Samacheer Kalvi Class 12 Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Exercise 8.2

Question 1. Mention two branches of statistical inference?
Answer: The two main parts of statistical inference are:
(i) Estimation
(ii) Testing of Hypothesis
Both branches help us make informed decisions about large groups based on smaller samples.
In simple words: Statistical inference has two parts: guessing values (estimation) and checking ideas (testing hypotheses).

🎯 Exam Tip: Remember these two branches are the core methods statisticians use to draw conclusions about populations from sample data.

Question 2. What is an estimator?
Answer: An estimator is a statistic from a sample that is used to guess an unknown value of a whole population. For example, the average height of students in a small group can be used as an estimator for the average height of all students in the school.
In simple words: An estimator is a tool (like a sample average) used to guess a value for a bigger group (population).

🎯 Exam Tip: Think of an estimator as a sample measurement that helps you make a good guess about a larger, unknown population characteristic.

Question 3. What is an estimate?
Answer: An estimate is a specific number we get when we use an estimator. It is a single, concrete value that comes from our sample calculations. For instance, if the average height of your sample of students is 160 cm, then 160 cm is the estimate.
In simple words: An estimate is the actual number you get when you calculate using your estimator.

🎯 Exam Tip: The estimate is the numerical result, while the estimator is the method or formula used to find it.

Question 4. What is point estimation?
Answer: Point estimation is when we use a single value from our sample to guess a single, exact value for a population characteristic. It provides just one number as the best possible guess. For instance, stating that the average age of a population is exactly 30 years old based on a sample.
In simple words: Point estimation means giving one specific number as your best guess for a population value.

🎯 Exam Tip: A point estimate is a single numerical value, like an exact number, that represents our best guess for a population parameter.

Question 5. What is interval estimation?
Answer: Interval estimation is when we provide a range of values within which a population characteristic is likely to fall, instead of just a single number. This range gives us a better idea of the uncertainty involved. This method is often preferred because it reflects the variability that naturally occurs in sampling.
In simple words: Interval estimation means guessing a range of numbers where a population value probably lies, not just one exact number.

🎯 Exam Tip: An interval estimate gives a "between this and that" answer, providing more realistic information than a single point estimate.

Question 6. What is confidence interval?
Answer: A confidence interval is a range of values calculated from sample data that is likely to contain an unknown population parameter. We choose a small value, `\( \alpha \)` (called the level of significance, usually 1% or 5%), and then find two constants, `\( c_1 \)` and `\( c_2 \)`, such that `\( P(c_1 < \Theta < c_2 | t) = 1 - \alpha \)`. These `\( c_1 \)` and `\( c_2 \)` values are known as confidence limits. The interval `\( [c_1, c_2] \)` is the confidence interval, and `\( (1 - \alpha) \)` is called the confidence coefficient, indicating the probability that the interval contains the true parameter. A higher confidence coefficient means a wider interval, suggesting more certainty but less precision.
In simple words: A confidence interval is a range of numbers from your sample that probably holds the true value of something in the whole population.

🎯 Exam Tip: Always specify the confidence level (e.g., 95% or 99%) when talking about a confidence interval, as it shows how sure you are that the interval contains the true population parameter.

Question 7. What is null hypothesis? Give an example.
Answer: According to Professor R. A. Fisher, the null hypothesis, denoted as `\( H_0 \)`, is the initial statement or assumption that we test for possible rejection. We assume it is true until proven otherwise. For example, if we want to check if the average height of students in a school is 160 cm, our null hypothesis would be `\( H_0: \mu = 160 \) cm`. This hypothesis typically states there is no effect or no difference between groups.
In simple words: The null hypothesis `\( H_0 \)` is the starting idea you want to test, often saying there is no change or difference.

🎯 Exam Tip: The null hypothesis always includes an equality sign (`=`), representing the status quo or no effect.

Question 8. Define alternative hypothesis.
Answer: The alternative hypothesis, denoted as `\( H_1 \)`, is the statement that contradicts the null hypothesis. It is what we conclude if we reject the null hypothesis, suggesting that there *is* an effect or a difference. For example, if our null hypothesis is `\( H_0: \mu = \mu_0 \)` (meaning the population mean equals a specific value `\( \mu_0 \)`), then the alternative hypothesis could be:
(i) `\( H_1: \mu \neq \mu_0 \)` (the mean is not equal to `\( \mu_0 \)`)
(ii) `\( H_1: \mu > \mu_0 \)` (the mean is greater than `\( \mu_0 \)`)
(iii) `\( H_1: \mu < \mu_0 \)` (the mean is less than `\( \mu_0 \)`).
The alternative hypothesis is what you are trying to prove.
In simple words: The alternative hypothesis `\( H_1 \)` is the opposite of the null hypothesis, saying there is a change or difference.

🎯 Exam Tip: The alternative hypothesis (`\( H_1 \)` ) always includes an inequality sign (`<`, `> `, or `\(\neq\) `), representing the claim or difference you are trying to find evidence for.

Question 9. Define critical region.
Answer: The critical region, also known as the region of rejection, is a specific part of the sample space where, if a test statistic falls into it, we decide to reject the null hypothesis `\( H_0 \)`. This region includes values that are extreme enough to be considered unlikely if the null hypothesis were true. The size of this region is determined by the chosen level of significance.
In simple words: The critical region is the area where, if your test result lands there, you will reject your main idea (null hypothesis).

🎯 Exam Tip: Understand that if your calculated test statistic falls into the critical region, it means your result is statistically significant enough to reject the null hypothesis.

Question 10. Define critical value.
Answer: The critical value is the point or boundary that separates the critical region (rejection region) from the acceptance region in a hypothesis test. If the calculated test statistic goes beyond this critical value, the null hypothesis is rejected. The critical value depends on two main things:
(i) The level of significance `\( (\alpha) \)` chosen for the test.
(ii) Whether the alternative hypothesis is two-tailed (testing for difference in both directions) or single-tailed (testing for difference in only one direction).
This value is a crucial threshold in decision-making.
In simple words: The critical value is a dividing line that tells you if your test result is strong enough to reject the main idea.

🎯 Exam Tip: The critical value acts as a benchmark; if your test statistic crosses it, your findings are considered strong enough to be statistically significant.

Question 11. Define level of significance.
Answer: The level of significance, denoted as `\( \alpha \)`, is the probability of making a Type I error. A Type I error happens when we incorrectly reject a null hypothesis that is actually true. Common levels of significance used in hypothesis testing are 5% (0.05) and 1% (0.01). This value is always chosen and fixed before any data is collected or analyzed. Setting `\( \alpha \)` helps to control the risk of drawing a wrong conclusion.
In simple words: The level of significance (`\( \alpha \)` ) is how much risk you are willing to take of being wrong when you say there's a difference.

🎯 Exam Tip: A smaller level of significance (e.g., 0.01) means you need stronger evidence to reject the null hypothesis, reducing the chance of a Type I error.

Question 12. What is type I error
Answer: A Type I error occurs when we wrongly reject the null hypothesis `\( H_0 \)` even though it is actually true. It is like saying a person is guilty when they are innocent. This type of error is controlled by the level of significance `\( \alpha \)`. For example, if `\( \alpha \)` is set to 0.05, there's a 5% chance of making a Type I error.
In simple words: A Type I error is when you decide there is a change or effect, but actually there isn't.

🎯 Exam Tip: Think of a Type I error as a "false positive" – you detected an effect, but it wasn't really there.

Question 13. What is single tailed test.
Answer: A single-tailed test (or one-tailed test) is a hypothesis test where the alternative hypothesis (`\( H_1 \)`) states that the population parameter is either greater than or less than the value in the null hypothesis (`\( H_0 \)`) but not both. This means the rejection region is entirely in one tail (either the right or left) of the sampling distribution. For example, if we are testing if a new medicine *increases* effectiveness, we would use a single-tailed test like `\( H_1: \mu > \mu_0 \)`. Similarly, if we test if it *decreases* effectiveness, we use `\( H_1: \mu < \mu_0 \)`.
In simple words: A single-tailed test checks for a difference in only one direction, like if something is strictly bigger or strictly smaller.

🎯 Exam Tip: Use a one-tailed test when you have a specific directional expectation (e.g., "greater than" or "less than"); otherwise, a two-tailed test is generally more appropriate.

Question 14. A sample of 100 items, draw from a universe 64 and S.D 3, has a mean value 63.5. Is the difference in the mean significant?
Answer:
Given:
Sample size \( n = 100 \)
Population mean \( \mu = 64 \)
Population SD \( \sigma = 3 \)
Sample mean \( \bar{x} = 63.5 \)

Null Hypothesis `\( H_0: \mu = 64 \)` (The sample comes from a population with mean 64 and SD 3)
Alternative Hypothesis `\( H_1: \mu \neq 64 \)` (two-tailed test) (The sample does not come from a population with mean 64 and SD 3)

Level of significance `\( \alpha = 5\% = 0.05 \)`

Test statistic:
`\( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)`
`\( z = \frac{63.5 - 64}{\frac{3}{\sqrt{100}}} \)`
`\( = \frac{-0.5}{\frac{3}{10}} \)`
`\( = \frac{-0.5}{0.3} \)`
`\( = -1.667 \)`

Absolute value of z: `\( |z| = |-1.667| = 1.667 \)`

Critical value at 5% level of significance: `\( Z_{\frac{\alpha}{2}} = 1.96 \)`

Since the calculated value `\( |z| = 1.667 \)` is less than the table critical value `\( Z_{\frac{\alpha}{2}} = 1.96 \)`, we accept the null hypothesis. This means there is not enough evidence to say that the sample mean is significantly different from the population mean of 64. The observed difference could be due to random chance. The data suggests the sample could reasonably come from the population with a mean of 64.
In simple words: We calculated a 'z' value of -1.667. Because this number is smaller than the special 'critical' value of 1.96, we cannot say the sample mean is truly different from the population mean. It seems to be the same.

🎯 Exam Tip: Always compare the absolute value of your calculated test statistic to the critical value. If `\( |z| < Z_{\text{critical}} \)`, accept `\( H_0 \)`; if `\( |z| \ge Z_{\text{critical}} \)`, reject `\( H_0 \)`. Remember to write the inference clearly.

Question 15. A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be reasonably regarded as a sample from a large population with mean height of 67.39 inches and standard deviation 1.30 inches?
Answer:
Given:
Sample size \( n = 400 \)
Sample mean \( \bar{x} = 67.47 \) inches
Population mean \( \mu = 67.39 \) inches
Population SD \( \sigma = 1.30 \) inches

Null Hypothesis `\( H_0: \mu = 67.39 \)` inches (The sample comes from a population with mean 67.39 inches)
Alternative Hypothesis `\( H_1: \mu \neq 67.39 \)` inches (two-tailed test) (The sample does not come from a population with mean 67.39 inches)

Level of significance `\( \alpha = 5\% = 0.05 \)`

Test statistic:
`\( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)`
`\( z = \frac{67.47 - 67.39}{\frac{1.30}{\sqrt{400}}} \)`
`\( = \frac{0.08}{\frac{1.30}{20}} \)`
`\( = \frac{0.08}{0.065} \)`
`\( = 1.23076 \)`
`\( \approx 1.2308 \)`

Critical value at 5% level of significance: `\( Z_{\frac{\alpha}{2}} = 1.96 \)`

Since the calculated value `\( |z| = 1.2308 \)` is less than the table critical value `\( Z_{\frac{\alpha}{2}} = 1.96 \)`, we accept the null hypothesis. This means that the sample of 400 individuals can reasonably be considered to come from a population with a mean height of 67.39 inches. The observed difference is not statistically significant and can be attributed to random sampling variation. This confirms the sample aligns with the larger population characteristics.
In simple words: Our calculated 'z' value (1.2308) is smaller than the critical value (1.96). This means the sample's average height is similar enough to the population's average height, so we can consider them from the same group.

🎯 Exam Tip: Clearly state the null and alternative hypotheses, then perform the calculation and compare the calculated z-value with the critical z-value for your conclusion.

Question 16. The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state's education system, the scores of 100 of the state's students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores.
Answer:
Given:
Population mean `\( \mu = 76 \)`
Population standard deviation `\( \sigma = 8 \)`
Sample size `\( n = 100 \)`
Sample mean `\( \bar{x} = 72 \)`

Null Hypothesis `\( H_0: \mu = 76 \)` (There is no significant difference; the state's average score is 76)
Alternative Hypothesis `\( H_1: \mu \neq 76 \)` (two-tailed test) (There is a significant difference; the state's average score is not 76)

Level of significance `\( \alpha = 0.05 \)`

Test statistic:
`\( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) \)`
`\( z = \frac{72 - 76}{\frac{8}{\sqrt{100}}} \)`
`\( = \frac{-4}{\frac{8}{10}} \)`
`\( = \frac{-4}{0.8} \)`
`\( = -5 \)`

Absolute value of z: `\( |z| = |-5| = 5 \)`

Critical value at 5% level of significance (two-tailed): `\( Z_{\frac{\alpha}{2}} = 1.96 \)`

Since the calculated value `\( |z| = 5 \)` is greater than the table critical value `\( Z_{\frac{\alpha}{2}} = 1.96 \)`, we reject the null hypothesis. This means there is a significant difference between the state's average scores and the national average scores. The state's education system's performance is notably different from the national standard. This significant difference suggests the state's students are not performing at the national average.
In simple words: Our calculated 'z' value (5) is much bigger than the critical value (1.96). This means the state's average score is very different from the national average. So, we reject the idea that there's no difference.

🎯 Exam Tip: Always note if it's a two-tailed or one-tailed test when looking up critical values, as this affects the value used for comparison. Large `\( |z| \)` values far from zero usually lead to rejecting the null hypothesis.

Question 17. The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance.
Answer:
Given:
Population mean `\( \mu = 1800 \)`
Population standard deviation `\( \sigma = 100 \)`
Sample size `\( n = 50 \)`
Sample mean `\( \bar{x} = 1850 \)`

Null Hypothesis `\( H_0: \mu = 1800 \)` (The mean breaking strength has not increased)
Alternative Hypothesis `\( H_1: \mu > 1800 \)` (one-tailed test) (The mean breaking strength has increased)
*Note: Although the problem states "has increased" implying a one-tailed test, the solution provided in the source uses a two-tailed critical value. To follow the provided solution's calculation approach, we will calculate `\( |z| \)` and compare it with `\( Z_{\frac{\alpha}{2}} \)`. In a strict one-tailed test for `\( H_1: \mu > 1800 \)`, we would compare `\( z \)` directly to `\( Z_{\alpha} \)`.*

Level of significance `\( \alpha = 0.01 \)`

Test statistic:
`\( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) \)`
`\( z = \frac{1850 - 1800}{\frac{100}{\sqrt{50}}} \)`
`\( = \frac{50}{\frac{100}{7.0711}} \)`
`\( = \frac{50 \times 7.0711}{100} \)`
`\( = \frac{353.555}{100} \)`
`\( = 3.53555 \)`
`\( \approx 3.536 \)`

Absolute value of z: `\( |z| = |3.536| = 3.536 \)`

Critical value at 1% level of significance (two-tailed): `\( Z_{\frac{\alpha}{2}} = 2.58 \)`

Since the calculated value `\( |z| = 3.536 \)` is greater than the table critical value `\( Z_{\frac{\alpha}{2}} = 2.58 \)`, we reject the null hypothesis. This means we can support the claim that the breaking strength of the cables has increased at the 0.01 level of significance. The new manufacturing technique has successfully improved the cable strength.
In simple words: Our calculated 'z' value (3.536) is higher than the special 'critical' value (2.58). This tells us that the cables are indeed stronger. So, we agree with the claim that the new method made them better.

🎯 Exam Tip: Be very careful when setting up the null and alternative hypotheses, especially for one-tailed versus two-tailed tests, as it affects the critical value you use for comparison.