Sampling Techniques and Statistical Inference Exercise 8.1

Dear Class 12 Samacheer Kalvi students, here are the Text Book Solutions for Exercise 8.2 Business Maths Chapter 8 Sampling Techniques and Statistical Inference for your reference and study.

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Sampling Techniques and Statistical Inference Important Formulas

Solutions for Sampling Techniques and Statistical Inference Exercise 8.1

Solutions for Sampling Techniques and Statistical Inference Exercise 8.3

Sampling Techniques and Statistical Inference Exercise 8.2

$\text{1. Mention two branches of statistical inference?}\\$

$\text{The two branches of statistical inference are Estimation and Testing of Hypothesis.}\\$

$\text{2. What is an estimator?}\\$

$\text{Any sample statistic which is used to estimate an unknown population} \\$
$\text{ parameter is called an estimator.}\\$

$\text{3. What is an estimate?} \\$

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

$\text{4. What is point estimation?} \\$

$\text{When a single value is used as an estimate, the estimate is called a point estimate}\\$
$\text{of the population parameter. In other words, an estimate of a population parameter} \\$
$\text{ given by a single number is called as point estimation.} \\$

$\text{5. What is interval estimation?}\\$

$\text{There are situations where point estimation is not desirable and we are } \\$
$\text{interested in finding limits within which the parameter would be expected } \\$
$\text{to lie is called an interval estimation.} \\$

$\text{6. What is confidence interval?} \\$

$\text{A confidence interval (CI) is a range of values that's used to estimate an unknown}\\$
$\text{population parameter (like a mean or proportion) based on sample data.}\\$
$\text{It gives you an idea of how confident you can be that your sample estimate is close} \\$
$\text{to the true population value.}\\$

$\text{7. What is null hypothesis? Give an example.}\\$
$\text{According to Prof. R.A.Fisher, ``Null hypothesis is the hypothesis which is tested } \\$
$\text{for possible rejection under the assumption that it is true''. It is denoted by } \\$
$H_{0} . \text{For example: If we want to find the population mean has a specified value }\\$
$\mu_{0} , \text{then the null hypothesis H_{0} is set as follows}\; H_{0} : \mu = \mu_{0}\\$

$\text{8. Define alternative hypothesis.}\\$
$\text{Any hypothesis which is complementary to the null hypothesis is called as}\\$
$\text{ the alternative hypothesis and is usually denoted by}\; H_{1}.}\\$

$\text{9. Define critical region..}\\$
$\text{A region corresponding to a test statistic in the sample space which tends}\\$
$\text{to rejection of} \; H_{0 } \;\text{is called critical region or region of rejection.} \\$

$\text{10. Define critical value.}\\$
$\text{The value of test statistic which separates the critical region and the}\\$
$\text{acceptance region is called the critical value or significant value. It }\\$
$\text{depends upon the level of significance and the alternative hypothesis}\\$
$\text{whether it is two-tailed or single tailed.}\\$

$\text{11. Define level of significance.}\\$
$\text{The probability of type I error is known as level of significance and it is}\\$
$\text{denoted by} \;\alpha.\; \text{The level of significance usually employed in testing}\\$
$\text{of hypothesis are} \; 5\% \;and \;1\%. \\$

$\text{12. What is type I error.}\\$
$\text{There is every chance that a decision regarding a null hypothesis may be}\\$
$\text{correct or may not be correct. The error of rejecting} \;H_{0} \text{when it is}\\$
$\text{true is called a Type I error.}\\$

$\text{13. What is single tailed test.}\\$
$\text{A single-tailed test is a type of hypothesis test used in statistics when}\\$
$\text{ you want to determine if there is evidence that a population parameter}\\$
$\text{ is greater than or less than a certain value — but not both.}\\$

$\text{14. A sample of 100 items, draw from a universe with mean value 4 and S.D 3,}\\$
$\text{ has a mean value 3.5. Is the difference in the mean significant at }\\$
$\text{0.05 level of significance?}\\$

$n=100\\$
$\mu=4\\$
$\sigma=3\\$
$\bar{x}=3.5\\$
$\text{The test statistic:}\\$
$H_{0}: \mu=4\\$
$H_{1}: \mu\neq 4\\$
$z=\dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\\$
$=\dfrac{3.5-4}{\frac{3}{\sqrt{100}}}\\$
$=\dfrac{-.5}{\frac{3}{10}}\\$
$=\dfrac{-.5}{.3}\\$
$z=-1.667\\$
$\left| z \right|=1.667\text{ at }\;5 \% \;\text{level of significance.}\\$
$\left| z \right|< z_{\frac{\alpha}{2}}=1.96\\$
$H_{0}\; \text{lies in the acceptance region.}\\$
$\text{Therefore}\; H_{0}\;\text{is accepted.}\\$

$\text{15. A sample of 400 individuals is found to have a mean height of 67.47 inches.}\\$
$\text{Can it be reasonably regarded as a sample from a large population with mean}\\$
$\text{height of 67.39 inches and standard deviation 1.30 inches at 0.05 level of significance?}\\$

$n=400\\$
$\mu=67.39\\$
$\sigma=1.3\\$
$\bar{x}=67.47\\$
$\text{The test statistic:}\\$
$H_{0}: \mu=67.39\\$
$H_{1}: \mu\neq 67.39\\$
$z=\dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\\$
$=\dfrac{67.47-67.39}{\frac{1.3}{\sqrt{400}}}\\$
$=\dfrac{0.08}{0.065}\\$
$z=-1.231\\$
$\left| z \right|=1.123\text{ at }\;5 \% \;\text{level of significance.}\\$
$\left| z \right|< z_{\frac{\alpha}{2}}=1.96\\$
$H_{0}\; \text{lies in the acceptance region.}\\$
$\text{Therefore}\; H_{0}\;\text{is accepted.}\\$

$\text{16. The average score on a nationally administered aptitude test was 76 and the}\\$
$\text{corresponding standard deviation was 8. In order to evaluate a state's education} \\$
$\text{system, the scores of 100 of the state's students were randomly selected. These } \\$
$\text{students had an average score of 72. Test at a significance level of 0.05 if there } \\$
$\text{is a significant difference between the state scores and the national scores.} \\$

$n=100\\$
$\mu=76\\$
$\sigma=8\\$
$\bar{x}=72\\$
$\text{The test statistic:}\\$
$H_{0}: \mu=76\\$
$H_{1}: \mu\neq 76\\$
$z=\dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\\$
$=\dfrac{72-76}{\frac{8}{\sqrt{100}}}\\$
$=\dfrac{-4}{0.8}\\$
$z=-5\\$
$\left| z \right|=5\text{ at }\;5 \% \;\text{level of significance.}\\$
$\left| z \right|> z_{\frac{\alpha}{2}}=1.96\\$
$H_{0}\; \text{lies in the critical region.}\\$
$\text{Therefore}\; H_{0}\;\text{is rejected.}\\$

$\text{17. The mean breaking strength of cables supplied by a manufacturer is 1,800 with}\\$
$\text{a standard deviation 100. By a new technique in the manufacturing process it is } \\$
$\text{claimed that the breaking strength of the cables has increased. In order to test }\\$
$\text{this claim a sample of 50 cables is tested. It is found that the mean breaking } \\$
$\text{strength is 1,850. Can you support the claim at 0.01 level of significance.} \\$

$n=50\\$
$\mu=1800\\$
$\sigma=100\\$
$\bar{x}=1850\\$
$\text{The test statistic:}\\$
$H_{0}: \mu=1800\\$
$H_{1}: \mu\neq 1800\\$
$z=\dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\\$
$=\dfrac{1850-1800}{\frac{100}{\sqrt{50}}}\\$
$=\dfrac{50}{14.1421}\\$
$z=3.536\\$
$\left| z \right|=3.536\text{ at }\;1 \% \;\text{level of significance.}\\$
$\left| z \right|> z_{\frac{\alpha}{2}}=2.58\\$
$H_{0}\; \text{lies in the critical region.}\\$
$\text{Therefore}\; H_{0}\;\text{is rejected.}\\$