In a paint-drying experiment, the drying time for a test specimen is normally di

Question

In a paint-drying experiment, the drying time for a test specimen is normally distributed with mean 75 min and standard deviation=9 min . The hypotheses Ho: mu=75 versus Ha: mu < 75 are to be tested using a random sample of n=25 observations.

a. How many standard deviations (of X ) below the null value x =72.3?

b. If x =72.3, what is the conclusion using =0.1 ?

c. What is for the test procedure that rejects Ho when z <=2.88 ?

d. If a level .01 test is used with n=100, will you accept or reject Ho when mu=76 ?

Explanation / Answer

the null hypothesis is H0:mu=75 vs the alternative is H1:mu<75

for that we have a random sample of size n=25 from a normal population with mean mu and standard deviation sigma=9

let X denotes the sample mean then X would follow a normal distribution with mean 75 and standard deviation=9/sqrt(25)=9/5=1.8

a)let X is k times standard deviation below the null value 72.3

so we must have P[X<75-1.8k]=P[X<72.3]

so 75-1.8k=72.3 or, k=(75-72.3)/1.8=1.5 [answer]

b) the test statistic is given as Z=(X-75)*/1.8 which under H0 follows a N(0,1) distribution.

now since the alternative hypothesis is less than type hence H0 is rejected iff z<-talpha

where alpha=0.1 and talpha is the upper alpha point of a N(0,1) distribution and z is the observed value of Z

so when X=72.3

then Z=(72.3-75)/1.8=-1.5 -t0.1=- 1.281552

so z<  -t0.1 hence we fail to reject H0 and the conclusion is mean is indeed <75 min [answer]

c) alpha=P[rejecting H0| H0 is true]=P[z<=2.88]=0.9980116 [answer]

d) now level=alpha=0.01

n=100 the standard deviation would be 9/sqrt(100)=9/10

then when mu=76 null hypothesis is H0: mu=76 vs H1: mu<76

so test statistic is Z=(X-76)/0.9 which under H0 follows N(0,1)

H0 is rejected iff z<-t0.01  

z=(72.3-76)/0.9=-4.111111 -t0.01=-2.326348

so z<-t0.01

hence H0 is rejected

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