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 σ = 0.32 and that x = 5.21. (Use α = 0.05.)(a) Does this indicate conclusively that the true average percentage differs from 5.5?State the appropriate null and alternative hypotheses.

Answer 1

The null and alternative hypothesis are:

`H_0: \mu=5.5\\\\H_a:\mu< 5.5`

At a significance level of 0.05, there is enough evidence to support the claim that the population percentage of SiO2 is signficantly different from 5.5%.

P-value = 0.000004.

Explanation:

This is a hypothesis test for the population mean.

The claim is that the population percentage of SiO2 is signficantly different from 5.5%.

Then, the null and alternative hypothesis are:

`H_0: \mu=5.5\\\\H_a:\mu< 5.5`

The significance level is 0.05.

The sample has a size n=16.

The sample mean is M=5.21.

The standard deviation of the population is known and has a value of σ=0.26.

We can calculate the standard error as:

`\sigma_M=(\sigma)/(√(n))=(0.26)/(√(16))=0.065`

Then, we can calculate the z-statistic as:

`z=(M-\mu)/(\sigma_M)=(5.21-5.5)/(0.065)=(-0.29)/(0.065)=-4.462`

This test is a left-tailed test, so the P-value for this test is calculated as:

`\text{P-value}=P(z<-4.462)=0.000004`

As the P-value (0.000004) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population percentage of SiO2 is signficantly different from 5.5%.