Question

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x = 52.5 and a sample standard deviation of s = 4.3. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude? (Use alpha = 0.05.) State the appropriate null and alternative hypotheses. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) State the conclusion in the problem context.

Answer 1

Answer:

Null hypothesis:$\mu \leq 50$  

Alternative hypothesis:$\mu > 50$  

$t=\frac{52.5-50}{\frac{4.3}{\sqrt{45}}}=3.90$    

$p_v =P(t_{(44)}>3.90)=0.0002$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 50 mils at 5% of signficance.  

Step-by-step explanation:

1) Data given and notation  

$\bar X=52.5$ represent the sample mean

$s=4.3$ represent the sample standard deviation

$n=45$ sample size  

$\mu_o =50$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t 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 true average penetration be at most 50 mils, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 50$  

Alternative hypothesis:$\mu > 50$  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-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:  

$t=\frac{52.5-50}{\frac{4.3}{\sqrt{45}}}=3.90$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=45-1=44$  

Since is a one side right tailed test the p value would be:  

$p_v =P(t_{(44)}>3.90)=0.0002$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 50 mils at 5% of signficance.