Question

A data set lists earthquake depths. The summary statistics arenequals=500500​,x overbarxequals=5.435.43​km,sequals=4.864.86km. Use a0.010.01significance level to test the claim of a seismologist that these earthquakes are from a population with a mean equal to5.005.00.Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.What are the null and alternative​ hypotheses?A.Upper H 0H0​:muμequals=5.005.00kmUpper H 1H1​:muμless than<5.005.00kmB.Upper H 0H0​:muμnot equals≠5.005.00kmUpper H 1H1​:muμequals=5.005.00kmC.Upper H 0H0​:muμequals=5.005.00kmUpper H 1H1​:muμnot equals≠5.005.00kmD.Upper H 0H0​:muμequals=5.005.00kmUpper H 1H1​:muμgreater than>5.005.00kmDetermine the test statistic.nothing​(Round to two decimal places as​ needed.)Determine the​ P-value.nothing​(Round to three decimal places as​ needed.)State the final conclusion that addresses the original claim.

Answer 1

Answer:

C. H 0​:μ=5.00km   H1​:μ≠5.00km

$z=\frac{5.43-5.00}{\frac{4.86}{\sqrt{500}}}=1.978$    

$p_v =2*P(z>1.978)=0.048$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is nto significantly different from 5.00.    

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Data given and notation    

$\bar X=5.43$ represent the average for the sample

$s=4.86$ represent the sample standard deviation for the sample    

$n=500$ sample size    

$\mu_o =5.00$ represent the value that we want to test  

$\alpha=0.01$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)    

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to check if the population mean is equal to 5.00, the system of hypothesis would be:    

Null hypothesis:$\mu = 5.00$    

Alternative hypothesis:$\mu \neq 5.00$    

Since the sample size >30 and large, the estimator for the population deviation would be s, and we can use the z test to compare the actual mean to the reference value, and the statistic is given by:    

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$z=\frac{5.43-5.00}{\frac{4.86}{\sqrt{500}}}=1.978$    

P-value    

Since is a two tailed test the p value would be:    

$p_v =2*P(z>1.978)=0.048$

In Excel we can us ethe following formula to find the p value "=2*(1-NORM.DIST(1.978;0;1;TRUE))"  

Conclusion    

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is nto significantly different from 5.00.