Question

In a random sample of 64 people, 48 are classified as "successful." If the population proportion is 0.70, determine the standard error of the proportion.

Answer 1

The standard error of the proportion can be calculated using the formula SE = √(p(1-p)/n), where p represents the population proportion and n represents the sample size. In this case, the population proportion is 0.70 and the sample size is 64. Substituting these values into the formula, the standard error of the proportion is approximately 0.0573.

Step-by-step explanation:

The standard error of the proportion can be calculated using the formula:

SE = √(p(1-p)/n)

where 'p' represents the population proportion and 'n' represents the sample size.

In this case, the population proportion is 0.70 and the sample size is 64. Substituting these values into the formula:

SE = √(0.70(1-0.70)/64) = √(0.21/64) = √0.00328125 ≈ 0.0573

So, the standard error of the proportion is approximately 0.0573.

Answer 2

`\hat p=(48)/(64)= 0.75` represent the estimated proportion successfull

The standard error for this case is given by this formula:

`SE= \sqrt{(\hat p(1-\hat p))/(n)}`

And replacing we got:

`SE= \sqrt{(0.75*(1-0.75))/(64)}= 0.0541`

Step-by-step explanation:

We have the following info:

`n= 64` represent the sample size

`X= 48` represent the number of people classified as successful

`\hat p=(48)/(64)= 0.75` represent the estimated proportion successfull

The standard error for this case is given by this formula:

`SE= \sqrt{(\hat p(1-\hat p))/(n)}`

And replacing we got:

`SE= \sqrt{(0.75*(1-0.75))/(64)}= 0.0541`