The objective of **point estimation of parameters** is to obtain a single number from the sample which will represent the unknown value of the parameter.

Practically we did not know about the population mean and standard deviation i.e. population parameters such as mean, standard deviation, etc. However, our goal is to measure (estimate) the mean and standard deviation of the population we are interested in from sample information to save time, cost, etc. This can be done by estimating the sample mean and standard deviation as the best guess for the true population mean and standard deviation. We can call this estimate a “best guess” and termed a “point estimate” as it is a single number summarized one.

A** Point Estimate** is a statistic (a statistical measure from the sample) that gives a plausible estimate (or possibly a best guess) for the value in question.

$\overline{x}$ is a point estimate for $\mu$ and *s* is a point estimate for $\sigma$.

Or we can say that

A statistic used to estimate a parameter is called a point estimator or simply an estimator. The actual numerical value which we obtain for an estimator in a given problem is called an estimate.

Generally symbol $\theta$ (unknown constant) is used to denote a population parameter which may be a proportion, mean, or some measure of variability. The available information is in the form of a random sample $X_1, X_2, \cdots, X_n$ of size *n* drawn from the population. We wish to formulate a function of the sample observations $X_1, X_2, \cdots, X_n$; that is, we look for a statistic such that its value computed from the sample data would reflect the value of the population parameter as closely as possible. The estimator of $\theta$ is commonly denoted by $\hat{\theta}$. Different random samples usually provide different values of the statistic $\hat{\theta}$ having its sampling distribution.

Note that Unbiasedness, Efficiency, Consistency, and Sufficiency are the criteria (statistical properties of the estimator) to identify whether a statistic is a “good” estimator.

**Application of Point Estimator Confidence Intervals**

We can build intervals with confidence as we are not only interested in finding the point estimate for the mean but also in determining how accurate the point estimate is. Here the Central Limit Theorem plays a very important role in building confidence interval. We assume that the sample standard deviation is close to the population standard deviation (which will almost always be true for large samples). The standard deviation of the sampling distribution of the estimator (here for mean) is

\[\sigma_x \approx \frac{\sigma}{\sqrt{n}}\]

Our interest is to find an interval around *$\overline{x}$* such that there is a large probability that the actual (true) mean falls inside the computed interval. This interval is called a *confidence interval* and the large probability is called the *confidence level*.

**Example **of Point Estimation of Parameters

**Example**of Point Estimation of Parameters

**Question:** Suppose that we check for clarity in 50 locations in Lake and discover that the average depth of clarity of the lake is 14 feet with a standard deviation of 2 feet. What can we conclude about the average clarity of the lake with a 95% confidence level?

**Solution:** Variable $x$ (depth of lack at 50 locations) can be used to provide a point estimate for $\mu$ and s to provide a point estimate for $s$. To answer how accurate is $x$ as a point estimate, we can construct a *95% confidence interval for $\mu$ *as follows.

Draw the picture given below and use the standard normal table to find the z-score associated with the probability of .025 (there is .025 to the left and .025 to the right i.e. two-tailed case).

The Z-score for a 95% confidence level is about $\pm 1.96$.

\begin{align*}

Z&=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\\

\pm 1.96&=\frac{\overline{x}-\mu}{\frac{2}{\sqrt{n}}}\\

\overline{x}-14&=\pm 0.5488

\end{align*}

Note that $Z\frac{\sigma}{\sqrt{n}}$ is called the *margin of error*.

The 95% confidence interval for the mean clarity will be (13.45, 14.55)

In other words, there is a 95% chance that the mean clarity is between 13.45 and 14.55.

In general, if $z$ is the standard normal table value associated with a given level of confidence then a $\alpha$% confidence interval for the mean is

\[\overline{x} \pm Z_{\alpha}\frac{\sigma}{\sqrt{n}}\]

See more at Wikipedia about Point Estimation of Parameters