However, after a month, you noticed that your month-to-month conversions have decreased. Hence state and verify relation between (a). Do you know how to calculate this? 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. Your email address will not be published. September 10 @ 12:25 pm, Draw all possible sample of size n = 3 with replacement from the population 3,6,9 and 12. $${\text{Var}}\left( {\bar X} \right) = \sum {\bar X^2}f\left( {\bar X} \right) – {\left[ {\sum \bar X\,f\left( {\bar X} \right)} \right]^2} = \frac{{887.5}}{{10}} – {\left( {\frac{{90}}{{10}}} \right)^2} = 87.75 – 81 = 6.75$$. Find the sample mean X ¯ for each sample and … For this population, mean = 2 and standard deviation = 2. How many samples of size three can be extracted from this population (sampling without replacement)? 6:05 pm. For b part you need to list all the samples you counted in part a. Please tell me this question as soon as possible, Aimen Naveed “Let’s say that you want to increase conversions on a banner displayed on your website. Please help me out with this question! Thus in total there are $$5 \times 5 = 25$$ samples or pairs which are possible. Example: If random samples of size three are drawn without replacement from the population consisting of four numbers 4, 5, 5, 7. (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, $\left( {\begin{array}{*{20}{c}} N \\ n \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right) = 10$. Form the complete set of samples of size three and for each sample, compute the sample mean and median. If we sample with replacement, then the probability of choosing a female on the first selection is given by 30000/50000 = 60%. Hence state and verify relation between (a). 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) $${\text{Var}}\left( {\bar X} \right) = \frac{{{\sigma ^2}}}{n}\left( {\frac{{N – n}}{{N – 1}}} \right) = \frac{{18}}{2}\left( {\frac{{5 – 2}}{{5 – 1}}} \right) = 6.75$$. The sampling distribution of the sample mean $$\bar X$$ and its mean and standard deviation are: $${\text{E}}\left( {\bar X} \right) = \sum \bar Xf\left( {\bar X} \right) = \frac{{90}}{{10}} = 9$$ September 18 @ Form the sampling distribution of sample means and verify the results. If any two bulbs are selected with replacement, there are $$25$$ possible samples, as listed in the table below: ... (a). Please I want samples of size 3 N=4 with replacement. {1, 6, 7}, I would start. September 10 @ Hi, Consider the population of the first seven integers: 1, 2, 3, 4, 5, 6, and 7; N=7. $${\sigma _{\bar X}} = \sqrt {\sum {{\bar X}^2}\,f\left( {\bar X} \right) – {{\left[ {\sum \bar X\,f\left( {\bar X} \right)} \right]}^2}} \,\,\,\, = \,\,\,\sqrt {\frac{{997}}{{36}} – {{\left( {\frac{{63}}{{12}}} \right)}^2}} = 0.3632$$. The mean of the population is 4 not 2. Draw all possible samples of size 2 without replacement from a population consisting of 3, 6, 9, 12, 15. 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