Question

Batteries A company that produces AA batteries tests the lifetime of a randomsample of 30 batteries using a special device designed to imitate real-world use. Based onthe 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%.”36

a. Determine the point estimate, margin of error, standard error, and sample standard

deviation.

b. A reporter translates the statistical announcement into “plain English” as follows: “95% of this company’s AA batteries last between 430 and 470 minutes.” Comment on this interpretation.

c. Your friend, who has just started studying statistics, claims that there is a 95% probability that the mean lifetime will fall between 430 and 470 minutes. Do you agree? Explain your reasoning.

d. Give a statistically correct interpretation of the confidence level that could be published in a newspaper report.

Short answer

Part a) $\overline{x}=450,E=20,SE=11.3095,s=61.9447$

Part b) The interpretation is incorrect

Part c) Not agree with the claim

Part d) we are $95\%$confident that the true population means is between $430$and $470$minutes.

Step-by-step solution

Part a) Step 1: Given information

The $95\%$confidence interval for the mean: $(430,470)$

Sample size (n)$=30$

Part a) Step 2: Explanation

We know, the formula of

Sample mean: $\overline{x}=\frac{\text{Lower limit}+U_{\text{pper limit}}}{2}$

The margin of error: $E=\frac{U_{pper}\text{limit-Lower limit}}{2}$

Standard error: $SE=\frac{s}{\sqrt{n}}$

Using the confidence interval, the Sample mean is,

$\overline{x}=\frac{430+470}{2}=450$

And the margin of error is,

$E=\frac{470-430}{2}=20$

First, find the critical-$t$-value:

$t_{\alpha /2}=2.042..\text{Using excel formula,}=TINV(0.05,29)$

The margin of error formula can be used to calculate the standard deviation.

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}} \\ \Rightarrow 20=2.042\times \frac{s}{\sqrt{30}} \\ \Rightarrow s=\frac{20}{0.3229} \\ \Rightarrow s=61.9447 \end{gathered}$$

standard deviation $=61.9447$

So, the standard error is$SE=\frac{61.9447}{\sqrt{30}}=11.3095$

Part b) Step 1: Given information

Comment from a reporter: $95\%$of this company's AA batteries last between $430$and $470$minutes.

Part b) Step 2: Explanation

We can't be certain about the confidence interval because the battery will either last between $430$and $470$minutes or it won't. The correct interpretation is that we are $95\%$certain that the average battery life is between $430$and$470$minutes.

Part c) Step 1: Given information

Claim: There are a $95\%$chance that the average life will be between $430$and $470$minutes.

Part c) step 2: Explanation

We can't be certain about the confidence interval because the battery will either last between $430$and $470$minutes or it won't. If the sample means were very unlikely to occur by chance for a given population mean, the probability of a different sample means being in the same confidence interval is less than $95\%$ As a result, I disagree with the claim

Part d) Step 1: Given information

The $95\%$confidence interval for the mean: $(430,470)$

Part d) Step 2: Explanation

We are $95\%$certain that the true population means is between $430$and$470$ minutes.