Consider a paint-drying situation in which drying time for a test specimen is no

Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with = 8. The hypotheses H0: = 73 and Ha: < 73 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 = 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.)

(c) For the test procedure with = 0.005, what is (70)? (Round your answer to four decimal places.)
(70) =   

(d) If the test procedure with = 0.005 is used, what n is necessary to ensure that (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 = 76? (Round your answer to four decimal places.)

Explanation / Answer

(a.)

Here the null value is 73 and standard deviation is 8. The number of standard deviations for 72.3 is (73-72.3)/8 = 0.88.

Answer: 0.88

(b.)

Z = (x -µ)/(s/Ön) = (72.3-73)/(8/Ö25) = -0.44

Answer : z = -0.44

P-value = P ( z < -0.44 )

By using normal table we get,

P ( z < -0.44 ) = 0.3300

Answer: p-value = 0.3300

(c.)

Here first we find the critical z for alpha = .005 and we get this from normal table as -2.576.

Now

X bar = µ+ ((s/Ön)*z) = 73 – ((8/sqrt(25))*2.576) = 68.8784

Now true mean is 70

P ( x bar < 68.8784)

Z =(68.8784-70)/(8/SQRT(25)) = -0.70

P (z<-0.70) ; by using normal table we get as = 0.2420

Answer: Type II Error = 1 – 0.242 = 0.7580

(d.) If type II error is .01 then the z-score for that is 2.326

Now we can write,

X bar = µ+ ((s/Ön)*z)

n = ( s*z)^2 / ( x bar - µ)^2

   = (8*2.236)^2 /(68.8784-70)^2 = 255

Answer: 255

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