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Estimation of Population Mean: Sampling Distribution, Confidence Intervals, and Sample Size Determination

A solved assignment on population mean estimation and confidence intervals.

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    Estimation of Population Mean: Sampling Distribution, Confidence Intervals, and Sample Size Determination 1. (Sampling distribution & confidence interval) A random variable has a population mean equal to 1,573 and population variance equal to 952,021. Your interest lies in estimating the population mean of this random variable. (Of course, you do not know what the population mean and population variance are.) With that in mind, you take a representative sample of size 85 from the population of the random variable. You then use this sample data to calculate the sample average as an estimate for the population mean. 1. (a) Using your knowledge about the central limit theorem (CLT), and assuming that the CLT has already “established itself” / “kicked in” when the sample size is 85, what is the probability that the sample average that you calculated will lie between 1,502 and 1,748? In this case, 𝑋 ത ~ 𝑁𝑜𝑟𝑚𝑎𝑙 ቆ 1573 , ට ଽ52021 85 ቇ Required probability = 𝑃 ( 1502 < 𝑋 ത < 1748 ) = 𝑃 ( 𝑋 ത < 1748 ) − 𝑃 ( 𝑋 ത < 1502 ) = 𝑃 ቌ 𝑍 < 1748 − 1573 ට వ52021 85 ቍ − 𝑃 ቌ 𝑍 < 1502 − 1573 ට వ52021 85 ቍ = P(Z<1.65) – P(Z< - 0.67) = 0.9505 - 0.2514 = 0.6991 2. (b) You then use your sample data to calculate a 92% confidence interval for the population mean. Assuming that you can estimate the variance of the underlying random variable very precisely with your sample data, what should be the width of this confidence interval? The width of this confidence interval would be twice the size of the margin of error which is = 2* 𝑍 0 . 04 ∗ ට ௏ ( 𝑋 ) ௡ = 2 ∗ 1 . 75 ∗ ට ଽ52021 85 = 370.4093

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