Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 7. the hypotheses h0: μ = 75 and ha: μ < 75 are to be tested using a random sample of n = 25 observations. (a) how many standard deviations (of x) below the null value is x = 72.3? (round your answer to two decimal places.) standard deviations (b) if x = 72.3, what is the conclusion using α = 0.01? (round your answers to two decimal places.) test statistic z = critical value z = what can you conclude? reject the null hypothesis. there is sufficient evidence to conclude that the mean drying time is less than 75. do not reject the null hypothesis. there is sufficient evidence to conclude that the mean drying time is less than 75. do not reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 75. reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 75. (c) what is α for the test procedure that rejects h0 when z ≤ −2.8? (round your answer to four decimal places.) α = (d) for the test procedure of part (c), what is β(70)? (round your answer to four decimal places.) β(70) = (e) if the test procedure of part (c) is used, what n is necessary to ensure that β(70) = 0.01? (round your answer up to the next whole number.) n = specimens (f) if a level 0.01 test is used with n = 100, what is the probability of a type i error when μ = 76? (round your answer to four decimal places.)

Answer 1

Part A:

The z score of the hypothesis testing of n samples of a normally distributed data set is given by:

$z= \frac{x-\mu}{\sigma/\sqrt{n}}$

Given that the population mean is 75 and the population standard deviation is 7, then the number of standard deviation of the mean for x = 72.3 is given by the z score:

$z= \frac{72.3-75}{7/\sqrt{25}} \\ \\ = \frac{-2.7}{7/5} = \frac{-2.7}{1.4} \\ \\ =-1.93$

Therefore, 72.3 is 1.93 standard deviations below the mean.



Part B:

The test statistics of the hypothesis testing of n samples of a normally distributed data set is given by:

$z= \frac{x-\mu}{\sigma/\sqrt{n}}$

Thus given that x = 72.3, μ = 75, σ = 7 and n = 25,

$z= \frac{72.3-75}{7/\sqrt{25}} \\ \\ = \frac{-2.7}{7/5} = \frac{-2.7}{1.4} \\ \\ =-1.93$

The p-value is given by:

$P(-1.93)=0.03$

Since α = 0.01 and p-value = 0.03, 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 75.



Part C

In order to reject the null hypothesis, the p-value must be less than or equal to α.

The p-value for z ≤ -2.8 is 0.00256.

Therefore, the α for the test procedure that rejects the null hypothesis when z ≤ −2.8 is 0.0026



Part D:

In general for the alternative hypothesis , $H_a :\mu\ \textless \ \mu_0$

$\beta(\mu') = P\left(X \ \textgreater \ \mu_0-z_{1-\alpha}\frac{\sigma}{\sqrt{n}}|\mu'\right) \\ \\ = 1-P\left(-z_{1-\alpha}+\frac{\mu_0-\mu'}{\sigma/\sqrt{n}}\right)$

So for the test procedure with α = 0.0026

$\beta(70) = 1 - P\left(-z_{0.9974}+\frac{75-70}{7/5}\right) \\ \\ =1 - P(-2.7949+3.5714)=1-P(0.7765) \\ \\ =1-0.78128\approx\bold{0.2187 }$



Part E:

For α = 0.0026, and a general sample size n we have that

$\beta(70) = 1 - P\left(-z_{0.9974}+\frac{75-70}{7/\sqrt{n}}\right) \\ \\ =1 - P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)$

Since, we want n so that β(70) = 0.01, thus

$1 - P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)=0.01 \\ \\ \Rightarrow P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)=1-0.01=0.99 \\ \\ \Rightarrow P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)=P(2.3262) \\ \\ \Rightarrow -2.7949+ \frac{5}{7/\sqrt{n}}=2.3262 \\ \\ \Rightarrow \frac{5}{7/\sqrt{n}}=5.1211 \\ \\ \Rightarrow \frac{7}{\sqrt{n}} = \frac{5}{5.1211} =0.9764 \\ \\ \Rightarrow \sqrt{n}= \frac{7}{0.9764} =7.1695 \\ \\ \Rightarrow n=(7.1695)^2=51.4$

so we need n = 52.



Part F:

p-value = $P-value=P(\bar{X}\leq\bar{x}) \\ \\ =P(\bar{X}\leq72.3)=P\left(z\leq \frac{72.3-76}{7/5} \right) \\ \\ =P\left(z\leq \frac{-3.7}{1.4} \right)=P(z\leq-2.643) \\ \\ =\bold{0.00411}$

Since p-value is larger that .01 we fail to reject µ = 76.