Question

To characterize random uncertainty of a pressure measuring technique, twelve pressure measurements were made of a certain constant pressure source, giving the following results in kPa: 125, 128, 129, 122, 126, 125, 125, 130, 126, 127, 124, and 123. (a) Estimate the 95% confidence interval of the next measurement obtained with this technique; (b) Estimate the 95% confidence interval of the average of next 5 measurements obtained with this technique; (c) Using only the 12 measurements available, how would you report the value of measured pressure, including the 95% confidence interval?

Answer 1

Let X denotes measured pressure.

a) Since sample size is small and population variance is unknown. We use t-distribution to find confidence interval.

95% confidence interval for next meausrement is

x	(x-xbar)^2
125	0.6944
128	4.6944
129	10.0278
122	14.6944
126	0.0278
125	0.6944
125	0.6944
130	17.3611
126	0.0278
127	1.3611
124	3.3611
123	8.0278
1510	61.6667

S =sqrt(S²) = 2.2669

alpha =level of significance = 0.05

d.f. = n-1 =11

t_11,0.025 = 2.2009.

Hence 95% confidence interval for next value measured pressure is

= ( 125.8333 - 2.2669 * 2.2009 , 125.8333 + 2.2669 * 2.2009)

=(120.8440, 130.8225)

95% next value of measured pressure lies between ( 120.8440, 130.8225)

b) n = 5

d.f. = 4

t4,0.025 = 2.7764

95% confidence interval for average of next measurement is

= ( 125.8333 - 2.7764 * 1.0138, 125.8333 +2.7764 * 1.0138 )

= (123.0186, 128.648)

95% the average value lies between ( 123.0186, 128.648).

c) Since 95 % confidence interval for measured pressure is (120.8440, 130.8225).

All the 12 values lies between these two value, there is no outlier find.