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2. in a random sample of 1200 persons aged 65 and over, the proportion with osteoarthritis was found to be 0.115 (or 11.5%) (a) what is the standard error for the estimate of the proportion of all person aged 65 and over with osteoarthritis?

Answer :

The standard error for the estimate of the proportion of all persons aged 65 and over with osteoarthritis is 0.0092.

The subset chosen from the larger set to make assumptions is known as a random sample. Standard Error can be defined as the term that measures the accuracy of the sample distribution which represents the population.

We know that,

                   Standard Error   =   √ ( P ( 1 - P ) / n )    → 1

As per the given question,

  • n   =   1200
  • P   =   0.115

Substitute the values in 1,

Standard Error ( SE )   =   √ ( P ( 1 - P ) / n )

                                    =   √ ( 0.115 ( 1 - 0.115 ) / 1200 )

                                    =   √ ( 0.115 ( 0.885 ) / 1200 )

                                    =   √ ( 0.101775 / 1200 )

                                    =   √ ( 0.0000848125)

Standard Error ( SE )   =   0.0092093702   ≅   0.0092

Therefore, 0.0092 is the standard error for the estimate of the proportion of all persons aged 65 and over with osteoarthritis.

To know more about Standard Error refer to:

https://brainly.com/question/28583871

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