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 and that . a. Does this indicate conclusively that the true average percentage differs from 5.5

Answer 1

Answer:

Step-by-step explanation:

This is a test of mean where σ, the population standard deviation is known. The null hypothesis H₀ : µ =  µ₀ = 5.5; the alternative H₁ : µ ≠ µ₀. Since σ is known, z =  (x − µ₀

)/(σ/√n)   is normally distributed for all n and so,  for a two-tailed test, we reject H₀ if z ≥ z(α/2) or z ≤ −z(α/2). In this case, α = 0.05 so z(α/2) = z(.025) = 1.960.

P = 2P (Z ≥ |z|) = 2P

*(Z ≥  (x − µ₀)/(

σ/√

n))

⇒  P = 2P

*(Z ≥  (5.25 − 5.5)/(

0.32/√16))  = 2P

*(Z ≥  -3.125)  = 2*(0.9991) = 1.9982

We reject the null hypothesis that the true average is 5.5.

Answer 2

Answer:

since p value < x(0.05) so we reject the null hypothesis

step 5 ( conclusion)

reject the null hypothesis and there is sufficient evidence to support the claim that the true average percentage differs from 5.5.

Step-by-step explanation:

   

check the attachment for detailed explanation