Question

show work please

Problem 4: Errors in medical prescriptions occur and a study examined whether electronic prescribing may help reduce errors. Two groups of doctors used written prescriptions and had similar error rates before the study. One group switched to e-prescriptions while the other continued to use written prescriptions, and error rates were measured one year later. The results are given in the following table. Electronic 254 354 38 3078 2370 a. Find a 99% confidence interval for the difference in the two proportions, P.-P., where is the proportion of errors using written prescriptions and is the proportion of errors using e- prescriptions. Round your answers to three decimal places. (5 points) Note: For a 99% confidence interval : -2.576. b. Provide an appropriate interpretation (in the context of this problem) of your 99% confidence interval. (3 points) 2

Answer 1

a).

We are given number of errors in written prescription

1478

Total number of written prescriptions

nus 3848

Therefore the proportion of errors using written prescriptions  

I w Pw 1478 3848 0.3841 nw

We are given number of errors in electronic prescription $x_{e}=254$

Total number of electronic prescriptions $n_{e}=3848$

Therefore the proportion of errors using electronic prescriptions  $\hat{p}_{e}=\frac{x_{e}}{n_{e}}=\frac{254}{3848}=0.0660$

The 99% confidence interval for the difference in population proportion for $p_{w}-p_{e}$ is given by

$CI=\hat{p}_{w}-\hat{p}_{e}\mp Z_{1-\alpha /2}\sqrt{\frac{\hat{p}_{w}(1-\hat{p}_{w})}{n_{w}}+\frac{\hat{p}_{e}(1-\hat{p}_{e})}{n_{e}}}$

The tabulated value is Z=2.576 which is given. Substituting, we get

$CI=0.3841-0.066\mp 2.576\sqrt{\frac{0.3841(1-0.3841)}{3848}+\frac{0.066(1-0.066)}{3848}}$

$CI=0.3181\mp 2.576\sqrt{\frac{0.3841(0.6159)}{3848}+\frac{0.066(0.934)}{3848}}$

$CI=0.3181\mp 2.576\sqrt{\frac{0.23656}{3848}+\frac{0.061644}{3848}}$

$CI=0.3181\mp 2.576\sqrt{0.00006148+0.0000160}$

$CI=0.3181\mp 2.576\sqrt{0.0000775}$

$CI=0.3181\mp 2.576*0.0088$

$CI=0.3181\mp 0.0227$

$CI=(0.2954,0.3408)$

The 99% confidence interval for the difference in population proportion for $p_{w}-p_{e}$ is given by (0.295,0.341)

b). Based on the data provided, the 99% confidence interval for the difference between the population proportions $p_{w}-p_{e}$ is 0.295 and 0.341, which indicates that we are 99% confident that the true difference between population proportions is contained by the interval (0.295, 0.341).