The length of a rectangular pool is 4 meters less than twice the width. if the pool's perimeter is 88 meters, what is the width?
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Answer:
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