Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

Question 1.

Explain the types of sampling.

Solution:

The different types of sampling are

- simple random sampling
- Stratified random sampling and
- Systematic sampling

(i) In simple random sampling, every item of the population has an equal chance for being selected. The sampling can be done with replacement (or) without replacement. A random sampling from a finite population with replacement is equivalent to sampling from an infinite population without replacement. This technique will give useful results only if the population is homogeneous. The following are some of the methods of selecting a random sample.

(a) Use of an unbiased die or coin: If we have to choose between two alternatives, a coin is tossed, and depending on the head or tail course of action is taken. A die can be employed if there are six different alternatives.

(b) Lottery sampling: Here a random sample is selected by identifying each element of the population by means of a card of a pack of uniform cards or (by writing the number on pieces of paper) and to select a required number of cards after thorough mixing of the cards.

(c) Random numbers: Random numbers are formed of ‘random digits’ and arranged in the form of a table having a number of rows and columns. Tippett’s numbers form one such table wherein 40,000 digits were selected at random from census reports and combined by groups of four into 10,000 numbers.

(ii) In stratified random sampling, a population of units is divided into L sub-populations of N_{1}, N_{2}, …… N_{L}. The sub-populations being non-overlapping and mutually exhaustive so that N = N_{1} + N_{2} + …….. + N_{L}. Each subpopulation is known as a stratum. If we select n_{1}, n_{2}, ……. n_{l} items, respectively, from these strata, we get a stratified sample. If a simple random sample is taken from each stratum, the whole procedure is referred to as stratified random sampling.

(iii) Systematic sampling is a form of restricted random selection which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in serial order and every ith element, starting from any of the first items is chosen. For example, suppose we require a 5% sample of students from a college where there are 2000 students, we select a random number from 1 to 20. If it is 12, then our sample consists of students with numbers 12, 32, 52, 72, …… 1992.

Question 2.

Write a short note on sampling distribution and standard error.

Solution:

sampling distribution:

The sampling distribution of a statistic 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.

For instance, if we draw a sample a size n from a given finite population of N, then the total number of possible samples is Nc_{n} = \(\frac { N! }{n!(N-n)!}\) = k(say)

Standard Error:

The standard deviation of the sampling distribution of a statistic is known as its standard Error abbreviated as S.E. The standard error (S.E) of some of the well-known statistics, for large samples, are given below, where n is the sample size, σ² is the population variance.

Question 3.

Explain the procedures of testing of hypothesis

Solution:

The following are the steps involved in hypothesis testing problems:

1. Null hypothesis: Set up the null hypothesis H_{0}

2. Alternative hypothesis: Set up the alternative hypothesis. This will enable us to decide whether we have to use two-tailed test or single-tailed test (right or left tailed)

3. Level of significance: Choose the appropriate level of significance (a) depending on the reliability of the estimates and permissible risk. This is to be fixed before the sample is drawn, i.e., a is fixed in advance.

4. Test statistic: Compute the test statistic

Z = \(\frac { t-E(t) }{\sqrt{var(t)}}\) = \(\frac { t-E(t) }{S.E(t)}\) N(0, 1) as n → ∞

5. Conclusion: We compare the computed value of Z in step 4 with the significant value or critical value or table value Zα at the given level of significance.

(i) If |Z | < Zα i.e., if the calculated value is less than the critical value we say it is not significant. This may due to fluctuations of sampling and sample data do not provide us sufficient evidence against the null hypothesis which may therefore be accepted.

(ii) If |Z |> Zα i.e., if the calculated value of Z is greater than critical value Zα then we say it is significant and the null hypothesis is rejected at the level of significance α.

Question 4.

Explain in detail the test of significance for a single mean.

Solution:

Let xi, (i = 1, 2, 3, …, n) is a random sample of size from a normal population with mean µ and variance σ² then the sample mean is distributed normally with mean and variance

\(\frac { σ^2 }{n}\), i.e \(\bar { x }\) N(µ, \(\frac { σ^2 }{n}\))

Thus for large samples, the standard normal variate corresponding to \(\bar { x }\) is

Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\) N (0, 1)

Under the null hypothesis that the sample has been drawn from a population with mean and variance σ², i.e., there is no significant difference between the sample mean (\(\bar { x }\)) and the population mean (α), the test statistic (for large samples) is:

Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\)

Question 5.

Determine the standard error of proportion for a random sample of 500 pineapples was taken from a large consignment and 65 were found to be bad?

Solution:

sample size n = 500

No. of bad pine apples = 65

sample proportion = P = \(\frac { 65 }{500}\) = 0.13

Q = 1 – p ⇒ Q = 1 – 0.13

∴ Q = 0.87

The S.E for sample proportion is given by

S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.13)(0.87) }{500}}\)

= \(\sqrt{\frac { 0.1131 }{500}}\) = \(\sqrt{0.0002262}\)

= 0.01504

∴ S.E = 0.015

Hence the standard error for sample proportion is S.E = 0.015

Question 6.

A sample of 100 students are drawn from a school The mean weight and variance of the sample are 67.45 kg and 9 kg respectively find (a) 95% and (b) 66% confidence intervals for estimating the mean weight of the students.

Solution:

sample size n = 100

The sample mean = \(\bar { x }\) = 67.45

The sample variance S² = 9

The sample standard deviation S = 3

S.E = \(\frac { S }{√n}\) = \(\frac { 3 }{\sqrt{100}}\) = \(\frac { 3 }{10}\) = 0.3

(a) The 95% confidence limits for µ are given by

\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E < µ < \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E

67.45 – (1.96 × 0.3) ≤ µ ≤ 67.45 + (1.96 × 0.3) 67.45 – 0.588 ≤ µ ≤ 67.45 + 0.588

66.862 ≤ µ ≤ 68.038

The confidential limits is (66.86, 68.04)

(b) The 99% confidence limits for estimating µ are given by

\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E ≤ µ ≤ \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E

67.45 – (2.58 × 0.3) ≤ µ ≤ 67.45 + (2.58 × 0.3)

67.45 – 0.774 ≤ µ ≤ 67.45 + 0.774

66.676 ≤ µ ≤ 68.224

∴ The 99% confidence limits is (66.68, 68.22)

Question 7.

The mean I.Q of a sample of 1600 children was 99. it is likely that this was a random sample from a population with a mean I.Q 100 and standard deviation of 15? (Test at 5% level of significance)

Solution:

sample size n = 1600

\(\bar { x }\) = 99

sample mean

Population mean µ = 100

population S.D σ = 15

under the Null hypothesis H_{0} : µ = 100

Alternative hypothesis H_{1} : µ = 100 (two tails)

Level of significance µ = 0.05

z = -2.666

z = -2.67

Calculated value |z| = 2.67

critical value at 5% level of significance is

z_{\(\frac { α }{2}\)} = 1.96

Inference:

Since the calculated value is greater than table value i.e z ⇒ z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is rejected. Therefore we conclude that the sample mean differs, significantly from the population mean.