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.)

(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
H_a: μ > 5.5H₀: μ = 5.5
H_a: μ < 5.5    H₀: μ = 5.5
H_a: μ ≠ 5.5H₀: 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.

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.    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.

(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

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

a) H₀: μ = 5.5
H_a: μ ≠ 5.5

z =(5.22-5.5)*sqrt(16)/0.32=-3.5

p value=0.0004 ( please try 0.0005 if this comes wrong)

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

b)

Hypothesized mean true mean standard deviation sample size standard error of mean _ơNn for 0.01 level and two tailed test critival value Za acceptance region: μ-Za*ox < <-l+Za*ơKZ type ll error-probability of not rejecting 5.5 0.32 16 0.0800 2.5758 5.7061 (5.7061-μα)/ox)) 1.3258) -0.9082-0.0001 0.9081 0.0919 5.2939 <IK <z- 3.8258 <Z< β P( (5.2939-ua)/ox)) = power of the test

from above: e probability of detecting this departure from H₀ =0.0919

c)

n=247