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 (sigma) = 0.3 and that x-bar = 5.25.
a) Does this indicate conclusively that the true average percentage differs from 5.5?
b) If the true average percentage is (mu) = 5.6 and a level a = 0.01 test based on n = 16 is used,
what is the probability of detecting this departure from H0?

Answer 1

A. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.

B. P(Z<-4.93333) = 0.0000

C. n = 5945 samples

Explanation:

A)

The rightnull and alternative hypotheses are given as below:

Null hypothesis H0: µ = 5.5

Alternative hypothesis Ha: µ ≠ 5.5

Conducting the one sample z test for population mean.

Test statistic formula is given as below:

Z = (Xbar - µ)/[σ/√(n)]

Given,

Xbar = 5.23

µ = 5.5

σ = 0.30

n = 16

Z = (5.23 – 5.5) / [ 0.30 /√(16)]

Z = -3.6000

P-value = 0.0003

α value = 0.05

P-value < α value

So, we reject the null hypothesis. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.

B

From the information given

Xbar = 5.23

µ = 5.6

σ = 0.30

n = 16

α value = 0.01

Z = (Xbar - µ)/[σ/√(n)]

Z = (5.23 - 5.6)/(0.30/√(16))

Z = -4.93333

P(Z<-4.93333) = 0.0000

Required probability = 0.0000

C

We are given α =0.01

Therefore, critical Z value = 2.57

E = 0.01

σ = 0.30

n = (Z*σ/E)^2

n = (2.57*0.30/0.01)^2 = 5944.41

n = 5945 samples