Question

The titanium content of an alloy is being studied in the hope of ultimately increasing the tensile strength. An analysis of six recent heats chosen at random produces the following titanium contents.6.6% 9.1% 7.0% 6.3% 8.5% 10.0%
What is the value of the test statistic, if the alternative hypothesis is the mean titanium content is greater than 9.5%? Round your final answer to two decimal places.

Answer 1

Answer:

$t=\frac{7.917-9.5}{\frac{1.501}{\sqrt{6}}}=-2.58$    

$p_v =P(t_{5}$

If we compare the p value with a significance level for example $\alpha=0.05$ we see that $p_v < \alpha$ so we can conclude that we can reject the null hypothesis, and there is enough evidence to conclude that the true mean is significantly lower than 9.5% at 0.05 of signficance

Step-by-step explanation:

Data given and notation    

Data: 6.6% 9.1% 7.0% 6.3% 8.5% 10.0%

We can calculate the mean and the sample deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X-i -\bar X)^2}{n-1}}$

$\bar X=7.917$ represent the sample mean

$s=1.501$ represent the sample standard deviation

$n=6$ sample size    

$\mu_o =9.5$ represent the value that we want to test  

$\alpha$ 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 determine if the true mean os greater or not than 9.5%, the system of hypothesis would be:    

Null hypothesis:$\mu \geq 9.5$    

Alternative hypothesis:$\mu < 9.5$    

We don't know the population deviation, so for this case we can use the 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:    

$t=\frac{7.917-9.5}{\frac{1.501}{\sqrt{6}}}=-2.58$    

Calculate the P-value    

The degrees of freedom are given by:

$df = n-1 = 6-1=5$

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

$p_v =P(t_{5}$

In Excel we can use the following formula to find the p value "=T.DIST(-2.58,5,TRUE)"  

Conclusion    

If we compare the p value with a significance level for example $\alpha=0.05$ we see that $p_v < \alpha$ so we can conclude that we can reject the null hypothesis, and there is enough evidence to conclude that the true mean is significantly lower than 9.5% at 0.05 of signficance