Answer :
Part A:
The z score of the hypothesis testing of n samples of a normally distributed data set is given by:
[tex]z= \frac{x-\mu}{\sigma/\sqrt{n}} [/tex]
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:
[tex]z= \frac{72.3-73}{9/\sqrt{25}} \\ \\ = \frac{-0.7}{9/5} = \frac{-0.7}{1.8} \\ \\ =-0.39[/tex]
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:
[tex]z= \frac{x-\mu}{\sigma/\sqrt{n}} [/tex]
Thus given that x = 72.3, μ = 73, σ = 9 and n = 25,
[tex]z= \frac{72.3-73}{9/\sqrt{25}} \\ \\ = \frac{-0.7}{9/5} = \frac{-0.7}{1.8} \\ \\ =-0.39[/tex]
The p-value is given by:
P(-0.39) = 0.3483
Since α = 0.005 and p-value = 0.3483, this means that the p-value is greater than the α, ant thus, we will faill to reject the null hypothesis.
Therefore, the conclussion is do not reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 73.
Part C:
In general for the alternative hypothesis, [tex]H_a :\mu\ \textless \ \mu_0[/tex]
[tex]\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) [/tex]
So for the test procedure with α = 0.005
[tex]\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 }[/tex]
Part D:
For α = 0.005, and a general sample size n we have that
[tex]\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)[/tex]
Since, we want n so that β(70) = 0.01, thus
[tex]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[/tex]
so we need n = 217.
Part E
[tex]P-value=P(\bar{X}\leq\bar{x}) \\ \\ =P(\bar{X}\leq72.3)=P\left(z\leq \frac{72.3-76}{9/10} \right) \\ \\ =P\left(z\leq \frac{-3.7}{0.9} \right)=P(z\leq-4.111) \\ \\ =\bold{0.00002}[/tex]
The z score of the hypothesis testing of n samples of a normally distributed data set is given by:
[tex]z= \frac{x-\mu}{\sigma/\sqrt{n}} [/tex]
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:
[tex]z= \frac{72.3-73}{9/\sqrt{25}} \\ \\ = \frac{-0.7}{9/5} = \frac{-0.7}{1.8} \\ \\ =-0.39[/tex]
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:
[tex]z= \frac{x-\mu}{\sigma/\sqrt{n}} [/tex]
Thus given that x = 72.3, μ = 73, σ = 9 and n = 25,
[tex]z= \frac{72.3-73}{9/\sqrt{25}} \\ \\ = \frac{-0.7}{9/5} = \frac{-0.7}{1.8} \\ \\ =-0.39[/tex]
The p-value is given by:
P(-0.39) = 0.3483
Since α = 0.005 and p-value = 0.3483, this means that the p-value is greater than the α, ant thus, we will faill to reject the null hypothesis.
Therefore, the conclussion is do not reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 73.
Part C:
In general for the alternative hypothesis, [tex]H_a :\mu\ \textless \ \mu_0[/tex]
[tex]\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) [/tex]
So for the test procedure with α = 0.005
[tex]\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 }[/tex]
Part D:
For α = 0.005, and a general sample size n we have that
[tex]\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)[/tex]
Since, we want n so that β(70) = 0.01, thus
[tex]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[/tex]
so we need n = 217.
Part E
[tex]P-value=P(\bar{X}\leq\bar{x}) \\ \\ =P(\bar{X}\leq72.3)=P\left(z\leq \frac{72.3-76}{9/10} \right) \\ \\ =P\left(z\leq \frac{-3.7}{0.9} \right)=P(z\leq-4.111) \\ \\ =\bold{0.00002}[/tex]