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Unit 29: Methods of Point Estimation and Interval Estimation

Example 11: A random sample of size 10 was drawn from a normal population with an unknown Notes
mean and a variance of 44.1 (inch) . If the observations are (in inches): 65, 71, 80, 76, 78, 82, 68, 72, 65
2
and 81, obtain the 95% confidence interval for the population mean.
2
Solution: We are given n = 10, σ = 44.1 and ∑x = 738.
i
738
∴ x = 10 = 73.8.

Since the population s.d. σ is known, using formula (9), 95% confidence limits for μ are given by
44.1
73.8 ± 1.96 = 73.8 1.96 ×
±
10 4.41
= 73.8 ± 1.96 × 2.1
= 73.8 ± 4.1 = 77.9 and 69.7.
The 95% confidence interval for μ is therefore 69.7 to 77.9 inches.
(7) for Mean m: (s.d. unknown)

In random samples from a Normal population ( μ σN , 2 )

x − μ
t = S/ n − 1
follows t distribution with (n – 1) degree of freedom, where S is the sample s.d. If t denotes
.025
the upper 2.5% point of t distribution with (n – 1) d.f., then the 95% confidence interval for μ is
obtained from
x − μ
− .025 ≤ t ≤ t .025
S/ n − 1
Hence, for the population mean μ

⎛ S ⎞
±
95% confidence limits = xt .025 ⎜ ⎟ (11 a)
⎝ − ⎠ n 1
⎛ S ⎞
±
99% confidence limits = xt .005 ⎜ ⎟ (11 b)
⎝ − ⎠ n 1
(8) for Difference of Means( μ − μ 2 ) 1 : (common s.d. unknown)
Assuming that two independent samples are drawn from two Normal populations with means
,
μ μ but a common unknown s.d. σ .
2
1
− x ) (μ − x μ − )
( 1 2 1 2
t =
s 1 + 1
n 1 n 1
follows t distribution with ( + n 2 − n 2 d.f., where
) 1

S
n
S
( 11 2 + n 2 2 2 )
s = ( + n 2 − n 2 ) 1
2

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