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 = 5.24. (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.H0: μ = 5.5Ha: μ > 5.5H0: μ = 5.5Ha: μ ≥ 5.5 H0: μ = 5.5Ha: μ < 5.5H0: μ = 5.5Ha: μ ≠ 5.5

Answer:The null and alternative hypothesis are: [tex]H_0: \mu=5.5\\\\H_a:\mu< 5.5[/tex] 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.Step-by-step 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: [tex]H_0: \mu=5.5\\\\H_a:\mu< 5.5[/tex] 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: [tex]\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.26}{\sqrt{16}}=0.065[/tex] Then, we can calculate the z-statistic as: [tex]z=\dfrac{M-\mu}{\sigma_M}=\dfrac{5.21-5.5}{0.065}=\dfrac{-0.29}{0.065}=-4.462[/tex] This test is a left-tailed test, so the P-value for this test is calculated as: [tex]\text{P-value}=P(z<-4.462)=0.000004[/tex] 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%.