85%

t-value corresponding to 9 degrees of freedom and 95% confidence level is 2.262

E = t × sd/√n = 2.262 × 0.956/√10 = 0.684 mm

Lower limit = mean - E = 21.226 - 0.684 = 20.542 mm

Upper limit = mean + E = 21.226 + 0.684 = 21.910 mm

95% confidence interval for the population mean is between 20.542 and 21.910 mm