Part A:
The z score of the hypothesis testing of n samples of a normally distributed data set is given by:
![z= \frac{x-\mu}{\sigma/\sqrt{n}}](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%20)
Given that the population mean is 73 and the population standard
deviation is 9, then the number of standard deviation below the null value of x =
72.3 is given by the z score:
![z= \frac{72.3-73}{9/\sqrt{25}} \\ \\ = \frac{-0.7}{9/5} = \frac{-0.7}{1.8} \\ \\ =-0.39](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7B72.3-73%7D%7B9%2F%5Csqrt%7B25%7D%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B-0.7%7D%7B9%2F5%7D%20%3D%20%5Cfrac%7B-0.7%7D%7B1.8%7D%20%20%5C%5C%20%20%5C%5C%20%3D-0.39)
Therefore, 72.3 is 0.39 standard deviations below the null value.
Part B:
The test statistics of the hypothesis testing of n samples of a normally distributed data set is given by:
![z= \frac{x-\mu}{\sigma/\sqrt{n}}](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%20)
Thus given that x = 72.3, μ = 73, <span>σ = 9 and n = 25,
</span><span>
![z= \frac{72.3-73}{9/\sqrt{25}} \\ \\ = \frac{-0.7}{9/5} = \frac{-0.7}{1.8} \\ \\ =-0.39](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7B72.3-73%7D%7B9%2F%5Csqrt%7B25%7D%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B-0.7%7D%7B9%2F5%7D%20%3D%20%5Cfrac%7B-0.7%7D%7B1.8%7D%20%20%5C%5C%20%20%5C%5C%20%3D-0.39)
</span><span>The p-value is given by:
P(-0.39) = 0.3483
</span><span>Since α = 0.005 and p-value = 0.3483, this means that the p-value is greater than the <span>α, ant thus, we will faill to reject the null hypothesis.
Therefore, the conclussion is </span>
do not reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 73.
</span>Part C:
In general for the alternative hypothesis,
![H_a :\mu\ \textless \ \mu_0](https://tex.z-dn.net/?f=H_a%20%3A%5Cmu%5C%20%5Ctextless%20%5C%20%5Cmu_0)
![\beta(\mu') = P\left(X \ \textgreater \ \mu_0-z_{1-\alpha}\frac{\sigma}{\sqrt{n}}|\mu'\right) \\ \\ = 1-P\left(-z_{1-\alpha}+\frac{\mu_0-\mu'}{\sigma/\sqrt{n}}\right)](https://tex.z-dn.net/?f=%5Cbeta%28%5Cmu%27%29%20%3D%20P%5Cleft%28X%20%5C%20%5Ctextgreater%20%5C%20%20%5Cmu_0-z_%7B1-%5Calpha%7D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7C%5Cmu%27%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%201-P%5Cleft%28-z_%7B1-%5Calpha%7D%2B%5Cfrac%7B%5Cmu_0-%5Cmu%27%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%5Cright%29%20)
So for the test procedure with α = <span>0.005
![\beta(70) = 1 - P\left(-z_{0.995}+\frac{73-70}{9/5}\right) \\ \\ =1 - P(-2.5755+1.6667)=1-P(-0.9088) \\ \\ =1-0.1817\approx\bold{0.8183 }](https://tex.z-dn.net/?f=%5Cbeta%2870%29%20%3D%201%20-%20P%5Cleft%28-z_%7B0.995%7D%2B%5Cfrac%7B73-70%7D%7B9%2F5%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D1%20-%20P%28-2.5755%2B1.6667%29%3D1-P%28-0.9088%29%20%5C%5C%20%20%5C%5C%20%3D1-0.1817%5Capprox%5Cbold%7B0.8183%20%7D)
</span>Part D:
For α = 0.005, and a general sample size n we have that
![\beta(70) = 1 - P\left(-z_{0.995}+\frac{73-70}{9/\sqrt{n}}\right) \\ \\ =1 - P\left(-2.5755+ \frac{3}{9/\sqrt{n}} \right)](https://tex.z-dn.net/?f=%5Cbeta%2870%29%20%3D%201%20-%20P%5Cleft%28-z_%7B0.995%7D%2B%5Cfrac%7B73-70%7D%7B9%2F%5Csqrt%7Bn%7D%7D%5Cright%29%20%5C%5C%20%5C%5C%20%3D1%20-%20P%5Cleft%28-2.5755%2B%20%5Cfrac%7B3%7D%7B9%2F%5Csqrt%7Bn%7D%7D%20%5Cright%29)
Since, we want n so that β(70) = 0.01, thus
![1 - P\left(-2.5755+ \frac{3}{9/\sqrt{n}} \right)=0.01 \\ \\ \Rightarrow P\left(-2.5755+ \frac{3}{9/\sqrt{n}} \right)=1-0.01=0.99 \\ \\ \Rightarrow P\left(-2.5755+ \frac{3}{9/\sqrt{n}} \right)=P(2.3262) \\ \\ \Rightarrow -2.5755+ \frac{3}{9/\sqrt{n}}=2.3262 \\ \\ \Rightarrow \frac{3}{9/\sqrt{n}}=4.9017 \\ \\ \Rightarrow \frac{9}{\sqrt{n}} = \frac{3}{4.9017} =0.6120 \\ \\ \Rightarrow \sqrt{n}= \frac{9}{0.6120} =14.7051 \\ \\ \Rightarrow n=(14.7051)^2=216.2](https://tex.z-dn.net/?f=1%20-%20P%5Cleft%28-2.5755%2B%20%5Cfrac%7B3%7D%7B9%2F%5Csqrt%7Bn%7D%7D%20%5Cright%29%3D0.01%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20P%5Cleft%28-2.5755%2B%20%5Cfrac%7B3%7D%7B9%2F%5Csqrt%7Bn%7D%7D%20%5Cright%29%3D1-0.01%3D0.99%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20P%5Cleft%28-2.5755%2B%20%5Cfrac%7B3%7D%7B9%2F%5Csqrt%7Bn%7D%7D%20%5Cright%29%3DP%282.3262%29%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20-2.5755%2B%20%5Cfrac%7B3%7D%7B9%2F%5Csqrt%7Bn%7D%7D%3D2.3262%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cfrac%7B3%7D%7B9%2F%5Csqrt%7Bn%7D%7D%3D4.9017%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%20%5Cfrac%7B9%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cfrac%7B3%7D%7B4.9017%7D%20%3D0.6120%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Csqrt%7Bn%7D%3D%20%5Cfrac%7B9%7D%7B0.6120%7D%20%3D14.7051%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20n%3D%2814.7051%29%5E2%3D216.2)
so we need n = 217.
Part E