Question

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.

Answer 1

Answer:

The 95% confidence interval would be given by $0.281 \leq \mu_1 -\mu_2 \leq 0.529$  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".  

$\bar X_1 =9.2$ represent the sample mean 1

$\bar X_2 =8.8$ represent the sample mean 2

n1=27 represent the sample 1 size  

n2=30 represent the sample 2 size  

$\sigma_1 =0.3$ population standard deviation for sample 1

$\sigma_2 =0.1$ population standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =9.2-8.8=0.4$

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: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$  

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

$0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281$  

$0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519$  

So on this case the 95% confidence interval would be given by $0.281 \leq \mu_1 -\mu_2 \leq 0.529$