Question

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

Brand X: Brand Y:
X1=9.2 volts, X2=8.8 volts,
σ1=0.3, σ2=0.1
n1=27, n2=30.
Find the 95% confidence interval of the true difference in the means of their voltage. Assume that both variables are normally distributed.

Answer 1

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

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{(\sigma^2_1)/(n_1)+(\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{(0.3^2)/(27)+(0.1^2)/(30)}=0.281`

`0.4+1.96\sqrt{(6^2)/(36)+(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`