Estimation refers to the process by which one makes inferences about a population, based on information obtained from a sample.
Point Estimate vs. Interval Estimate
Statisticians use sample statistics to estimate population parameters. For example, sample means are used to estimate population means; sample proportions, to estimate population proportions.
An estimate of a population parameter may be expressed in two ways:
Confidence Intervals
Statisticians use a confidence interval to express the precision and uncertainty associated with a particular sampling method. A confidence interval consists of three parts.
The confidence level describes the uncertainty of a sampling method. The statistic and the margin of error define an interval estimate that describes the precision of the method. The interval estimate of a confidence interval is defined by the sample statistic + margin of error.
For example, suppose we compute an interval estimate of a population parameter. We might describe this interval estimate as a 95% confidence interval. This means that if we used the same sampling method to select different samples and compute different interval estimates, the true population parameter would fall within a range defined by the sample statistic + margin of error 95% of the time.
Confidence intervals are preferred to point estimates, because confidence intervals indicate (a) the precision of the estimate and (b) the uncertainty of the estimate.
Confidence Level
The probability part of a confidence interval is called a confidence level. The confidence level describes the likelihood that a particular sampling method will produce a confidence interval that includes the true population parameter.
Here is how to interpret a confidence level. Suppose we collected all possible samples from a given population, and computed confidence intervals for each sample. Some confidence intervals would include the true population parameter; others would not. A 95% confidence level means that 95% of the intervals contain the true population parameter; a 90% confidence level means that 90% of the intervals contain the population parameter; and so on.
Margin of Error
In a confidence interval, the range of values above and below the sample statistic is called the margin of error.
For example, suppose the local newspaper conducts an election survey and reports that the independent candidate will receive 30% of the vote. The newspaper states that the survey had a 5% margin of error and a confidence level of 95%. These findings result in the following confidence interval: We are 95% confident that the independent candidate will receive between 25% and 35% of the vote.
Note: Many public opinion surveys report interval estimates, but not confidence intervals. They provide the margin of error, but not the confidence level. To clearly interpret survey results you need to know both! We are much more likely to accept survey findings if the confidence level is high (say, 95%) than if it is low (say, 50%).
Consider the following results of 10 tosses of a coin: H, T, T, T, T, H, T, H, T, T a) Estimate the probability of head (H) for this coin. b) Estimate the standard error of your estimate.
Let X denote the toss of a single coin. Further, let X = 1 if a head results, and X = 0 if a tail results. This X is a Bernoulli (p) random variable, where p denotes the probability of head. Let pˆ denote the estimator of p.
a) The estimated value of p is pˆ = (1 + 0 + 0 + . . . + 1 + 0 + 0)/10 = 0.3.
b) The estimated standard error of pˆ is √pˆ(1 − pˆ)/n) = √0.3(0.7)/10 = 0.14.
Suppose the following data shows the number of the problems from the Practice Problems Set attempted in the past week by 10 randomly selected students: 2, 4, 0, 7, 1, 2, 0, 3, 2, 1.
a) Find the sample mean.
b) Find the sample variance.
c) Estimate the mean number of practice problems attempted by a student in the past week.
d) Estimate the standard error of the estimated mean.
a) X = Pn i=1 Xi/n = (2 + 4 + . . . + 2 + 1)/10 = 2.2
b) S^2=∑n i=1(Xi − X)^2/(n − 1) = (2 − 2.2)^2 + (4 − 2.2)^2 + . . . + (2 − 2.2)^2 + (1 − 2.2)^2/(10−1) = 4.4
c) The estimate is X = 2.2
d) Estimated standard error of X is S/√ n = √ 4.4/10 = 0.66
CAIIB Paper 1 Study Material |
CAIIB Paper 2 Study Material |
CAIIB Paper 3 Study Material |