Question

R8.3. Batteries A company that produces AA batteries tests the lifetime of a random sample of 40 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes,
and our confidence in that interval is 95%.
(a) Determine the sample mean and standard deviation.
(b) A reporter translates the statistical announcement into “plain English” as follows: “If you buy one of this company’s AA batteries, there is a 95% chance that it will last between 430 and 470 minutes.” Comment on this interpretation.
(c) Your friend, who has just started studying statistics, claims that if you select 40 more AA batteries at random from those manufactured by this company, there is a 95% probability that the mean life time will fall between 430 and 470 minutes. Do you agree? Explain.
(d) Give a statistically correct interpretation of the confidence interval that could be published in a newspaper report.

Short answer

(a) The mean is $450$and standard deviation is $61.9447$.

(b) The interpretation is incorrect application.

(c) No, a confidence interval provides a statement about an unknown population mean, not another sample mean.

(d) A $95\%$confident that the population mean is between $430$and $470$ minutes.

Step-by-step solution

Part (a) Step 1: Given information

To determine the sample mean and standard deviation.

Part (a) Step 2: Explanation

Since,$n=40$.

And the $95\%$confidence interval is$430$to $470$minutes.

Determine the sample mean falls in the middle of the confidence interval as:
$\overline{x}=\frac{430+470}{2}$
$=450$

Determine the margin of error as follows:

Margin of error$=450-430$

$=20$

Then, using table B and the degree of freedom as follows, get the critical value as:

$df=40-1$

$=39>30$

$t_{\alpha /2}=t_{0.025}$

$=2.042$

Part (a) Step 3: Explanation

A confidence interval's margin of error for the population mean as:

Margin of error $=t_{\alpha /2}\times \frac{s}{\sqrt{n}}$

$=2.042\times \frac{s}{\sqrt{40}}$

$=0.3229s$

Finally, from the above:

$20=0.3229s$

$\Rightarrow s=\frac{20}{0.3229}$

$=61.9447$

As a result, the mean is $450$ and standard deviation is $61.9447$.

Part (b) Step 1: Given information

A $95\%$chance that it will last between $430$and $470$ minutes. To comment on this interpretation.

Part (b) Step 2: Explanation

Since the battery will either last between $430$ and $470$ minutes or it will not, the likelihood is either $0$or $1$.
A more accurate translation would be: We are $95\%$ confident that the average battery life time is between $430$ and $470$ minutes.
As a result, this is incorrect application.

Part (c) Step 1: Given information

To explain that agree with there is a $95\%$probability that the mean life time will fall between $430$and $470$ minutes.

Part (c) Step 2: Explanation

No, since if the sample mean (on which the confidence interval is based) is extremely improbable to occur by chance for a given population mean, the likelihood of an alternative sample mean being in the same confidence interval is less than $95$percent.

As a result, no, a confidence interval provides a statement about an unknown population mean, not another sample mean.

Part (d) Step 1: Given information

To find that statistically correct interpretation of the confidence interval that could be published in a newspaper report.

Part (d) Step 2: Explanation

Let, $n=40$.

Then, $95\%$confidence interval is $430$to $470$minutes
The standard deviation is $61.9447$, while the mean is $450$.
Hence, a statistically correct interpretation of the confidence interval that might be stated in a newspaper story might be that the population mean is between $430$and $470$ minutes.