Question

an appliance manufacturer stockpiles washers and dryers in a large warehouse for shipment to retail stores. sometimes handling them the appliances get danged even though the damage may be minor the company must sell those machines at drastically reduced prices one day an day inspector randomly checks 60 washers and finds taht 5 of them have scratches or dents. compute a 95% confidence interval for the proportion of appliances from the manufacturer that get damaged during shipment

Answer 1

Given: $n=60, X=5, \alpha=0.05$

The sample proportion is computed as follows, based on the sample size $n=60$ and the number of favorable cases $X=5:$

$$\hat{p}={X \over n}={5 \over 60}={1 \over12}\approx0.083333$$

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$$CI(proportion)=\big(\hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p}) \over n}}, \hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p}) \over n}} \big)=$$

$$=\big({1 \over12}-1.96\sqrt{{{1 \over12}(1-{1 \over12}) \over 60}}, {1 \over12}+1.96\sqrt{{{1 \over12}(1-{1 \over12}) \over 60}}\big)=$$

$$=(0.013,0.153)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.013<p<0.153,$ which indicates that we are 95% confident that the true population proportion p is contained in the interval $(0.013,0.153).$