Page 318 - Fundamentals of Probability and Statistics for Engineers

Parameter Estimation 301

f (t)
T

1– α

α/ 2 α/ 2
t
–t n, α / 2 t n, α / 2

Figure 9.8 [100(1 )]% confidence limits for T with n degrees of freedom

Example 9.19. Problem: let us assume that the annual snowfall in the Buffalo
area is normally distributed. Using the snowfall record from 1970–79 as given
in Problem 8.2(g) (Table 8.6, page 257), determine a 95% confidence interval
for mean m.
Answer: for this example, ˆ 0:05, n ˆ 10, the observed sample mean is

1
x ˆ 120:5 ‡ 97:0 ‡ ‡ 97:3†ˆ 112:4;
10

and the observed sample variance is

1 2 2 2
2
s ˆ ‰ 120:5 112:4† ‡ 97:0 112:4† ‡ ‡ 97:3 112:4† Š
9
ˆ 1414:3:

Using Table A.4, we find that t 9, 0:025 ˆ 2:262. Substituting all the values given
above into Equation (9.141) gives

P 85:5 < m < 139:3†ˆ 0:95:

It is clear that this interval would be different if we had incorporated more
observations into our calculations or if we had chosen a different set of yearly
snowfall data.

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