When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as one tailed test.

The error of rejecting H_o when it is true is type I error.

The probability of type I error is known as level of significance denoted by \(\alpha\). It is always fixed in advance before collecting the sample information.

The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value.

A region corresponding to a test statistic in the sample space which tends to rejection of Ho is called critical region.

Any hypothesis which is complementary to the null hypothesis is called as the alternative hypothesis and it is usually denoted by H₁.

Null hypothesis is the hypothesis which is tested for possible rejection under the assumption that it is true and it is denoted by H₀.
Example: If we want to find the population mean has a specific value μ_o, then the null hypothesis H_o is H_o : μ = μ₀

The interval within which the unknown value of parameter is, expected to lie is called confidence interval. It indicates the probability that the population parameter lies within a specilied range. If o is the population paramcter, then we choose a small value a, known as level of significance (1% or 5%) and determine 2 constants c, and c,such that p(c₁ < 0 < c₂/t) = 1- \(\alpha\). When t is the value of statistic. The quantities c₁ and c₂ are determined as confidence limits and the interval [c₁, c₂]  within which the unknown value of the population parameter is expected to lieis known as confidence interval.

Generally, there are situations where point estimation is not desirable and we are interested in finding limits within which the parameter would be expected to lie is called an interval estimation.

When a single value is used as an estimate, the estimate is called a point estimate of the population parameter In other words, an estimate of a population parameter given by a single numberis called as point estimation.

When we observe a specific numerical value, of our estimator, we call that value is an estimate. In other words, an estimate is a specific observed value of statistic.

Any sample statistic which is used to estimate an unknown population parameter is called an estimator (i.e.,). an estimator is a sample statistic used to estimate a population parameter.

The two branches of statistical inference are
(i) Estimation and
(ii) Testing of hypotheses.

1. This method is simple and convenient.
2. This method distributes the sample more evenly over the entire listed population.

1. Systematic samples are not random samples.
2. If N is not a multiple of n, then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

1. This method is economical as it saves time, money and labour.
2. Personal bias is completely eliminated.

The standard deviation of the sampling distribution of a statistic is known as its Standard Error.

It is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.

The statistical constants of the population like mean (μ), variance (σ²) are referred as population parameters.

Any statistical measure computed from sample is known as statistic.

A selection of a group of individuals from a population in such a way that it represents the population is called as sample.

The group of individuals considered under study is called as population. It refers not only to people but to all items that have been chosen for the study.

Give n = 100, \(\bar x\) =162 cm, s = 8 cm is known in this problem
since σ is unknown , so we consider \(\hat{\sigma}\) = s and \(\varphi\) = 160 cm
\(S.E = \frac { \hat{\sigma} }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 8 }{ \sqrt { 100 } } =0.8\)
Therefore the standard error for the average height of large group of students of 160 cm is 0.8.

Givens \(\sigma^2\) = 4 which implies \(\sigma\) = 2, n = 1 hour = 60 min, \(\bar { X } \) = 20/min
Standard Error \(=\frac { \sigma }{ \sqrt { n } } =\frac { 2 }{ \sqrt { 60 } } =0.2582\)

There many ways to select 15 random samples from the given Tippet’s random number table. Since the population size is 83(two-digit number). Here the door numbers are assigned from 1 to 83. Assume that at random we first choose 2^nd column. So the first sample is 66 and other 14 samples are 74, 52, 39, 15, 34, 11, 14, 13, 27, 61, 79, 72, 35, and 60. If the numbers are above 83, choose the next number ranging from 1 to 83.

There many ways to select 5 random samples from the given Kendall-Babington Smith - Random number table. Assume that at random we select 3^rd column 1^st value. This location gives the digit to be 75. So the first sample will be 75, and then the next choices can be in the same 3^rd column which follows as 55, 16, 67, and 26. Therefore 75, 55,16,67, and 26 will be used as random samples. The various shaded numbers can be taken as 5 random sample numbers. Apart from this, one can select any 5 random sample numbers as they like.