Question

In a recent survey of 79 heads of household, 46 said they did online banking on their cell phones. Find the standard error for the sample proportion of heads of household who do online banking on their cell phones. Enter your answer as a decimal rounded to three decimal places.

Answer 1

`SE_p = 0.055`

Explanation:

Given

`n = 79` --- Heads

Proportion, p = 46 out of 79

Required

Determine the standard error for sample proportion (SEp)

This is calculated using the following formula

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

In this case:

`p = (46)/(79)`

`p = 0.5823`

Substitute values for p and n in:

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

`SE_p = \sqrt{(0.5823 * (1-0.5823))/(79)}`

`SE_p = \sqrt{(0.5823 * 0.4177)/(79)}`

`SE_p = \sqrt{(0.24322671)/(79)}`

`SE_p = √(0.00307881911)`

`SE_p = 0.0554871076`

`SE_p = 0.055` ---- Approximated