Question 6( Multiple Choice Worth 1 points)
1 answer:
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Answer:
The null and alternative hypothesis are:
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:
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:
Then, we can calculate the z-statistic as:
This test is a left-tailed test, so the P-value for this test is calculated as:
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%.