Question

The mean drying time of a certain paint in a certain application is 12 minutes. A new additive will be tested to see if it reduces the drying time. One hundred specimens will be painted, and the sample mean drying time will be computed. Let μ be the mean drying time for the new paint. Assume the population standard deviation of drying times is 2 minutes. Assume that, in fact, the true mean drying time of the new paint is 11.5 minutes.

a) What are the null and alternative hypotheses?

b) For what values of should Ho be rejected so that the power of the test is .85? What will the level then be?

c) How large of a sample is needed so that a 5% level test has power .85?

Answer 1

H0 : µ = 12

Ha : µ < 12

SD=2

Variance =4

N( µ, 4/ )

n=100

N( µ, 4/100 )

Standard error of = 2/10

To reject a null hypothesis when true mean is 11.5 .( this is power)

P ( < c) =0.85

P( ( - 11.5 ) / ( 2/10 ) > (c- 11.5) / (2/100 ) ) =0.85

P( Z< (c- 11.5) / (1/5)) =0.85

P(Z < (c- 11.5)*5) =0.85

((c- 11.5)*5) =0.85

(c- 11.5)*5) = 1.04 [ from standard normal table]

c= 11.708

Z value when null hypothesis is true,Z0 = (11.708 - 12 ) / ( 2/10) = - 1.46

level= P( Z< Z0)= ( -1.46) = 0.07215.

When sample mean is less than 11.708 thrnr H0 can be rejected so that power is .85, Then significance level will be 0.07215 or 7.215 % .

c)

significance level = 0.05 , power =.0.85

P ( > c) =0.95 [ When null hypothesis is true]

P( ( - 12) / (2/ ) > (c- 12) / (2 / ) ) =0.95

P( Z < (c- 12) / ( 2 / )) =0.05

((c- 12) / ( 2 / )) =0.05

(c- 12) / ( 2 / )= -1.645 [ from standard normal table]

(c-12) * = - 3.29 --------- 1st eq

To reject a null hypothesis when true mean is 11.5 ( this is power)

P ( < c) =0.85 [ When null hypothesis is true]

P( ( - 11.5) / (2/ ) < (c- 11.5) / (2 / ) ) =0.85

P( Z < (c- 11.5) / ( 2 / )) =0.85

((c- 11.5) / ( 2 / )) =0.85

(c- 11.5) / ( 2 / )= 1.04 [ from standard normal table]

(c-11.5) * = 2.08 ---------2nd equation

2nd eqation - 1st eqation

0.5 * = 2.08 + 3.29

.5 = 5.37

or, n= 115.35 . We can select sample size of 115.

putting value of n in equation 2 we get

c = 11.693

So sample size should be 115.

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