Basic Statistics · Unit 7
Sampling Distributions and Estimation
Exam-focused notes for Sampling Distributions and Estimation (Basic Statistics, STA154): what the TU syllabus asks and how it has actually been tested, with 7 solved past questions from this unit.
What this unit covers
- Sampling distribution of mean
- Central limit theorem
- Point estimation
- Interval estimation and confidence intervals
- Confidence interval for population mean
- Confidence interval for population proportion
- Interpretation of confidence intervals
Confidence interval for population proportion
A sample survey of 400 customers shows that 350 are satisfied with ABC company providing internet service. Estimate the proportion of satisfied customers in the market with 95% and 99% confidence interval. [5]
- Sample size: $n = 400$ - Number of satisfied customers: $x = 350$ - Confidence levels required: 95% and 99% $$\hat{p} = \frac{x}{n} = \frac{350}{400} = 0.875$$ $$\hat{q} = 1 - \hat{p} = 0.125$$ $$SE = \sqrt{\frac{\hat{p}\,\hat{q}}{n}} = \sqrt{\frac{0.875 ...
Full solved answer →A sample survey of 80 customers shows that 56 are satisfied with an IT service. Estimate the proportion of satisfied customers in the population with a 95% confidence interval. $n = 80, \quad x = 56$ [5]
- Sample size: $n = 80$ - Number satisfied: $x = 56$ - Confidence level: $95\%$, so $z{\alpha/2} = 1.96$ $$\hat{p} = \frac{x}{n} = \frac{56}{80} = 0.70$$ $$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.70 \times 0.30}{80}} = \sqrt{\frac{0.21}{80}...
Full solved answer →Confidence interval for population mean
The fuel consumption of a new model of cars is being tested. In one trial, 50 cars chosen at random were driven under the identical conditions and the distances, x km, covered on 1 liter of petrol were recorded. The results gave the following totals: Σ x = 525, Σ x² = 5625. Calculate the 99% confidence interval for the mean petrol consumption, in km per liter. Interpret the result. [5]
- Sample size: $n = 50$ - $\Sigma x = 525$ - $\Sigma x^2 = 5625$ - Confidence level: 99% $$\bar{x} = \frac{\Sigma x}{n} = \frac{525}{50} = 10.5 \text{ km/liter}$$ Using the sample estimate of variance (with $n$ divisor, common at this level): $$s^2 = \frac{...
Full solved answer →The mass of vitamin E in a capsule manufactured by a certain drug company is normally distributed with standard deviation 0.042 mg. A random sample of 36 capsules was analyzed and the mean mass of vitamin E was found to be 5.12 mg. Calculate a symmetric 95% confidence interval for the population mean mass of vitamin E per capsule. [5]
- Population standard deviation: $\sigma = 0.042$ mg (known) - Sample size: $n = 36$ - Sample mean: $\bar{x} = 5.12$ mg - Confidence level: $95\%$ - Distribution: Normal Since $\sigma$ is known and the population is normal, use the $z$-distribution: $$\text...
Full solved answer →Assuming that the population is normally distributed, construct 95% confidence interval for the population mean using the following sample data. 1, 2, 3, 4, 5, 20. Again in the same data set, replace the value of 20 by 6, then compute the confidence interval for the mean. Explain why there is considerable difference in the confidence interval? [5]
- Dataset A: $1, 2, 3, 4, 5, 20$ with $n = 6$ - Dataset B (20 replaced by 6): $1, 2, 3, 4, 5, 6$ with $n = 6$ - Confidence level: 95%, so $\alpha = 0.05$ - Population assumed normal, $\sigma$ unknown, small sample $\Rightarrow$ use $t$-distribution with $df...
Full solved answer →Interval estimation and confidence intervals
Define interval estimation. In an examination, the random sample of 10 students selected from examination and their marks obtained in an examination was 45, 55, 20, 54, 40, 55, 51, 35, 43 and 46. Find 99% confidence interval for population mean. Assuming that the population from which samples are drawn is normally distributed. [5]
Interval estimation is a method of statistical inference in which we estimate a population parameter by constructing a range (interval) of values, together with a stated confidence level, within which the true parameter value is expected to lie. It is expre...
Full solved answer →Sampling distribution of mean
The following data represent the number of days absent of IT faculty per semester in a population of 4 faculties in an academic institute. 1, 3, 6, 7. a.) Select all possible samples of size n = 2 with replacement, and construct the sampling distribution of mean.b.) Compare the population mean and mean of all sample means. Are they equal?c.) Compare the shape of the population data and shape of the sampling distribution. Do you find any differences? Comment.d.) Compare the population standard deviation and standard deviation of sample means and explain your observation.[2.5+2.5+2.5+2.5]
- Population values: 1, 3, 6, 7 - Population size: $N = 4$ - Sample size: $n = 2$, sampling with replacement --- Total samples $= N^n = 4^2 = 16$. Sample Mean Sample Mean ---------------------------- (1,1) 1.0 (6,1) 3.5 (1,3) 2.0 (6,3) 4.5 (1,6) 3.5 (6,6) 6...
Full solved answer →Make Unit 7 stick
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