Question

Out of 54 randomly selected patients of a local hospital who were surveyed, 49 reported that they were satisfied with the care they recieved. Construct and interpret a 95% confidence interval for the percentage of all patients satisfied with their care at the hospital.

Answer 1

The 95% confidence interval is given below:

`\hat{p} \pm z_{(0.05)/(2) } \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

Where:

`\hat{p} = (49)/(54) =0.9074`

`z_{(0.05)/(2) }=1.96` is the critical value at 0.05 significance level

`n=54` is the sample size

`\therefore 0.9074 \pm 1.96 \sqrt{(0.9074(1-0.9074))/(54)}`

`0.9074 \pm 0.077315056`

`\left( 0.830, 0.985 \right)`

Therefore, the 95% confidence interval for the percentage of all patients satisfied with their care at the hospital is
`\left( 0.830, 0.985 \right)`