Statistics II · Unit 1 · 6 hrs
Sampling Distribution and Estimation
Exam-focused notes for Sampling Distribution and Estimation (Statistics II, STA215): what the TU syllabus asks and how it has actually been tested, with 12 solved past questions from this unit.
What this unit covers
- Sampling distribution
- sampling distribution of mean and proportion
- Central Limit Theorem
- Concept of inferential Statistics
- Estimation
- Methods of estimation
- Properties of good estimator
- Determination of sample size
- Relationship of sample size with desired level of error
- Problems and illustrative examples related to computer Science and IT
Estimation
Define confidence level in estimation. A quality control inspector collected a random sample of 400 tubes of toothpaste from the production line and found that 20 of the tubes had leaks from the tail end. Construct 96% confidence interval for the percentage of all the toothpaste tubes that had leakage and interpret the result. [5]
The confidence level $(1-\alpha)$ is the probability that the interval estimate constructed from sample data will contain the true population parameter. A 96% confidence level means that if repeated random samples were taken and an interval built from each,...
Full solved answer →An effort to estimate the mean amount per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers and sample mean is found at 24.80. Assume population standard deviation is 5. a. Compute standard error of mean. b. Find 95% confidence interval estimate for the population mean. [5]
Parameter Value ------------------ Sample size ($n$) 49 Sample mean ($\bar{x}$) 24.80 Population standard deviation ($\sigma$) 5 Confidence level 95% Since $\sigma$ is known, we use the Z-distribution. --- $$SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{49}...
Full solved answer →A survey was conducted among 70 students studying B.Sc. CSIT in some colleges randomly. Among them, 50 students secured more than 80% marks in statistics. Compute 99% and 95% confidence intervals for the population proportion of students who secured more than 80% marks in subject statistics, and comment on the results. [5]
Parameter Value ------------------ Sample size $n$ 70 Students scoring 80% ($X$) 50 Sample proportion $p = X/n$ $50/70 = 0.7143$ $q = 1 - p$ $0.2857$ $Z$ for 99% $2.576$ $Z$ for 95% $1.96$ All required data present. $$S.E.(p) = \sqrt{\frac{pq}{n}} = \sqrt{\...
Full solved answer →A machine produce metal rods used in an automobile suspension system. A random sample of 6 rods is selected and diameter is measured. The measuring data (in millimeters) are as follows. Assuming that the sample drawn from the normally distributed population. Find 95% two sided confidence interval on the mean rod diameter, and interpret the result with reference to the given problem.
| 8.24 | 8.26 | 8.20 | 8.28 | 8.21 | 8.23 |
[5]
- Sample size: $n = 6$ - Data (mm): $8.24, 8.26, 8.20, 8.28, 8.21, 8.23$ - Population: normally distributed - Population standard deviation $\sigma$: unknown - Confidence level: $95\%$, so $\alpha = 0.05$ Since $\sigma$ is unknown and $n$ is small, use the ...
Full solved answer →A manufacturer of computer paper has a production process that operates continuously throughout an entire production shift. The paper is expected to have an average length of 11 inches and standard deviation is known to be 0.01 inch. Suppose random sample of 100 sheets is selected and the average paper length is found to be 10.68 inches. Set up 95% and 90% confidence interval estimate of the population average paper length. [5]
Parameter Value ------------------ Expected/target mean 11 inches (target, not used in CI) Population standard deviation ($\sigma$) 0.01 inch Sample size ($n$) 100 Sample mean ($\bar{X}$) 10.68 inches Since $\sigma$ is known and $n = 100$ (large), we use th...
Full solved answer →Central Limit Theorem
Write short note on: a) Central limit theorem. b) Determination of required sample size to estimate population proportion. [5]
--- The Central Limit Theorem states that if a sufficiently large random sample of size n is drawn from any population (regardless of its shape) with population mean μ and population standard deviation σ, then the sampling distribution of the sample mean (X...
Full solved answer →Define Central limit theorem. The life of a certain brand of an electric bulb may be considered a random variable with mean 1350 hours and standard deviation 550 hours. Using central limit theorem, find the probability that the average life time of 100 bulbs exceeds 1440 hours. [5]
Given data: - Population mean: $\mu = 1350$ hours - Population standard deviation: $\sigma = 550$ hours - Sample size: $n = 100$ bulbs - Value tested: $\bar{X} = 1440$ hours - Required: $P(\bar{X} 1440)$ All data present. The Central Limit Theorem (CLT) sta...
Full solved answer →Determination of sample size
What do you understand by estimation? If we want to determine average mechanical aptitude of a large group of workers, how large a random sample is needed to be able to assert with probability 0.95 that the sample mean will not differ from the true mean by more than 2.0 points? Assume that population standard deviation is 30. [5]
Parameter Value ------------------ Confidence level $(1-\alpha)$ 0.95 Maximum allowable error $(E)$ 2.0 points Population standard deviation $(\sigma)$ 30 Critical value $Z{\alpha/2}$ at 95% 1.96 Estimation is the statistical procedure of using sample data ...
Full solved answer →A study of 1000 computer engineers conducted by their professional organization reported that 300 stated that their firms’ greatest concern was to uplift the professional quality of work. In order to conduct a follow up study to estimate the population proportion of computer engineers to fulfill their greatest concern within ±0.01 with 99% confidence interval, how many computer engineers would be required to be surveyed? [5]
Parameter Value ------------------ Preliminary sample size $n0 = 1000$ Number with greatest concern $X = 300$ Sample proportion $\hat{p} = 300/1000 = 0.30$ Margin of error $E = 0.01$ Confidence level $99\%$ --- Step 1: Determine p and q $$p = 0.30, \quad q ...
Full solved answer →Determine the minimum sample size required so that the sample estimate lies within 10% of the true value 95% level of confidence, when coefficient of variation is 60% [5]
Parameter Value ------------------ Allowable relative error, $e$ 10% of true value = 0.10 Confidence level 95% Coefficient of variation, $CV$ 60% = 0.60 Formula For estimating the mean within a relative error $e$ (fraction of the true mean): $$n = \left(\fr...
Full solved answer →sampling distribution of mean and proportion
Explain the sample distribution of mean with reference to some numerical example. Illustrate the practical implications of the Central Limit Theorem (CLT) in inferential statistics.[10]
When all possible random samples of size n are drawn from a population of size N, and the mean of each sample is computed, the probability distribution formed by these sample means is called the sampling distribution of the sample mean. Let a population hav...
Full solved answer →Describe the concept of sampling distribution of mean with reference to the population data (20, 21, 22 & 23) of size 4. In order to explain this, perform simple random sampling with replacement taking all possible samples with sample size n = 2. While describing the sampling distribution following issues will be covered: a. population mean & population variance, and its distribution b. Sample mean & sample variance, and its distribution c. Comparison of population mean and sample mean; population variance and sample variance; population distribution and sampling distribution based on the given data. d. Standard error of mean e. Final comments based on your result[10]
- Population values: $X = \{20, 21, 22, 23\}$ - Population size: $N = 4$ - Sampling: Simple Random Sampling With Replacement (SRSWR) - Sample size: $n = 2$ - All possible samples taken All required data present. --- $$\mu = \frac{\sum X}{N} = \frac{20+21+22...
Full solved answer →Make Unit 1 stick
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