Question

Consider a pa nt drying situation In which drying time or a test spec n-25 observations. en Is normally distributed with o 6 TI e ypotheses 0: ·74 and μく are to be tested using aan om sample of (a) How many standard devlations (of X) below the null value is x 72.3? (Round your answer to two declmal places.) 1.42 standard deviations (b) If x = 72.3, what is the conclusion using α-0.0067 Calculate the test statistic and detrmine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) P-value0.00 State the concluslon In the problem context. z =-1.42 ● Reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 74. Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 74 Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 74 Reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 74 (c) For the test procedure with α-0.006, what is β(70)? (Round your answer to four decimal places.) β(70)-0.1587 | X (d l the test procedure with r = n-49 006 s used what n s necessary to ensure that ופ 70 : 0.017 Round your answer up to the next whole number. x specimens (e) If a level 0.01 test is used with 0.0000 = 100, what is the probability of a type 1 error when μ = 767 (Round your answer to four decimal places.)

Answer 1

We have to test (Left tailed Z-test)

Here $n=25,\sigma =6$

a) We have to find the number of standard deviations below is the null value (74) is $\overline{x}=72.3$

$\frac{74-72.3}{6}={\color{Blue} 0.28333333}$

b)The test statistic

$z=\left (\overline{x_1}-74\right )/\sqrt{\frac{s _1^2}{n_1}}=-1.7/\sqrt{6^2/25}\\ {\color{Blue} z=-1.41667}$

The P-value is

$\textup{P-value}= \Phi\left ( -1.41667 \right )\\ {\color{Blue} \textup{P-value}= 0.078}$

Since P-value is greater than $0.078>\alpha =0.006$

Do not reject null hypothesis. There is no sufficient evidence.

c) The power of the test is

$\beta =P\left ( \textup{ do not reject H0}| \textup{ H0 false} \right )$

For a Left- tailed test the Type II error is

$\beta\left ( \mu _2 \right ) =P\left ( Z>-z_{1-\alpha } +\frac{ \mu _1-\mu _2 }{\sigma /\sqrt{n}}\right )\\ \beta \left ( 70 \right )=1-\Phi \left ( -z_{1-\alpha } +\frac{ 74-70}{\sqrt{6^2/25}}\right )\\ {\color{Blue}\beta \left ( 70 \right )=0.20577}\\$
c) When $\beta =0.01$ ,

$1-\Phi \left ( -z_{1-\alpha } +\frac{74-70 }{\sqrt{36/n}}\right )<0.01\\ \Phi \left ( -z_{1-\alpha } +\frac{74-70 }{\sqrt{36/n}}\right )>0.99\\ -z_{1-\alpha } +\frac{74-70 }{\sqrt{36/n}}>\Phi ^{-1}\left ( 0.99 \right )\\$

$n\geqslant 52.67$

The minimum sample size is ${\color{Blue} 53}$

e) Type I error

${\color{Blue} \alpha =0.01}$