Question

The following table is the result of a survey of 1109 households showing the number of cars they have:

​​Number of Cars​​ 0 ​1​2​3​≥4

​​Percent of households​ 19​​13​33​26​9

a- Determine the sample proportion of households that own two or more cars.

b- If the proportion of households that own two or more cars is equal to 0.5, determine the probability that in a sample of 999 households at least 509 would own two or more cars.

c- The margin of error of the survey was ± 3 percentage points. This means that the sample proportion would be within 3 percentage points of the population proportion. If the population proportion of the households that had two or more cars was 0.5, determine the probability that the margin of error would be ± 3 percentage points.

Answer 1

​​Number of Cars	​​Percent of households​
0	19
1	13
2	33
3	26
<4	9
Sum	100

a)

sample proportion of households that own two or more cars.=

favourable cases / total number

= [ 33 + 26 + 9 ] / 100

= 0.68

b)

The following information about the Binomial distribution is provided:

The population proportion of success is p = 0.5

and the sample size is n= 999.

We need to compute Pr(X≥509):

The population mean is computed as:

and the population standard deviation is computed as:

Therefore, we get that

c)

we know that distribution of sample proportion will follow normal distribution .

also we are given that the margin of error of the survey is +- 3% ~0.03.

which means sample proportion would lie between

[0.5 – 3*(0.03) , 0.5 + 3*(0.03) ] = [ 0.41 , 0.59 ]

Now what we want is Prob[ 0.41 < p < 0.59 ] ?

Since we know p ~ normal distribution and by its empirical rule 99.7% of the data lies between 3 S.D.

Hence Prob[ 0.41 < p < 0.59 ] = 0.997

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