Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 6. 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.) Incorrect: Your answer is incorrect. standard deviations (b) If x = 72.3, what is the conclusion using α = 0.005? Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Reject the null hypothesis. There is not 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. Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 75. Reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 75. Correct: Your answer is correct. (c) For the test procedure with α = 0.005, what is β(70)? (Round your answer to four decimal places.) β(70) = (d) If the test procedure with α = 0.005 is used, what n is necessary to ensure that β(70) = 0.01? (Round your answer up to the next whole number.) n = specimens (e) 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

a)

Standard Error , SE = σ/√n =   6.0000   / √    25   =   1.2000      
Z-test statistic= (x̅ - µ )/SE = (   72.300   -   75   ) /    1.2000   =   -2.250

answer: 2.25

b)

test stat = -2.25

p-Value   =   0.0122
Decision:   p-value>α, Do not reject null hypothesis   

Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 75.

c)

true mean ,    µ =    70              
                      
hypothesis mean,   µo =    75              
significance level,   α =    0.005              
sample size,   n =   25              
std dev,   σ =    6.0000              
                      
δ=   µ - µo =    -5              
                      
std error of mean,   σx = σ/√n =    6.0000   / √    25   =   1.20000

Zα =       -2.5758   (left tail test)

P(type II error) , ß =   P(Z > Zα - δ/σx)                  
= P(Z >    -2.576   - (   -5   /   1.2000   ))
=P(Z>    1.591   ) =   0.0558   [ Excel function: =1-normsdist(z) ]      

d)

True mean   µ =    70                              
hypothesis mean,   µo =    75                              
                                      
Level of Significance ,    α =    0.005                              
std dev =    σ =    6                              
ß=       0.01                              
δ=   µ - µo =    -5                              
                                      
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   +   2.576   ) *   6.0   /   -5   ) ² =   34.61
                                      
so, sample size =        35                              

e)

Z =   (X - µ )/(σ/√n) = (   72.3   -   76.00   ) / (   6.000   / √   100   ) =   -6.167  
                                          
P(X ≤   72.3   ) = P(Z ≤   -6.167   ) =   0.0000          

so, P(type I error) = 0 (answer)