Samacheer Kalvi Class 12 Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Exercise 8.4

Question 1. Explain the types of sampling.
Answer: There are two main types of sampling methods:
(1) **Non-Random Sampling (or Non-Probability Sampling):** In this method, the chance of each item in the population being selected for the sample is not equal.
(2) **Random Sampling (or Probability Sampling):** Here, every single item in the entire population has an equal and fair chance of being chosen for the sample. This makes the sample more representative. As Dr. Yates noted, "Every member of a parent population has had equal chances of being included." Harver also explained that "A random sample is a sample selected in such a way that every item in the population has an equal chance of being included."
The different types of probability sampling include:
(1) Simple random sampling
(2) Stratified random sampling
(3) Systematic sampling
In simple words: Sampling is how you pick items for a study. You can pick them randomly (where everyone has an equal chance) or non-randomly (where some have a higher chance). Random picking helps make sure your small group truly represents the bigger group.

🎯 Exam Tip: When explaining sampling types, always define the two main categories (random and non-random) and briefly mention their sub-types to show comprehensive understanding.

Question 2. Write short note on sampling distribution and standard error.
Answer:
**Sampling Distribution:**
A sampling distribution of a statistic is like a special frequency distribution. It is created by taking many different samples of the same size from a large population. For each sample, we calculate a statistic (like the mean or proportion), and then we make a distribution of all these calculated statistic values. This helps us understand how stable our statistic is.
For example, if we draw a sample of size \( n \) from a finite population of \( N \), the total number of possible samples, \( k \), can be found using the combination formula:
\( N C_n = \frac{N!}{n!(N-n)!} = k \) (say)
**Standard Error:**
Standard Error, often abbreviated as S.E., measures how much the sample statistic is likely to vary from the actual population parameter. It is the standard deviation of the sampling distribution of a statistic. A smaller standard error means that the sample statistic is a more accurate estimate of the population parameter. For large samples, the standard errors of some well-known statistics are provided, with \( n \) being the sample size and \( \sigma^2 \) representing the population variance.
In simple words: A sampling distribution shows all the possible results you can get from many small groups taken from a big group. Standard error tells you how much your small group's average might be different from the big group's true average.

🎯 Exam Tip: Remember that sampling distribution describes the variability of a statistic across multiple samples, while standard error quantifies this variability, making it crucial for confidence intervals and hypothesis testing.

Question 3. Explain the procedures of testing of hypothesis
Answer: The following steps are used when testing hypotheses:
1. **Null Hypothesis (H₀):** First, we set up the null hypothesis. This is a statement that there is no effect or no difference, and we usually try to find evidence against it.
2. **Alternative Hypothesis (H₁):** Next, we set up the alternative hypothesis. This is the statement that we want to prove or accept if the null hypothesis is rejected. It helps us decide if we need a two-tailed test (checking for differences in both directions) or a single-tailed test (checking for a difference in one specific direction).
3. **Level of Significance ( \( \alpha \) ):** We choose an appropriate level of significance ( \( \alpha \) ), which is usually 0.05 or 0.01. This value represents the maximum risk we are willing to take of rejecting the null hypothesis when it is actually true. This level is decided before any sample data is collected.
4. **Test Statistic:** We then calculate the test statistic from our sample data. This value helps us to decide whether to reject or fail to reject the null hypothesis. For example, for a large sample, the test statistic \( Z \) is computed as:
\( Z = \frac{t-E(t)}{\sqrt{var(t)}} = \frac{t-E(t)}{S.E(t)} \sim N(0, 1) \)
5. **Conclusion:** Finally, we compare the calculated test statistic (e.g., \( Z \)) with the critical value (or table value \( Z_\alpha \)) at our chosen level of significance.
(i) If \( |Z| < Z_\alpha \), meaning the calculated value is less than the critical value, we say it is not significant. This suggests that any observed difference might just be due to random fluctuations in sampling, and the data does not provide enough evidence to reject the null hypothesis, so it may be accepted.
(ii) If \( |Z| > Z_\alpha \), meaning the calculated value is greater than the critical value, we say it is significant. This means there is enough evidence to reject the null hypothesis at the chosen level of significance \( \alpha \).
In simple words: To test an idea, first, we guess there's no change (null hypothesis). Then, we guess there is a change (alternative hypothesis). We set a "risk" level for being wrong. We do some math with our data to get a test score. Finally, we compare our score to a special number to decide if our first guess was wrong.

🎯 Exam Tip: Clearly state all five steps of hypothesis testing. For each step, mention its purpose and how it contributes to the final decision regarding the null hypothesis.

Question 4. Explain in detail about the test of significance for single mean.
Answer: The test of significance for a single mean helps us determine if a sample mean is significantly different from a hypothesized population mean.
Let's say \( x_i \) (where \( i = 1, 2, 3, ..., n \)) represents a random sample of size \( n \) taken from a normal population. This population has a mean of \( \mu \) and a variance of \( \sigma^2 \). When we take many such samples, the sample mean, denoted as \( \bar{x} \), will also follow a normal distribution. Its mean will be \( \mu \) and its variance will be \( \frac{\sigma^2}{n} \). This can be written as \( \bar{x} \sim N(\mu, \frac{\sigma^2}{n}) \).
For large samples, the standard normal variable corresponding to \( \bar{x} \) is given by the test statistic \( Z \):
\( Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \sim N(0, 1) \)
Under the null hypothesis, which states that the sample comes from a population with a mean of \( \mu \) and variance \( \sigma^2 \), there is no significant difference between the sample mean \( (\bar{x}) \) and the population mean \( (\mu) \). So, the test statistic for large samples is:
\( Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \)
This Z-score helps us decide if the difference we see between the sample mean and the population mean is just by chance or if it's a real, significant difference.
In simple words: This test checks if the average of a small group is truly different from the average of the whole big group it came from. We use a special formula to get a 'Z' score, which helps us compare and decide if the difference is real or just a coincidence.

🎯 Exam Tip: When explaining the test of significance for a single mean, clearly state the null hypothesis, the formula for the Z-statistic, and how it helps compare the sample mean to the population mean.

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?
Answer:
Given:
Sample size \( n = 500 \)
Number of bad pineapples \( = 65 \)
First, calculate the sample proportion of bad pineapples, \( P \):
\( P = \frac{65}{500} = 0.13 \)
Next, calculate \( Q \), which is \( 1 - P \):
\( Q = 1 - P \implies Q = 1 - 0.13 = 0.87 \)
Now, we can find the Standard Error (S.E.) for the sample proportion using the formula:
\( S.E. = \sqrt{\frac{PQ}{N}} \)
Substitute the values of \( P \), \( Q \), and \( N \) (which is \( n \) here for the sample size):
\( S.E. = \sqrt{\frac{(0.13)(0.87)}{500}} \)
\( S.E. = \sqrt{\frac{0.1131}{500}} \)
\( S.E. = \sqrt{0.0002262} \)
\( S.E. \approx 0.01504 \)
Rounding to three decimal places, the standard error for the sample proportion is:
\( S.E. = 0.015 \)
In simple words: We found out what percentage of pineapples were bad in our sample. Then, we calculated a number called Standard Error, which tells us how much this percentage might change if we took another sample. It's like a measure of how precise our estimate is.

🎯 Exam Tip: Remember to calculate both \( P \) (proportion of success) and \( Q \) (proportion of failure) before applying the standard error formula for proportion. Rounding should be done at the final step, not intermediate steps, to maintain accuracy.

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.
Answer:
Given:
Sample size \( n = 100 \)
Sample mean \( \bar{x} = 67.45 \) kg
Sample variance \( S^2 = 9 \) kg
Calculate the sample standard deviation \( S \):
\( S = \sqrt{9} = 3 \) kg
Now, calculate the Standard Error (S.E.) of the mean:
\( S.E. = \frac{S}{\sqrt{n}} = \frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3 \)
**(a) 95% Confidence Limits for \( \mu \):**
For a 95% confidence interval, the critical value \( Z_{\frac{\alpha}{2}} \) is 1.96.
The formula for confidence limits is:
\( \bar{x} - Z_{\frac{\alpha}{2}} S.E. < \mu < \bar{x} + Z_{\frac{\alpha}{2}} S.E. \)
Substitute the values:
\( 67.45 - (1.96 \times 0.3) \leq \mu \leq 67.45 + (1.96 \times 0.3) \)
\( 67.45 - 0.588 \leq \mu \leq 67.45 + 0.588 \)
\( 66.862 \leq \mu \leq 68.038 \)
So, the 95% confidence interval is \( (66.86, 68.04) \). This means we are 95% confident that the true average weight of students falls between these values.
**(b) 99% Confidence Limits for \( \mu \):**
For a 99% confidence interval, the critical value \( Z_{\frac{\alpha}{2}} \) is 2.58.
Using the same formula:
\( \bar{x} - Z_{\frac{\alpha}{2}} S.E. \leq \mu \leq \bar{x} + Z_{\frac{\alpha}{2}} S.E. \)
Substitute the values:
\( 67.45 - (2.58 \times 0.3) \leq \mu \leq 67.45 + (2.58 \times 0.3) \)
\( 67.45 - 0.774 \leq \mu \leq 67.45 + 0.774 \)
\( 66.676 \leq \mu \leq 68.224 \)
So, the 99% confidence interval is \( (66.68, 68.22) \). This wider interval shows that to be more confident (99%), we need a broader range for the true average weight.
In simple words: We took a group of 100 students and found their average weight and how spread out the weights were. Then, we calculated two ranges. The first range (95% confident) means we are quite sure the real average weight of all students is within those numbers. The second, wider range (99% confident) means we are even more sure the real average is in that larger set of numbers.

🎯 Exam Tip: Remember that a higher confidence level (e.g., 99%) will always result in a wider confidence interval than a lower confidence level (e.g., 95%) because you need a larger range to be more certain.

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 mean I.Q 100 and standard deviation 15? (Test at 5% level of significance)
Answer:
Given:
Sample size \( n = 1600 \)
Sample mean \( \bar{x} = 99 \)
Population mean \( \mu = 100 \)
Population standard deviation \( \sigma = 15 \)
Level of significance \( \alpha = 0.05 \)
1. **Null Hypothesis (H₀):** The sample is from a population with mean IQ 100. (i.e., \( \mu = 100 \))
2. **Alternative Hypothesis (H₁):** The sample is not from a population with mean IQ 100. (i.e., \( \mu \neq 100 \)). This is a two-tailed test.
3. **Test Statistic \( Z \):**
\( Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \)
Substitute the values:
\( Z = \frac{99 - 100}{15/\sqrt{1600}} \)
\( Z = \frac{-1}{15/40} \)
\( Z = \frac{-1}{0.375} \)
\( Z = -2.666... \)
Rounding to two decimal places, \( Z = -2.67 \)
The calculated absolute value \( |Z| = |-2.67| = 2.67 \)
4. **Critical Value:**
For a 5% level of significance in a two-tailed test, the critical value \( Z_{\frac{\alpha}{2}} = Z_{0.025} \) is 1.96.
5. **Inference (Conclusion):**
Compare the calculated \( |Z| \) with the critical value \( Z_{\frac{\alpha}{2}} \):
\( 2.67 > 1.96 \)
Since the calculated value of \( |Z| \) (2.67) is greater than the critical table value (1.96) at the 5% level of significance, we reject the null hypothesis.
Therefore, we conclude that the sample mean IQ of 99 differs significantly from the population mean IQ of 100. It is unlikely that this sample came from the stated population.
In simple words: We checked if a group of 1600 children with an average IQ of 99 could have come from a larger group whose average IQ is 100. We did some math and found that the difference is too big to be just by chance. So, it's very likely that this group of children does not truly represent the larger group with an average IQ of 100.

🎯 Exam Tip: For hypothesis testing questions, always clearly state the null and alternative hypotheses, show the calculation of the test statistic, identify the critical value for the given significance level, and provide a clear conclusion based on the comparison.