Question

The desired percentage of Silicon Dioxide (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 and a sample mean of 5.25 was obtained. Suppose that the percentage of SiO2 in a sample is normally distributed with a sigma of 0.3. Does this indicate conclusively that the true average is smaller than 5.5? Carry the procedure at a 0.01 significance level. Use only the P-Value approach. State H0 and Ha

Answer 1

a) Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

|t| = |-3.33| = 3.33

c) P - value lies between in these intervals

0.001 < P < 0.005

Explanation:

Step( i ):-

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

Level of significance ∝ = 0.01

Step(ii):-

The test statistic

`t = (x^(-) -mean)/((S.D)/(√(n) ) )`

`t = (5.25 -5.5)/((0.3)/(√(16) ) )`

On calculation , we get

t = -3.33

|t| = |-3.33| = 3.33

Step(iii):-

P - value

The degrees of freedom γ = n-1 = 16-1 =15

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

Condition(i)

P - value < ∝ then reject H₀

Condition(ii)

P - value > ∝ then Accept H₀

we observe that 0.001 < P < 0.005

P- value < 0.01

we rejected H₀

(or)

The tabulated value = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

Conclusion:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5