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 (20 pts)

Answer 1

The true average percentage of Silicon Dioxide (SiO2) is less than 5.5.

Explanation:

In this case we need to test whether the true average percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is smaller than 5.5.

The hypothesis can be defined as follows:

H₀: The true average percentage of Silicon Dioxide (SiO2) is 5.5, i.e. μ = 5.5.

Hₐ: The true average percentage of Silicon Dioxide (SiO2) is less than 5.5, i.e. μ < 5.5.

The information provided is:

`\bar x=5.25\\\sigma=0.30\\n=16\\\alpha =0.01`

As the population standard deviation is provided, we will use a z-test for single mean.

Compute the test statistic value as follows:

`z=(\bar x-\mu)/(\sigma/√(n))=(5.25-5.5)/(0.30/√(16))=-3.33`

The test statistic value is -3.33.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

`p-value=P(Z<-3.33)=0.00043`

*Use a z-table for the probability.

The p-value of the test is 0.00043.

p-value = 0.00043 < α = 0.05

The null hypothesis will be rejected at 5% level of significance.

Thus, it can be concluded that the true average percentage of Silicon Dioxide (SiO2) is less than 5.5.

Answer 2

We conclude that the true average percentage of Silicon Dioxide is smaller than 5.5.

Explanation:

We are given that the desired percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is 5.5.

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.

Let
`\mu` = true average percentage of Silicon Dioxide.

So, Null Hypothesis,
`H_0` :
`\mu`
`\geq` 5.5 {means that the true average is greater than or equal to 5.5}

Alternate Hypothesis,
`H_A` :
`\mu` < 5.5 {means that the true average is smaller than 5.5}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean percentage of Silicon Dioxide = 5.25

σ = population standard deviation = 0.3

n = sample size = 16

So, the test statistics =
`(5.25-5.5)/((0.3)/(√(16) ) )`

= -3.33

The value of z test statistics is -3.33.

Now, the P-value of the test statistics is given by;

P-value = P(Z < -3.33) = 1 - P(Z
`\leq` 3.33)

= 1 - 0.9996 = 0.0004

Since, the P-value of the test statistics is less than the level of significance as 0.0004 < 0.01, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true average percentage of Silicon Dioxide is smaller than 5.5.