Question

The desired percentage of SiO₂ in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of SiO₂ in a sample is normally distributed with σ = 0.32 and that x = 5.22. (Use α = 0.05.)

* Does this indicate conclusively that the true average percentage differs from 5.5?
State the appropriate null and alternative hypotheses.

A-) H₀: μ = 5.5
H_a: μ ≥ 5.5

B-) H₀: μ = 5.5
H_a: μ < 5.5

C-) H₀: μ = 5.5
H_a: μ ≠ 5.5

D-) H₀: μ = 5.5
H_a: μ > 5.5

Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z	=
P-value	=

State the conclusion in the problem context.

*Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.

*Reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage.    

*Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage.

*Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.

(b) If the true average percentage is

μ = 5.6 and a level α = 0.01 test based on n = 16 is used, what is the probability of detecting this departure from H₀? (Round your answer to four decimal pl

(c) What value of n is required to satisfy α = 0.01 and β(5.6) = 0.01? (Round your answer up to the next whole number.)
n = ? samples

Answer 1

C-) H₀: μ = 5.5
H_a: μ ≠ 5.5

zscore=(x̅-μ)*√n/σ =					-3.5
p value =					0.0004

*Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.

b)

rejection region : x̅ <=μ+z*σ/√n or x̅ >=μ+z*σ/√n=						x̅ <=5.3136 or x̅ >=5.6864
hence probability of detecting this departure: P(x̅ <=5.3136 or x̅ >=5.6864)=1-P(-3.83<z<1.33)=									0.0919

c)

required sample size   n=

=(Zα+Zβ)2σ2/(μo-μa)2=

247