Question

Dullco Manufacturing claims that its alkaline batteries last at least 40 hours on average in a certain type of portable CD player. But tests on a random sample of 18 batteries from a day's large production run showed a mean battery life of 37.8 hours with a standard deviation of 5.4 hours. In a left-tailed test at α = .05, which is the most accurate statement?

We would strongly reject the claim.

We would clearly fail to reject the claim.

We would face a rather close decision.

We would switch to α = .01 for a more powerful test.

Not sure what they mean. Can you please explain why it is that answer please?

Answer 1

Answer:

$p_v =P(t_{17}<-1.728)=0.051$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best option for this case would be:

We would switch to α = .01 for a more powerful test.

Step-by-step explanation:

Previous concepts and data given  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X=37.8$ represent the sample mean  

$s=5.4$ represent the sample standard deviation  

n=18 represent the sample selected  

$\alpha=0.05$ significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 40, the system of hypothesis would be:    

Null hypothesis:$\mu \geq 40$    

Alternative hypothesis:$\mu < 40$    

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:    

$t=\frac{37.8-40}{\frac{5.4}{\sqrt{18}}}=-1.728$      

P-value

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

$df=n-1=18-1=17$

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

$p_v =P(t_{17}<-1.728)=0.051$    

Conclusion    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best option for this case would be:

We would switch to α = .01 for a more powerful test.

Since the p values is just a little higher than th significance level 0.051>0.05  but the values are two close. If we change the value of the significance by 0.01, we have that 0.051>0.01 and it's a more powerful test.