Amedical researcher wants to construct a 99% confidence interval for the proportion of knee replacement surgeries that result in complications. an article in the journal of bone and joint surgery suggested that approximately 8% of such operations result in complications. using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.082 ?

 Sample size of 73  is needed so that the confidence interval will have a margin of error of 0.082

Step-by-step explanation:

The formula to calculate the margin of error for population proportion is given by :-

Given : The proportion of operations result in complications :

Significance level :

Critical value =      [By standard normal distribution table]

Margin of error : E= 0.082

Substitute all the value in the above formula, we get

Squaring both sides , we get

Hence, the required sample size = 73

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

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step-by-step explanation: