The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in ¯x = 94.32. Assume that the distribution of the melting point is normal with σ = 1.20. (a) Test H0 : µ = 95 versus Ha : µ 6= 95 using a two-tailed level .01 test. (b) If a level .01 test is used, what is β(94), the probability of a type II error when µ = 94? (c) What value of n is necessary to ensure that β(94) = .1 when α = .01?

Given:

Mean, x' = 94.32

s.d = 1.20

n = 16

a) For null hypothesis:

H0 : u= 95

For alternative hypothesis:

Ha : u≠95

Level of significance, a = 0.01

(Two tailed test)

We reject null hypothesis, h0, if P<0.01 level of significance.

Calculating the test statistic, we have:

= -2.266

= -2.27

Calculating the P value:

P value = 2P(Z < -2.27)

Using the standard normal table,

NORMSDIST(-2.27)

= 0.0116

P value= 2(0.0116)

= 0.0232

Since P value(0.0232) is greater than level of significance (0.01), we fail to reject the null hypothesis H0.

We can say that there is enough evidence to conclude the data does not support that average melting point differs from the level of 95 at the level of 0.01 significance level.

b) B(u = 94)

= P(when u=94, do not reject H0)

Using the standard nkrmal table, the z-score corresponding to

Z(0.01/2)= 0.005 will be

= ∅(5.91)-∅(0.75)

P(Z≤5.91)-P(Z≤0.75)

From standard normal table, we have:

P(Z≤0.75)=0.7734, P(Z≤5.91)=1

= 1 - 0.7734

= 0.2266

Probability of making type II error when u=94 is 0.2266

c) Probability of committing type II error when u= 94 at a significance level of 0.01 will be =0.10.

B(94) = 0.10

Finding sample size, n for a two tailed test:

Using standard normal table, Z score corresponding to a/2 = 0.005 cummulative area(1-0.005 = 0.995) is Z= 2.58

Z score corresponding to 0.10 cummulative area(1-0.10 = 0.90) is Z = 1.28

Our sample size n, wil be=

= 21.46

= 22

Sample size = 22