# Statistics Indian Statistical Service Part 3 Question-3 Solved Solution

**3.(a) (i) Distinguish between sampling and non-sampling errors.**

Sampling error and non-sampling error are two types of errors that can occur in statistical analysis. Sampling error occurs when the sample that is used to estimate the population parameter is not a perfect representation of the population. It is the difference between the sample statistic and the population parameter, and it is due to chance. Sampling error can be reduced by increasing the sample size or using a more representative sample. Common examples of sampling error include random variation, bias in sample selection, and incomplete coverage of the population.

Non-sampling error, on the other hand, is an error that occurs in the data collection and analysis process that is not related to sampling. Non-sampling errors are systematic errors, meaning that they are not due to chance, but rather to a flaw in the data collection or analysis process. Non-sampling errors can be due to many factors such as measurement errors, non-response bias, errors in data processing, and errors in estimation techniques. They can be reduced by improving data collection methods, using appropriate statistical techniques, and careful data validation and processing. In summary, sampling error is due to chance and occurs when the sample used to estimate a population parameter is not a perfect representation of the population. Non-sampling error is systematic and occurs due to flaws in the data collection and analysis process that are not related to sampling. Both types of errors can affect the accuracy and reliability of statistical analysis and should be minimized as much as possible.

**(ii) Describe the effect of non-response on the bias of the sample estimate of the population mean under simple random sampling.**

** **
Non-response occurs when some individuals or units selected in a sample do not participate in a survey or study. It can affect the accuracy and representativeness of the sample, and therefore, impact the bias of the sample estimate of the population mean under simple random sampling.
In simple random sampling, each individual in the population has an equal probability of being selected for the sample. If non-response occurs, it means that some selected individuals do not participate in the survey, which reduces the sample size and can introduce bias into the sample estimate. This is because non-respondents may differ systematically from respondents in terms of the characteristics being measured, and this can lead to a biased estimate of the population mean.
For example, suppose we are conducting a survey to estimate the average income of a population using simple random sampling. If a certain group of high-income individuals refuses to participate in the survey, the sample may not be representative of the population and may underestimate the true population mean. This is because the non-response of high-income individuals may result in a sample that is biased towards low-income individuals.
In general, the impact of non-response on the bias of the sample estimate of the population mean depends on the characteristics of the non-respondents and the relationship between these characteristics and the variable being measured. To minimize the bias due to non-response, survey researchers may use techniques such as follow-up surveys, incentives for participation, and weighting adjustments to account for non-response and adjust the sample estimates accordingly.
**3.(b) We are interested in the average age of a large population of employees in a **
**particular service sector. The population is stratified based on the information of their age. A simple random sample was then taken from each strata with a total size 100. Under standard notations, using the information given in the table and ignoring fpc**
*Stratum ageYh**N***< 4050%25164040**** â€”**** 5030%451020> 5020%582040**
**(i) Give the stratified estimator of the population mean. Is this estimator different from simple mean calculated over all the sample?**
The stratified estimator of the population mean can be calculated as follows:
y_h = Y_h / n_h (mean of the h-th stratum)
t_s = âˆ‘h=1^L (N_h / N) * y_h (weighted mean of all stratum means)
T = âˆ‘h=1^L Y_h (total of all strata)
N = âˆ‘h=1^L N_h (total population size)
n = âˆ‘h=1^L n_h (total sample size)
Therefore, the stratified estimator of the population mean is:
yÌ‚_st = t_s = (0.5 * 25/40) + (0.3 * 45/20) + (0.2 * 58/40) = 42.1
The stratified estimator of the population mean can be calculated as follows:
**(ii) Give the variance of the estimator in (i)."**
To find the average age of the population of employees, we need to first calculate the weighted mean of the sample means of each stratum.
The formula for the weighted mean is:
$\bar{y}{w} = \frac{\sum{h=1}^{H} w_h \bar{y}{h}}{\sum{h=1}^{H} w_h}$
where:
$\bar{y}_{w}$ is the weighted mean
$\bar{y}_{h}$ is the sample mean of stratum $h$
$w_h$ is the weight for stratum $h$, which is proportional to the size of the stratum
The sample mean of each stratum is calculated as:
$\bar{y}{h} = \frac{\sum{i=1}^{n_h} y_{hi}}{n_h}$
where:
$\bar{y}_{h}$ is the sample mean of stratum $h$
$y_{hi}$ is the age of the $i$-th employee in stratum $h$
$n_h$ is the sample size of stratum $h$
Using the information given in the table, we can calculate the weighted mean as follows:
$\bar{y}_{w} = \frac{(0.5)(25)+(0.3)(45)+(0.2)(58)}{25+45+58} = \frac{12.5+13.5+11.6}{128} \approx 12.9$
Therefore, the average age of the population of employees in the service sector is approximately 12.9 years.
**(iii) What would have been the gain in precision had you resorted to proportional **
**Allocation. **
To calculate the gain in precision had proportional allocation been used, we first need to calculate the variances of the sample means using the current method (equal allocation) and the proportional allocation method.
For the current method, the variance of the sample mean of each stratum is given by:
$V(\bar{y}_h) = \frac{S_h^2}{n_h}$
Using the information given in the table, we can calculate the variances of the sample means as follows:
$V(\bar{y}_1) = \frac{16}{40} = 0.4$
$V(\bar{y}_2) = \frac{10}{45} \approx 0.222$
$V(\bar{y}_3) = \frac{20}{58} \approx 0.345$
The weighted variance of the sample means under equal allocation is then:
$V(\bar{y}w) = \frac{\sum{h=1}^{H} w_h V(\bar{y}h)}{\sum{h=1}^{H} w_h} = \frac{(0.5)(0.4)+(0.3)(0.222)+(0.2)(0.345)}{1} \approx 0.3126$
For proportional allocation, the variance of the sample mean of each stratum is given by:
$V(\bar{y}_h) = \frac{S_h^2}{n_h} \frac{N_h-n_h}{N_h-1}$
Using the information given in the table, we can calculate the variances of the sample means under proportional allocation as follows:
$V(\bar{y}_1) = \frac{16}{40} \frac{1}{0.5(100)-40} \approx 0.0133$
$V(\bar{y}_2) = \frac{10}{45} \frac{1}{0.3(100)-20} \approx 0.0049$
$V(\bar{y}_3) = \frac{20}{58} \frac{1}{0.2(100)-58} \approx 0.0088$
The weighted variance of the sample means under proportional allocation is then:
$V(\bar{y}w) = \frac{\sum{h=1}^{H} w_h V(\bar{y}h)}{\sum{h=1}^{H} w_h} = \frac{(0.5)(0.0133)+(0.3)(0.0049)+(0.2)(0.0088)}{1} \approx 0.0083$
The gain in precision is then given by:
$\text{gain in precision} = \frac{V(\bar{y}w){\text{equal allocation}}}{V(\bar{y}w){\text{proportional allocation}}} = \frac{0.3126}{0.0083} \approx 37.7$
Therefore, using proportional allocation would have resulted in a gain in precision of approximately 37.7 times compared to the equal allocation method.

**3.(c) For estimating mean of the population, show that systematic sampling is more**

**effective in removing the effect of a suspected linear trend in the population than **
**simple random sampling. 15**

Systematic sampling is a method of selecting a sample from a population by selecting every kth element from a list or sequence. Simple random sampling, on the other hand, involves randomly selecting elements from the population without any pattern or order.

When there is a suspected linear trend in the population, systematic sampling can be more effective than simple random sampling in removing its effect. This is because systematic sampling ensures that the sample is evenly spread throughout the population, while simple random sampling can result in a sample that is clustered in certain areas of the population.

Let's consider an example to illustrate this. Suppose we have a population of 1000 individuals and we suspect that there is a linear trend in the population. We want to estimate the mean height of the population. We can use systematic sampling by selecting every 10th individual from the population list. Alternatively, we can use simple random sampling by randomly selecting 100 individuals from the population.

If there is a linear trend in the population, the systematic sampling method will ensure that the sample is spread evenly throughout the population, including the areas where the trend is present. This can help to reduce the effect of the trend on the sample mean estimate.

In contrast, simple random sampling can result in a sample that is not evenly spread throughout the population, and may miss some of the areas where the trend is present. As a result, the sample mean estimate may be biased by the trend.

Therefore, in situations where a suspected linear trend may affect the population mean estimate, systematic sampling can be a more effective method of sampling than simple random sampling.

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