Question

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

Answer 1

Identify the sample statistics $n=54, X=49.$

Find the point estimates

$$\hat{p}={49 \over 54}\approx0.9074, \hat{q}\approx0.0926$$

Verify that the sampling distribution of $\hat{p}$ can be approximated by the normal distribution

$$n\hat{p}=49>5, n\hat{q}=5\geq5$$

Find the critical value $z^*$ that corresponds to the given level of significance

$$\alpha=0.05, z^*=z_{\alpha/2}=z_{0.05/2}=1.96$$

The 95% confidence interval is given below:

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

$$0.9074\pm 1.96\sqrt{{0.9074(1-0.9074) \over 54}}$$

$$0.9074\pm 0.0773$$

$$(0.8301,0.9847)$$

$Low: 0.8301$

$High: 0.9847$

Therefore, the 95% confidence interval for the percentage of all patients satisfied with their care at the hospital is 

$$(0.8301,0.9847)$$

We are $95\%$ confident that the percentage of all patients satisfied with their care at the hospital is between $83.01\%$

and $98.47\%$