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Len [333]
3 years ago
12

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

Mathematics
1 answer:
Reil [10]3 years ago
4 0

Answer:

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

Step-by-step explanation:

<u><em>Step( i )</em></u>:-

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.

<u><em> Null hypothesis : H₀ : μ = 5.5</em></u>

<u><em>  Alternative Hypothesis : H₁ : μ < 5.5</em></u>

 Level of significance ∝ = 0.01

<u><em>Step(ii)</em></u>:-

 The test statistic

                              t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

                             t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }

On calculation , we get

                            t = -3.33

                           |t| = |-3.33| = 3.33

<u><em>Step(iii)</em></u>:-

<u><em>P - value</em></u>

<u><em>The degrees of freedom γ = n-1 = 16-1 =15</em></u>

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

0.001 < P < 0.005

<u>Condition(i)</u>

P - value < ∝ then reject H₀

<u>Condition(ii)</u>

P - value > ∝ then Accept H₀

we observe that  0.001 < P < 0.005

P- value < 0.01

we rejected  H₀

<em>(or)</em>

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

<u><em>Conclusion</em></u>:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

<u><em></em></u>

             

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