Question

Please help find the Standard Error and the last question about confidence interval

"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. "Wrinkle recovery angle" measures how well a fabric recovers from wrinkles. Higher is better. Here are data on the wrinkle recovery angle (in degrees) for the same fabric specimens discussed in the previous exercise:

A manufacturer wants to know how large is the difference in mean wrinkle recovery angle.

a. The mean and the standard deviation for the Permafresh are:

b. The mean and the standard deviation for the Hylite are:

c. The standard error is:

SE=-34.65 Round to 3 decimal places.

The critical value from the distribution for a confidence interval of 99.8% is:

d. Using the the answer from above (all rounded to 3 decimal places), calculate by hand a 99.8% confidence interval for the difference in mean wrinkle recovery angle:

Answer 1

From the given information, for Permafresh: \(\bar{x}_{1}=129.60, s_{1}=13.01, n_{1}=5\), for Hylite: \(\bar{x}_{2}=8.91, s_{2}=17.82\), and \(n_{2}=4\)

Standard Error is \(S_{\left(\pi_{1}-x_{2}\right)}=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{\frac{13.01^{2}}{5}+\frac{17.82^{2}}{4}}=10.6414\)

\(100(1-\alpha) \%\) confidence interval for difference of true means is \(\left(\left(\bar{x}_{1}-\bar{x}_{2}\right)-t_{\alpha / 2} \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}},\left(\bar{x}_{1}-\bar{x}_{2}\right)+t_{\alpha / 2} \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}\right)\)

\(99.8 \%\) confidence interval for the difference in mean wrinkle recovery angle is \(\left(\left(\bar{x}_{1}-\bar{x}_{2}\right)-t_{0.01 / 2} \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}},\left(\bar{x}_{1}-\bar{x}_{2}\right)+t_{0.01 / 2} \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}\right)\)

\(=\left((129.6-164.3)-3.499 \sqrt{\frac{(13)^{2}}{5}+\frac{(17.8)^{2}}{4}},(129.6-164.3)+3.499 \sqrt{\frac{(13)^{2}}{5}+\frac{(17.8)^{2}}{4}}\right)\)

\(=(-71.8965,2.4965)\)

\(99.8 \%\) confi dence interval is \(-71.8965 \leq \mu_{1}-\mu_{2} \leq 2.4965 .\)