Question

A computer company claims that the batteries in its laptops last 4 hours on average. A consumer report firm gathered a sample of 16 batteries and conducted tests on this claim. The sample mean was 3 hours 50 minutes, and the sample standard deviation was 20 minutes. Assume that the battery time distribution as normal.
a) Test if the average battery time is shorter than 4 hours at a = 0.05.b) Construct a 95% confidence interval of the mean battery time.c) If you were to test H0: µ =240 minutes vs. H1: µ ≠ 240 minutes, what would you conclude from your result in part (b)?

Answer 1

Answer:

a) $t=\frac{230-240}{\frac{20}{\sqrt{16}}}=-2$    

$p_v =P(t_{(15)}<-2)=0.0320$  

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 reject the null hypothesis, so we can conclude that the true mean is significantly less than 4 hours= 240 minutes.

b) $230-2.131\frac{20}{\sqrt{16}}=219.345$    

$230+2.131\frac{20}{\sqrt{16}}=240.655$    

So on this case the 95% confidence interval would be given by (219.345;240.655)    

c) H0: µ =240 minutes vs. H1: µ ≠ 240

For this case since the confidence interval contain the value 240. Then we don't have enough evidence to reject the null hypothesis on this case at 5% of significance.

Step-by-step explanation:

Data given and notation  

$\bar X=230$ represent the sample mean on minutes

$s=20$ represent the sample standard deviation

$n=16$ sample size  

$\mu_o =240$ represent the value that we want to test  on minutes (because 4*60=240 min)

$\alpha=0.05$ 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)  

Part a

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 240 minutes, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 240$  

Alternative hypothesis:$\mu < 240$  

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:  

$t=\frac{230-240}{\frac{20}{\sqrt{16}}}=-2$    

P-value

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

$df=n-1=16-1=15$  

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

$p_v =P(t_{(15)}<-2)=0.0320$  

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 reject the null hypothesis, so we can conclude that the true mean is significantly less than 4 hours= 240 minutes.  

Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.131$

Now we have everything in order to replace into formula (1):

$230-2.131\frac{20}{\sqrt{16}}=219.345$    

$230+2.131\frac{20}{\sqrt{16}}=240.655$    

So on this case the 95% confidence interval would be given by (219.345;240.655)    

Part c

H0: µ =240 minutes vs. H1: µ ≠ 240

For this case since the confidence interval contain the value 240. Then we don't have enough evidence to reject the null hypothesis on this case at 5% of significance.