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From a recent company survey, it is known that the proportion of employees older than 55 and considering retirement is 8% . For a random sample of size 110 , what is standard deviation for the sampling distribution of the sample proportions, rounded to three decimal places?

Respuesta :

Answer:

[tex] Sd(p)= \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.08*(1-0.08)}{110}}= 0.026[/tex]

Step-by-step explanation:

For this case we are interested in the true population proportion [tex] p[/tex] and the distribution for this parameter is given by:

[tex] p \sim N (p, \sqrt{\frac{p(1-p)}{n}})[/tex]

We know that n = 110 and p =0.08

Solution to the problem

For this case the standard deviation is given by:

[tex] Sd(p)= \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.08*(1-0.08)}{110}}= 0.026[/tex]

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