how many samples of size 3 can be drawn from s={2,4,8,10,12,}, (a) with replacement: 5^3 = 125----- (b) without replacement: 5P3 = 5*4*3 = 60-----Cheers, Stan H. The mean and variance of the population are: $$\mu = \frac{{\sum X}}{N} = \frac{{45}}{5} = 9$$ and $${\sigma ^2} = \frac{{\sum {X^2}}}{N} – {\left( {\frac{{\sum X}}{N}} \right)^2} = \frac{{495}}{5} – {\left( {\frac{{45}}{5}} \right)^2} = 99 – 81 = 18$$, (i) $$E\left( {\bar X} \right) = \mu = 9$$ (ii) $${\text{Var}}\left( {\bar X} \right) = \frac{{{\sigma ^2}}}{n}\left( {\frac{{N – n}}{{N – 1}}} \right) = \frac{{18}}{2}\left( {\frac{{5 – 2}}{{5 – 1}}} \right) = 6.75$$. September 10 @ Thus in total there are $$5 \times 5 = 25$$ samples or pairs which are possible. The sampling distribution of the sample mean $$\bar X$$ and its mean and standard deviation are: $${\text{E}}\left( {\bar X} \right) = \sum \bar Xf\left( {\bar X} \right) = \frac{{90}}{{10}} = 9$$ Verification: (i) E ( X ¯) = μ = 9 (ii) Var ( X ¯) = σ 2 n ( N – n N – 1) = 18 2 ( 5 – 2 5 – 1) = 6.75. Please help me out with this question! Calculate the mean and standard deviation of this sampling distribution. Draw all possible samples of size 2 without replacement from a population consisting of 3, 6, 9, 12, 15. The mean and standard deviation of the population are: $$\mu = \frac{{\sum X}}{N} = \frac{{21}}{4} = 5.25$$ and $${\sigma ^2} = \sqrt {\frac{{\sum {X^2}}}{N} – {{\left( {\frac{{\sum X}}{N}} \right)}^2}} = \sqrt {\frac{{115}}{4} – {{\left( {\frac{{21}}{4}} \right)}^2}} = 1.0897$$, $$\frac{\sigma }{{\sqrt n }}\sqrt {\frac{{N – n}}{{N – 1}}} = \frac{{1.0897}}{{\sqrt 3 }}\sqrt {\frac{{4 – 3}}{{4 – 1}}} = 0.3632$$, Hence $${\mu _{\bar X}} = \mu$$ and $${\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }}\sqrt {\frac{{N – n}}{{N – 1}}}$$, Pearl Lamptey Hence state and verify relation between (a). Form a sampling distribution of sample means. September 10 @ In that case, sampling with replacement isn't much different from sampling without replacement. Variance of the sampling distribution of the mean and the population variance. Do you know how to calculate this? For that to work out, you’ve planned on adding an image to see if it increases conversions or not.You start your A/B test running a control version (A) against your variation (B) that contains the image. Find the sample mean X ¯ for each sample and … Discuss the relevance of the concept of the two types of errors in following case. a. Find the sample mean $$\bar X$$ for each sample and make a sampling distribution of $$\bar X$$. {1, 3, 4}, {1, 3, 5}, {1, 3, 6}, {1, 3, 7}, Variance of the sampling distribution of the mean and the population variance. b. Mean of the sampling distribution of the mean and the population mean; (b). How bias can be eliminated? When we sample without replacement, and get a non-zero covariance, the covariance depends on the population size. If random samples of size three are drawn without replacement from the population consisting of four numbers 4, 5, 5, 7. ... Required fields are marked *. ... For each sample in the list find the mean and median. {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, { 1, 2, 6}, {1, 2, 7}, 6:05 pm. September 18 @ Hi, Please tell me this question as soon as possible Form the complete set of samples of size three and for each sample, compute the sample mean and median. (b) what is a biased sample? Thus, the number of possible samples which can be drawn without replacement is $$\left( {\begin{array}{*{20}{c}} N \\ n \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) = 4$$, $${\mu _{\bar X}} = \sum \bar X\,f\left( {\bar X} \right)\,\,\,\, = \,\,\,\frac{{63}}{{12}} = 5.25$$ For b part you need to list all the samples you counted in part a. If any two bulbs are selected with replacement, there are $$25$$ possible samples, as listed in the table below: After 5 days, the variation (B) outperforms the control version by a staggering 25% increase in conversions with an 85% level of confidence.You stop the test and implement the image in your banner. Form the complete set of samples of size three and for each sample, compute the sample mean and median. (a). Please I want samples of size 3 N=4 with replacement. 12:25 pm, Draw all possible sample of size n = 3 with replacement from the population 3,6,9 and 12. b. Suppose a container contains $$3$$ good bulbs denoted by $${G_1},{G_2}$$ and $${G_3}$$ and $$2$$ defective bulbs denoted by $${D_1}$$ and $${D_2}$$. Example: If random samples of size three are drawn without replacement from the population consisting of four numbers 4, 5, 5, 7. Form the sampling distribution of sample means and verify the results. The probability of both people being female is 0.6 x 0.6 = 0.36. For part a the number of samples of size 3 is the number of ways of choosing 3 things from 7 things, sometimes called 7 choose 3 and written 7C3. Take all possible samples of size 3 with replacement from population comprising 10 12 14 16 18 make sampling distribution and verify, Aimen Naveed However, after a month, you noticed that your month-to-month conversions have decreased. We have population values 4, 5, 5, 7, population size $$N = 4$$ and sample size $$n = 3$$. How many samples of size three can be extracted from this population (sampling without replacement)? Consider the population of the first seven integers: 1, 2, 3, 4, 5, 6, and 7; N=7. Hi Liz, You need to check your calculations again. The mean of the population is 4 not 2. tell this question, Your email address will not be published. Thank you, Liz. If we sample with replacement, then the probability of choosing a female on the first selection is given by 30000/50000 = 60%. Please tell me this question as soon as possible, Aimen Naveed The mean of … (i) $${\text{E}}\left( {\bar X} \right) = \mu$$, (ii) $${\text{Var}}\left( {\bar X} \right) = \frac{{{\sigma ^2}}}{n}\left( {\frac{{N – n}}{{N – 1}}} \right)$$, We have population values 3, 6, 9, 12, 15, population size $$N = 5$$ and sample size $$n = 2.$$ Thus, the number of possible samples which can be drawn without replacement is, $\left( {\begin{array}{*{20}{c}} N \\ n \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right) = 10$. How many samples of size three can be extracted from this population (sampling without replacement)? Your email address will not be published. Hence state and verify relation between (a). For this population, mean = 2 and standard deviation = 2. The probability of a female on the second selection is still 60%. Form a sampling distribution of sample means. “Let’s say that you want to increase conversions on a banner displayed on your website. $${\sigma _{\bar X}} = \sqrt {\sum {{\bar X}^2}\,f\left( {\bar X} \right) – {{\left[ {\sum \bar X\,f\left( {\bar X} \right)} \right]}^2}} \,\,\,\, = \,\,\,\sqrt {\frac{{997}}{{36}} – {{\left( {\frac{{63}}{{12}}} \right)}^2}} = 0.3632$$. I would start. Mean of the sampling distribution of the mean and the population mean; (b). If the population is very large, this covariance is very close to zero. Do so in an orderly way. Compare your calculations with the population parameters. {1, 6, 7}, $${\text{Var}}\left( {\bar X} \right) = \sum {\bar X^2}f\left( {\bar X} \right) – {\left[ {\sum \bar X\,f\left( {\bar X} \right)} \right]^2} = \frac{{887.5}}{{10}} – {\left( {\frac{{90}}{{10}}} \right)^2} = 87.75 – 81 = 6.75$$. You need to check your calculations again. {2, 3, 4}, {2, 3, 5}, {2, 3, 6}, {2, 3, 7}, 12:23 pm, Draw all possible sample of size n = 3 with replacement from the population 3,6,9 and 12.