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

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.

2952	6641	3992	9792	7969	5911	3170	5624
4167	9524	1545	1396	7203	5356	1300	2693
2670	7483	3408	2762	3563	1089	6913	7991
0560	5246	1112	6107	6008	8125	4233	8776
2754	9143	1405	9025	7002	6111	8816	6446

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

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 Sample size n = 400
Sample mean \(\\ \bar { X } =67.47\)inches
Population mean μ = 67.39 & σ = 1.30 inches)
Null Hypotheses H_o:
μ = 67.39 inches (i.e., the sample has been drawn from the population with μ = 67.39 & σ = 1.30 inches)
Alternative Hypotheses H₁:
μ ≠ 67.39 inches (two tail test)
(i.e., the sample has not been drawn from the population with μ = 67.39 & σ = 1.30 inches) The level of significance a = 5% = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 67.47-67.39 }{ \frac { 1.30 }{ \sqrt { 400 } } } =\frac { 0.08 }{ 0.065 } =1.2308\)
\(\therefore |Z|=1.2308\)
The significant value \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here \(Z={ Z }_{ \frac { \alpha }{ 2 } }i.e.,1.2308<1.96\)
Inference: Since \({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance the null hypothesis H_o is accepted.
Hence, we conclude that the sample has been drawn from the population with mean height 67.39 inches and standard deviation 1.30 inches.

Given, sample size = 100, \(\bar x\) = 7.4, since σ is unknown but s = 1.2 is known.
In this problem, we consider \(\breve { \sigma } =s\quad { Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(S.E=\frac { \breve { \sigma } }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 1.2 }{ \sqrt { 100 } } =0.12\)
Hence 95% confidence limits for the population mean are
\(\bar { x } -{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } <\mu <\bar { x } +{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } \)
\(7.4-(1.96\times 0.12)\le \mu \le 7.4+(1.96\times 0.12)\)
\(7.4-0.2352\le \mu \le 7.4+0.2352\)
\(7.165\le \mu \le 7.635\)
This implies that the probability that the true value of the population mean breaking strength of the cotton threads will fall in this interval (7.165,7.635) at 95%.