Question

73 and 73 are to be tested using a random sample Consider a part-drying situation in which drying time for a test specimen is normally distributed with . The hypotheses o of n - 25 observations (a) How many standard deviations (of x below the null value is x 72.37 (Round your answer to two decimal places.) standard deviations (b) If x= 72.3, what is the conclusion using a = 0.0057 Calculate the test statistic and determine the value. (Round your test statistic to two decimal places and your fuvalue to four decimal places.) Pevalue- State the condusion in the problem context Reject the di vypothesis. There is not suficient evidence to conclude that the mean drying time is less than 73. Reject the null hypothesis. There is suficient evidence to conclude that the mean drying time is less than 73 Do not reject the nut othesis. There is not sufficient evidence to conclude that the mean drying time is less than 73. Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 73 (c) For the test procedure with B(70) - = 0.005, what is (70)? (Round your answer to four decimal places.) = 0.005 is used, what is necessary to ensure that (70) 0.017 (Round your answer up to the next whole number.) (d) in the test procedure with specimens (e) If a level 0.01 test is used with n - 100, what is the probability of a type 1 error when 767 (Round your answer to four decimal places.)

Answer 1

a)

Standard Error , SE = σ/√n =   9.0000   / √    25   =   1.8000      
Z-test statistic= (x̅ - µ )/SE = (   72.300   -   73   ) /    1.8000   =   -0.389

answer: 0.39 std dev

b)

Standard Error , SE = σ/√n =   9.0000   / √    25   =   1.8000      
Z-test statistic= (x̅ - µ )/SE = (   72.300   -   73   ) /    1.8000   =   -0.39
                          
  
p-Value   =   0.3487   [ Excel formula =NORMSDIST(z) ]              

do not reject null hypothesis.there is not sufficient evidence...................

c)

true mean ,    µ =    70              
                      
hypothesis mean,   µo =    73              
significance level,   α =    0.005              
sample size,   n =   25              
std dev,   σ =    9              
                      
δ=   µ - µo =    -3              
                      
std error of mean,   σx = σ/√n =    9.0000   / √    25   =   1.80000

Zα =       -2.5758   (left tail test)          
                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic >               -2.576      
this Z-critical value corresponds to X critical value( X critical), such that                      
                      
(x̄ - µo)/σx ≥ Zα                      
x̄ ≥ Zα*σx + µo                      
x̄ ≥    -2.576   *   1.8000   +   73  
       x̄ ≥    68.3635   (acceptance region)      
                      
now, type II error is ,ß =    P( x̄ ≥    68.364   given that µ =   70   )  
                      
   = P ( Z > (x̄-true mean)/σx )                   
=P(Z > (   68.364   -   70   ) /   1.8000   )
                      
   = P ( Z >    -0.909   ) =   0.8184 [ Excel function: =1-normsdist(z) ]  
                      

d)

True mean   µ =    70                              
hypothesis mean,   µo =    73                              
                                      
Level of Significance ,    α =    0.005                              
std dev =    σ =    9.000                              
power =    1-ß =    0.99                              
ß=       0.01                              
δ=   µ - µo =    -3                              
                                      
Z ( α ) =       2.5758   [excel function: =normsinv(α)                          
                                     
Z (ß) =        2.3263   [excel function: =normsinv(ß)                         
                                      
sample size needed =    n = ( ( Z(ß)+Z(α) )*σ / δ )² = ( (   2.3263   +   2.5758   ) *   9.0   /   -3   ) ² =   216.28
                                      
so, sample size =        217

e)

type I error is probability of rejecting Ho when it is true.

P(X<73|µ=76)

Z =   (X - µ )/(σ/√n) = (   73   -   76.00   ) / (   9.000   / √   100   ) =   -3.33
                                      
P(X ≤   73   ) = P(Z ≤   -3.333   ) =   0.0004 (answer)