Question 8) (3 pts) A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test Ho : u =175 millimeters versus H : u>175 millimeters, using the results of n = 10 samples. a) Find the type I error probability a if the critical region is # > 185. b) What is the probability of type II error if the true foam height is 185 millimeters? c) Find ß for the true mean of 195 millimeters.

Here,Null hypothesis : H_{0}: \mu=175Alternate hypothesi s: H_{1}: \mu>175n=10 and \sigma=20a) HereRejection region \omega=\left\{x:\right. Reject H_{0} when \left.\bar{x}>185\right\}Acceptence region \bar{\omega}=\left\{\mathrm{x}:\right. Accept \mathrm{H}_{0} when \left.\overline{\mathrm{x}} \leq 185\right\}\alpha=P\left\{\right. Rej ecting \mathrm{H}_{0} when it is true \}\begin{array}{l}=\mathrm{P}\{\overline{\mathrm{x}}>185 / \text { When } \mu=175\} \\=\mathrm{P}\left\{\frac{(\overline{\mathrm{x}}-175)}{20 / \sqrt{10}}>\frac{(185-75)}{20 / \sqrt{10}}\right\} \\=\mathrm{P}\{Z>1.58\} \\=1-\mathrm{P}\{Z \leq 1.58\} \\=1-0.9429 \\=0.0571\end{array}b) HereRejection region \omega=\left\{x:\right. Reject H_{0} when \left.\bar{x}>185\right\}Acceptence region \bar{\omega}=\left\{\mathrm{x}:\right. Accept \mathrm{H}_{0} when \left.\overline{\mathrm{x}} \leq 185\right\}Alternate hypothesis: \mathrm{H}_{\mathrm{a}}: \mu=185\beta=P\left\{\right. Accepting H_{0} when it is false \}\begin{array}{l}=\mathrm{P}\{\mathrm{x} \leq 185 / \text { When } \mu=185\} \\=\mathrm{P}\left\{\frac{(\overline{\mathrm{x}}-185)}{20 / \sqrt{10}} \leq \frac{(185-185)}{20 / \sqrt{10}}\right\} \\=\mathrm{P}\{Z \leq 0\} \\=0.5\end{array}c) HereRejection region \omega=\left\{x:\right. Reject H_{0} when \left.\bar{x}>185\right\}Acceptence region \bar{\omega}=\left\{\mathrm{x}:\right. Accept H_{0} when \left.\overline{\mathrm{x}} \leq 185\right\}Alternate hypothesis: \mathrm{H}_{1}: \mu=195\begin{aligned}\beta & =\mathrm{P}\left\{\text { Accepting } \mathrm{H}_{0} \text { when it is false }\right\} \\& =\mathrm{P}\{\overline{\mathrm{x}} \leq 185 / \text { When } \mu=195\} \\& =\mathrm{P}\left\{\frac{(\overline{\mathrm{x}}-195)}{20 / \sqrt{10}} \leq \frac{(185-195)}{20 / \sqrt{10}}\right\} \\& =\mathrm{P}\{Z \leq-1.58\} \\& =0.0571\end{aligned}         Plzzz don't give dislike.....plzzz comment if you have any problem i will try to solve your problem.....plzzz give thumbs up i am in need   Thank you ...