Question

Joe believes the average battery life of a vehicle is equal to 9 years. A sample of 5 vehicles was taken and the following data on battery lives was obtained. Assume the distribution of the population is normally distributed. What is the value of the test statistic?
Col1 Vehicle 1 2 3 4 5
Col2 Years 6 7 8 7 6
2 1 5 6 9

Answer 1

Answer:

$t=\frac{7.26-9}{\frac{0.856}{\sqrt{5}}}=-4.54$    

$p_v =2*P(t_{4}$    

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we ccan say that the true mean is different from 9 years at 5% of significance.  

Step-by-step explanation:

Data given and notation    

Data : 6.2, 7.1, 8.5, 7.6,6.9

The sample mean and devation can be founded with the following formulas:

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

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

$\bar X=7.26$ represent the average

$s=0.856$ represent the sample standard deviation    

$n=5$ sample size    

$\mu_o =76$ 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 apply a two tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  $\mu = 9$    

Alternative hypothesis :$\mu \neq 9$    

Compute the test statistic  

The statistic for this case 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:    

$t=\frac{7.26-9}{\frac{0.856}{\sqrt{5}}}=-4.54$    

Give the appropriate conclusion for the test  

First we need to find the degrees of freedom given by:

$df=n-1=5-1=4$

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

$p_v =2*P(t_{4}$    

Conclusion    

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we ccan say that the true mean is different from 9 years at 5% of significance.