Question

An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly more than 10 years. In order to test this claim, 9 car transmissions are randomly selected and their useful lifetimes are recorded. The sample mean lifetime is 13.5 years and the sample standard deviation is 3.2 years. Assuming that the useful lifetime of a typical car transmission has a normal distribution, based on these sample result, the correct conclusion at 1% significance level for this testing hypotheses problem is:
a. none of these answers.b. Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).c. Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).d. Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).e. Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).

Answer 1

Answer:

a. none of these answers

Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.281)

Step-by-step explanation:

Data given and notation  

$\bar X=13.5$ represent the mean height for the sample  

$s=3.2$ represent the sample standard deviation for the sample  

$n=9$ sample size  

$\mu_o =10$ represent the value that we want to test

$\alpha=0.01$ 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 mean is higher than 10 years, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 10$  

Alternative hypothesis:$\mu > 10$  

If we analyze the size for the sample is < 30 and 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:  

$t=\frac{13.5-10}{\frac{3.2}{\sqrt{9}}}=3.28$    

P-value

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

$df=n-1=9-1=8$  

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

$p_v =P(t_{(8)}>3.281)=0.00558$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that we have a mean higher than 10 years at 1% of significance.  

a. none of these answers