Given sample size n = 500
Probability of bad apples in the sample =\(\frac{65}{500}\)
∴ P=0.13
∴ Probability of good apples in the sample
⇒ q = 1 - P = 1 - 0.13 = 0.87
Standard error of proportion \(=\sqrt { \frac { pq }{ n } } \)
\(=\sqrt { \frac { (0.13)(0.87) }{ 500 } } =\sqrt { \frac { 0.1131 }{ 500 } } =\sqrt { 0.000262 } \)
∴ S.E=0.015

Hypothesis testing is also referred as "Statistical Decision Making".
There are two types of statistical hypothesis
(i) Null hypothesis: which is tested for possible rejection under the assumption that it is true
(ii) Alternative hypothesis which is complementary to the null hypothesis
For example: H_o:μ=μ₀
i) H₁:μ≠μ₀ is know as two tailed alternative test
ii) H₁:μ>μ₀ and H₁:μ<μ₀ are known as one tailed alternative
iii) H₁:μ>μ₀ is said to be right tailed test where the critical region lies entirely on the right tail of the normal curve.
iv) H₁:μ<μ0 is said to the left tailed test where the critical region lies entirely on the left tail of the normal curve.

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

S.No	Statistic	Standard Error
1	Sample mean	σ/√n
2	Observed sample proportion	\(\sqrt { PQ/n } \)
3	Sample standard deviation	\(\sqrt { { \sigma }^{ 2 }/2n } \)
4	Sample variance	\({ \sigma }^{ 2 }\sqrt { 2/n } \)
5	Sample quartiles	\(1.36263\sigma /\sqrt { n } \)
6	Sample median	\(1.25331\sigma /\sqrt { n } \)
7	Sample correlation coefficient	\((1-{ \rho }^{ 2 })/\sqrt { n } \)

Sample size n = 100,
Sample mean \(\\ \bar { X } =3.5\)
Population mean μ = 4
Population standard deviation σ = 3
Null Hypotheses: There is no significant difference in the mean. i.e., H_o : μ = 4
Alternative Hypotheses : There is Significant difference in the mean.
i.e., H₁ : μ ≠ H
The level of significance ∝ = 5% = 0.05
Applying the test statistic,\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 3.5-4 }{ \frac { 3 }{ \sqrt { 100 } } } =\frac { -.5 }{ .3 } =-1.667\)
\(\Rightarrow |Z|=1.667\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here Z < \({ Z }_{ \frac { \alpha }{ 2 } }\)i.e., 1.667<1.96
Inference: Since Z<\({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance, the null hypothesis H₀ is accepted. Hence there is no Significant difference in the mean.

\( \text { Given } n=1000, \bar{X}=119, \sigma=30\)
\(S . E=\sigma / \sqrt{n}\)
\(=30 / \sqrt{1000} \)
\(=30 / 31.623=0.9487 \)

Sample size N = 600
Population proportion P = 4% = .04
Q = 1-P = 1-.04
Q = 0.96
∴ The standard error for sample proportion is given by
\(S.E=\sqrt { \frac { PQ }{ N } } =\sqrt { \frac { (.04)(.96) }{ 600 } } \)
\(=\sqrt { \frac { 0.0384 }{ 600 } } =\sqrt { .000064 } \)
S.E=.008

There are many ways to select a sample of 10 3-digit even numbers. From the table, start from the first number and move along the column. Select the first three digits as the number. If it is an odd number, move to the next number. The selected sample is 416, 664, 952, 748, 524, 914, 154, 340, 140, 276.
\(\begin{array}{|l|l|l|l|l|l|l|l|} \hline 2952 & \underline{6641} & 3992 & 9792 & 7969 & 5911 & 3170 & 5624 \\ \hline \mathbf{4 1 6 7} & \underline{9524} & \mathbf{1 5 4 5} & 1396 & 7203 & 5356 & 1300 & 2693 \\ \hline 2670 & \underline{7483} & \underline{3408} & \underline{2762} & 3563 & 1089 & 6913 & 7991 \\ \hline 0560 & \underline{5246} & 1112 & 6107 & 6008 & 8125 & 4233 & 8776 \\ \hline 2754 & \underline{9143} & \underline{1405} & 9025 & 7002 & 6111 & 8816 & 6446 \\ \hline \end{array}\)

1. It can be kept small in size without losing its accuracy.
2. It is easy to administer, if the population under study is sub-divided.
3. It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had divided the geographical areas.

The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments are called Non-Sampling errors.
It may arise in the following ways:
a) Due to negligence and carelessness of the part of either investigator or respondents.
b) Due to lack of trained and qualified investigators.
c) Due to framing of a wrong questionnaire.
d) Due to applying wrong statistical measure.
e) Due to incomplete investigation and sample survey.

Errors, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors. Sampling Errors arise due to the following reasons:
a) Faulty selection of the sample instead of correct sample by defective sampling technique.
b) The investigator substitutes a convenient sample if the original sample is not available while investigation.
c) In area surveys, while dealing with border lines it depends upon the investigator whether to include them in the sample or not. This is known as Faulty demarcation of sampling units.

Sample size n = 50
Sample S.D s = 6.3
Population S.D \(\sigma\) = 6
The standard error for sample S.D is given by 
\(S.E=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } =\frac { 6 }{ \sqrt { 2(50) } } =\frac { 6 }{ \sqrt { 100 } } =0.6\)
Thus standard error for sample S.D = 0.6.

If the occurrence of 3 or 4 on the die is called a success, then
Sample size = 9000; Number of Success = 3240
Sample proportion =\(P=\frac { 3240 }{ 9000 } =0.36\)
Population proportion (P) = Prob(getting 3 or 4 when a die is thrown)
\(=\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } =0.3333\)
Thus P = 0.3333 and Q = 1–P = 1– 0.3333 = 0.6667
The S.E for sample proportion is given by
\(S.E=\sqrt { \frac { PQ }{ N } } =\sqrt { \frac { (0.3333)(0.6667) }{ 9000 } } =0.00496\)
Hence the standard error for sample proportion is S.E = 0.00496.

Given \(\sigma\) = 10, S.E. \(\bar { X } \) = 3
We know that S.E = \(\frac { \sigma }{ \sqrt { n } } \)
Therefore, \(3=\frac { 10 }{ \sqrt { n } } \Rightarrow \sqrt { n } =\frac { 10 }{ 3 } \)
Taking Squaring on both sides we get
\(n=\left(\frac{10}{3}\right)^{2}=\frac{100}{9}=11.11 \cong 11\), 
The required sample size is 11.

Stratum	Homogenous group	Percentage from population	Number of people in each strata	Random Samples
Category 1	Lower income class	39	\(\frac{39}{100}\) \(\times\) 500 = 195	195 \(\times\)\(\frac{68}{500}\) = 26.5~26
Category 2	Middle income class	38	\(\frac{38}{100}\) \(\times\) 500 = 190	190 \(\times\)\(\frac{68}{500} \) = 2.6 ~26
Category 3	Upper income class	23	\(\frac{23}{100}\) \(\times\) 500 = 115	115 \(\times\) \(\frac{68}{500}\) = 15.6~16
Total		100	500	68

Merits :
(a) A random stratified sample is superior to a simple random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled.
(b) A stratified random sample can be kept small in size without losing its accuracy.
(c) It is easy to administer, if the population under study is sub-divided.
(d) It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had divided the geographical areas.
Demerits :
(a) To divide the population into homogeneous strata (if not divided), it requires more money, time and statistical experience which is a difficult one.
(b) If proper stratification is not done, the sample will have an effect of bias.
(c) There is always a possibility of faulty classification of strata and hence increases variability.

Here, we have to select 10 random numbers ranging from 1550 to 8000 but the given random number table has only 2 digit numbers. To solve this, two - 2 digit numbers can be combined together to make a four- figure number. Let us select the 5^th and 6^th column and combine them to form a random number, then select the random number with given range. This gives 5 random numbers, similarly, 8^th and 9^th is selected and combined to form a random numbers, then select the random number with given range. This gives 5 random numbers, totally 10 four- figure numbers have been selected. The following table shows the 10 random numbers which are combined and selected.

Therefore the selected 10 random numbers are

5901	4310	5032	5194	1899
7259	7435	4362	1758	5308