Higher Engineering Mathematics

588 STATISTICS AND PROBABILITY

This indicates that the actual diameter is likely

to lie between 1.843 cm and 1.857 cm and that

this prediction stands a 90% chance of being

correct.

(b) The percentile value corresponding tot 0. 70 and
toν=11 is obtained from Table 61.2, and is
0.540, that is,tc= 0 .540.
The estimated value of the 70% confidence
limits is given by:

x±

tcs

√

(N−1)

= 1. 850 ±

(0.540)(0.016)

√

11

= 1. 850 ± 0 .0026 cm

Thus,the 70% confidence limits are 1.847 cm

and 1.853 cm, i.e. the actual diameter of the bar

is between 1.847 cm and 1.853 cm and this result

has an 70% probability of being correct.

Problem 10. A sample of 9 electric lamps are

selected randomly from a large batch and are

tested until they fail. The mean and standard

deviations of the time to failure are 1210 hours

and 26 hours respectively. Determine the confi-

dence level based on an estimated failure time

of 1210± 6 .5 hours.

For the sample: sample size,N=9; standard devia-
tion,s=26 hours; mean,x=1210 hours. The confi-
dence limits are given by:

x±

tcs

√

(N−1)

and these are equal to 1210± 6. 5
Sincex=1210 hours,

then ±

tcs

√

(N−1)

=± 6. 5

i.e. tc=±

6. 5

√

(N−1)

s

=±

(6.5)

√

8

26

=± 0. 707

From Table 61.2, atcvalue of 0.707, having aνvalue
ofN−1, i.e. 8, gives atpvalue oft 0. 75
Hence,the confidence level of an estimated fail-
ure time of 1210±6.5 hours is 75%, i.e. it is likely
that 75% of all of the lamps will fail between 1203.5
and 1216.5 hours.

Problem 11. The specific resistance of some

copper wire of nominal diameter 1 mm is esti-

mated by determining the resistance of 6 sam-

ples of the wire. The resistance values found (in

ohms per metre) were:

2 .16, 2.14, 2.17, 2.15, 2.16 and 2. 18

Determine the 95% confidence interval for the

true specific resistance of the wire.

For the sample: sample size,N=6, and mean,

x=

2. 16 + 2. 14 + 2. 17 + 2. 15 + 2. 16 + 2. 18

6

= 2. 16 m−^1

standard deviation,

s=

√ √ √ √ √ √ √ √ √ √ √

⎧

⎪⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎪

⎩

(2. 16 − 2 .16)^2 +(2. 14 − 2 .16)^2

+(2. 17 − 2 .16)^2 +(2. 15 − 2 .16)^2

+(2. 16 − 2 .16)^2 +(2. 18 − 2 .16)^2

6

⎫

⎪⎪

⎪⎪

⎪⎪

⎬

⎪⎪

⎪⎪

⎪⎪

⎭

=

√

0. 001

6

= 0. 0129 m−^1

The percentile value corresponding to a confidence

coefficient value oft 0. 95 and a degree of freedom

value ofN−1, i.e. 6− 1 =5 is 2.02 from Table 61.2.

The estimated value of the 95% confidence limits is

given by:

x±

tcs

√

(N−1)

= 2. 16 ±

(2.02)(0.0129)

√

5

= 2. 16 ± 0. 01165 m−^1

Thus,the 95% confidence limits are 2.148 m−^1

and 2.172 m−^1 which indicates that there is a 95%

chance that the true specific resistance of the wire lies

between 2.148m−^1 and 2.172m−^1.

Now try the following exercise.

Exercise 222 Further problems on estimat-

ing the mean of population based on a small

sample size

1. The value of the ultimate tensile strength of
   a material is determined by measurements on