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We shall concentrate on estimation of population proportions and population means. If we use a single value of a statistic ˆΘ, say ˆθ, to estimate a population ...
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Semester 1
Estimating population parameters from sample statistics is a central theme in statistical inference. We shall concentrate on estimation of population proportions and population means. If we use a single value of a statistic ˆΘ, say ˆθ, to estimate a population parameter θ, then we refer to this as a point estimate. For example, if we use the mean height (say 172cm) of a random sample of 200 students to estimate the mean height in a population of 100,000, this is a point estimate.
We wish to construct interval estimates for the population mean μ. To begin with, we assume that: (i) the population standard deviation σ is known; (ii) the size of our random sample n is greater than or equal to 30. Under the above assumptions, the Central Limit Theorem tells us that the random variable X¯ − μ σ/
n is approximately a standard normal random variable. This allows us to use the table of standard normal probabilities to make inferences about how “close” values of ¯X “typically” are to the true mean μ.
Let z 0. 025 denote the value such that 0.025 or 2.5% of the Z-distribution lies in the tail to the right of z 0. 025. Then P(−z 0. 025 ≤ Z ≤ z 0. 025 ) = 0. 95. It can be checked from the table of standard normal probabilities that z 0. 025 = 1.96. Thus
X¯ − μ σ/
n
σ √ n
) ≤ X¯ − μ ≤ 1 .96( σ √ n
Definition Given a value ¯x of X¯ , the large sample 95% confidence interval for μ is given by
¯x − 1 .96(
σ √ n ) ≤ μ ≤ x¯ + 1.96(
σ √ n
In practice, we calculate a single value ¯x of X¯ and use it to construct a single confidence interval. Once such an interval is constructed, we can assert with 95% confidence that μ lies in the interval.
In the same way as we calculated a 95% confidence interval for μ, we can construct intervals with different levels of confidence. For example, for a standard normal random variable Z
P(− 2. 58 ≤ Z ≤ 2 .58) = 0. 99.
In the same way as before, we say that
x¯ − 2. 58
σ √ n ≤ μ ≤ ¯x + 2. 58
σ √ n
is a 99% confidence interval for μ.
Similarly the 99% confidence interval for μ is
1023 − 2. 58
≤ μ ≤ 1023 + 2. 58
1023 − 9. 12 ≤ μ ≤ 1023 + 9. 12
It is possible to construct confidence intervals for levels of confidence other than 99% and 95%. Let zα/ 2 denote the value such that
P(−zα/ 2 ≤ Z ≤ zα/ 2 ) = 1 − α
. Then the 100(1 − α)% confidence interval for the mean μ is
¯x − zα/ 2 σ √ n
≤ μ ≤ ¯x + zα/ 2 σ √ n
In fact, we can be 95% confident that the error |¯x − μ| ≤ E provided that
σ √ n
which after rearrangement becomes
n ≥
Example The light bulb manufacturer in our earlier example wants to choose a random sample and to be 95% confident that the error in estimating the mean lifetime of a bulb with the sample mean is less than 5 hours. How large a sample is required?
Solution: E = 5, σ = 20, so we need to choose
n ≥
So we need a sample of size 62 or larger.
If the population standard deviation is not known, we need to use the sample standard deviation
(X 1 − X¯ )^2 + · · · + (Xn − X¯ )^2 n − 1
instead of σ. Provided our sample is large (n ≥ 30 as a rule of thumb) it is reasonable to use the value of S as an estimate of σ and to construct confidence intervals as above. If our sample size is small however, we cannot use the normal distribution to generate confidence intervals.
However if we know that the population is normal, then for a sample of size n the random variable
t = X¯ − μ S/
n
follows a student’s t distribution with n − 1 degrees of freedom. As with the standard normal distribution, tables have been calculated for this distribution and the key percentage points have been calculated.
Example The assembly times for a piece of machinery follow a normal distribution. A quality control engineer randomly selects 8 pieces and measures the assembly time for each one. The assembly times in minutes for the 8 pieces were:
Solution: As the size of the sample is small and we need to estimate the standard deviation from the sample data, we use the t-distribution to construct the confidence intervals.
¯x =
The sample variance s^2 is 0.1041 so the sample standard deviation is s = 0.323. Our sample size is n = 8, so we use the t-distribution with 8 − 1 = 7 degrees of freedom to construct confidence intervals.