Answer to Question #294633 in Statistics and Probability for Yaku

Question

a) A manufacturer has developed a new shing line, which the company claims has mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilograms. To test the claim, a random sample of 50 lines will be tested. The rejection region to be used is dened to be \bar{X} < 14:9.

i) State the null and alternative hypotheses.

ii) Find type I error probability, $\alpha$ ,

iii) Evaluate $\beta$ or the alternative $\mu$ = 14.8.

iv) Evaluate power function for the alternative $\mu$ = 14.8.

b) Suppose that we wish to test the hypothesis H₀ : $\mu$ = 68 kg versus H₁ : $\mu$ > 68 kg, for weights of male students at a certain college, using $\alpha$ = 0.05 level of significance, when it is known that $\sigma$ = 5.

i) Find the sample size required if the power of the test is to be 0.95

when the true mean is 69 kg.

ii) For what values of \bar{X} will the null hypothesis be rejected?

Answer 1

i) State the null and alternative hypotheses.

H₀: The mean is equal to 15 kilograms ( null hypothesis )

H₁: The mean is less than 15 kilograms ( alternative hypothesis )

ii) Find type I error probability, $\alpha$ ,

we define type I error as the probability of rejecting the null hypothesis when it is true

we compute the probability of committing type I error using the z score

Given population mean = 15, sample mean = 14.9, population standard deviation = 0.5 and sample size = 50, we have

z = ( 14.9 - 15) / ( 0.5/50^1/2) = -1.41

using the normal table, we obtain p ( z=-1.41) = 0.07927 which is the probability of committing type I error.

iii) Evaluate $\beta$ or the alternative $\mu$ = 14.8.

we use z score

z= (14.8 - 15 ) / ( 0.5/50^1/2 ) = -2.83

using the normal table, we obtain p(z=-2.83 ) = 0.00233

but type II error = 1 - type one error which equals 1 - 0.00233 = 0.99767 which is the required solution.

iv) Evaluate power function for the alternative $\mu$ = 14.8.

we use z score

define z = (14.9 - 14.8 ) / (0.5/50^1/2 ) = 1.41

using normal table, we have p(z=1.41 ) = 0.92073 which is the required solution.

b)

i) Find the sample size required if the power of the test is to be 0.95

when the true mean is 69 kg

we define the following at 0.05 level of significance

{ x^- > 68 +( ( 5/n^1/2) * ( z_0.05) ) } = { x^- > 68 +( ( 5/n^1/2) * ( 1.645) ) }

To find the sample size n such that power = 0.95,we define

$\implies$ PH₁:u =69( X^- > 68 + ( (5/n^1/2) * (1.645) ) ) = 0.95

$\implies$ PH₁:u =69( ( (X^- - 69) / (5/n^1/2) ) > ( ( 68-69) / (5/n^1/2) ) + ( 1.645 ) ) = 0.95

$\implies$ P( Z > ( ( 68-69 ) / (5/n^1/2) ) + 1.645 ) = 0.95

$\implies$ 1- P( Z ≤ ( ( 68-69 ) / (5/n^1/2) ) + 1.645 ) = 0.95

$\implies$1- P( Z ≤  -(n^1/2)/5 + 1.645 ) = 0.95

$\implies$ P( Z ≤  -(n^1/2)/5 + 1.645 ) = 1 - 0.95 = 0.05

$\implies$ P( Z ≤  -(n^1/2)/5 + 1.645 ) = 0.05

using normal tables we obtain the below

$\implies$ -(n^1/2)/5 + 1.645 ) = -1.645

$\implies$ n^1/2 = ( 2 * 1.645 * 5 )

$\implies$n^1/2 = 16.45

$\implies$n = ( 16.45 )²Â = 270.6025 which is approximately 271 which is the required sample size.

ii)

For 0.05 level of significance, the critical z value is 1.645.

since we have been given that u = 68 from the null hypothesis, n= 271 and standard deviation =5, we have

1.645 =( (X^- - 68) / ( 5/271^1/2) )

solving for X^- from the above equation we obtain that X^- = 68.4996 which is the required solution