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 = 5.21. (Use α = 0.05.) (a) Does this indicate conclusively that the true average percentage differs from 5.5? State the appropriate null and alternative hypotheses. H0: μ = 5.5 Ha: μ ≥ 5.5 H0: μ = 5.5 Ha: μ < 5.5 H0: μ = 5.5 Ha: μ ≠ 5.5 H0: μ = 5.5 Ha: μ > 5.5 Correct: Your answer is correct. 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. 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. Reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage. Correct: Your answer is correct. (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 places.) (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

a) zscore=(x̅-μ)*√n/σ =					-3.63
p value =					0.0002

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