A certain town has 25,000 families. These families own 1.6 cars, on the average; the SD is 0.90. And 10% of them have no cars at all. As part of an opinion survey, a simple random sample of 1,500 families is chosen. What is the chance that between 9% and 11% of the sample families will not own cars? So I'm looking at the solution of this question it shows that to get the Standard Error of the Sum is SD= sqrt(0.9⋅ 0.1) However, I thought the formula to get SE is: SE= SD of the Box * sqrt(NumberofDraws)In this case, isn't the Box of the SD given at 0.9? Why do we still use the bootstrap method to get SD= sqrt(0.9⋅ 0.1)=0.3 ?? Many thanks in advance!

albiguguiismx 2022-09-11 Answered
A certain town has 25,000 families. These families own 1.6 cars, on the average; the SD is 0.90. And 10% of them have no cars at all. As part of an opinion survey, a simple random sample of 1,500 families is chosen. What is the chance that between 9% and 11% of the sample families will not own cars?
So I'm looking at the solution of this question it shows that to get the Standard Error of the Sum is SD= 0.9   0.1 However, I thought the formula to get SE is:
SE= SD of the Box N u m b e r o f D r a w s
In this case, isn't the Box of the SD given at 0.9? Why do we still use the bootstrap method to get SD= 0.9   0.1 =0.3 ??
Many thanks in advance!
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Answers (1)

shosautesseleol
Answered 2022-09-12 Author has 16 answers
One is a normal distribution SD (zero.9) and the other binomial (zero.3). The hassle has taken a ordinary distribution, the quantity of cars, and is asking a query of a binomial nature, yes or no to ownership.

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