Page 357 - DMTH404_STATISTICS

Unit 24: Sampling Distributions

Solution. Notes

80 4 4
E
It is given that p = = . Therefore, ( ) p = p = and
500 25 25
4 21
(a) S.E.(p) = np (1 p- ) = 10 ´ ´ = 1.159 (srswr)
25 25

-
500 80 4 21
(b) S.E.(p) = ´ 10 ´ ´ = 1.063 (srswor)
499 25 25

Example 3: 20% under graduates of a large university are found to be smokers. A sample
of 100 students is selected at random. Construct the sampling distribution of the number of
smokers. Also find the probability that the number of smokers in the sample is greater than 25.

Solution.
20 1
It is given that p = = . Since sample size, n = 100, is large, the number of successes X will
100 5

1 1 4
be distributed normally with mean 100 ´ = 20 and standard error 100 ´ ´ = 4 .
5 5 5
-
æ 25 20ö
Further, (X > 25 ) = P z > ÷ = P (z 1.25> ) = 0.1056.
P
ç
è 4 ø
24.3.2 Sampling Distribution of the Difference of two Proportions

Let p be proportion of successes in a random sample of size n from a population with proportion
1 1
of successes = p and p be the proportion of successes in a random sample of size n from second
1 2 2
population with proportion of successes = p . Assuming that the sample sizes are large, we can
2
write
æ p (1 p- ) ö æ p (1 p- ) ö
p ~ N p , 1 1 ÷ and p ~ N p , 2 2 ÷
ç
ç
1 1 2 2
ç n 1 ÷ ç n 2 ÷
è ø è ø
Thus, their difference (p - p ) will be distributed normally with mean = p - p and standard error
1 2 1 2
p 1 (1 p- 1 ) p 2 (1 p- 2 )
+ .
n 1 n 2
Note: The above result will hold when we ignore fpc and the sample size, n and n , is greater
1 2
than 5 divided by the minimum of p , (1 - p ), p and (1 - p ).
1 1 2 2
Some other Sampling Distributions

We have seen that the sampling distributions of mean and proportion of successes are normal.

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