Chapter PVI, Problem 35RE

(a)

To determine

To explain to him the problem with doing that.

Explanation of Solution

It is given in the question that they have been asked to look at a battery company that claims its batteries last an average of $100$ hours under normal use. One of the editor of the newsletter says that $97$ hours is a lot less than the advertised $100$ hours so we should reject the company’s claim. Thus, the problem in doing that will be tat different samples give different means and this is fairly small sample. The difference may be due to natural sampling variation.

(b)

To determine

To find out what are the null and alternative hypotheses.

Answer to Problem 35RE

  $\begin{array}{l}{H}_0:\mu =100\\ H_a:\mu <100\end{array}$

Explanation of Solution

It is given in the question that they have been asked to look at a battery company that claims its batteries last an average of $100$ hours under normal use. The given claim is that the mean is less than $100$ . The null hypothesis states that the average mean is equal to $100$ . So, the null hypothesis support the claim then the alternative hypothesis states opposite to the null hypothesis. So, we have,

  $\begin{array}{l}{H}_0:\mu =100\\ H_a:\mu <100\end{array}$

(c)

To determine

To explain what assumptions must we make in order to proceed with inference.

Answer to Problem 35RE

All the assumptions are satisfied.

Explanation of Solution

It is given in the question that they have been asked to look at a battery company that claims its batteries last an average of $100$ hours under normal use. So, we have,

  $\begin{array}{l}{H}_0:\mu =100\\ H_a:\mu <100\end{array}$

Let us check the assumptions in order to proceed with inference as,

Random condition: It is satisfied as batteries selected are a random sample.

  $10\%$ condition: It is satisfied as the sample is less than $10\%$ of all the company’s batteries.

Normal condition: It is satisfied as the lifetimes are approximately normal in nature.

So, all the assumptions are satisfied.

(d)

To determine

To explain what do you conclude.

Answer to Problem 35RE

Explanation of Solution

It is given in the question that they have been asked to look at a battery company that claims its batteries last an average of $100$ hours under normal use. So, we have,

  $\begin{array}{l}{H}_0:\mu =100\\ H_a:\mu <100\end{array}$

Thus, to test the hypothesis for mean we will use the calculator $\text{TI}-89$ . In the STAT Tests menu, choose $2:$ T -Test. You must specify whether you are using data stored in a list or whether you will enter the mean, standard deviation, and size of your sample. You must also specify the hypothesized model mean and whether the test is to be two-tail, lower-tail, or upper-tail. Select whether the test is to be simply computed or whether to display the distribution curve and highlight the area corresponding to the P-value of the test.Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}P=0.1666\\ t=-1\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that this sample does not show that the average life of the batteries is significantly less than $100$ hours.

(e)

To determine

To explain has an error been made in part (d) and if so find out what kind.

Answer to Problem 35RE

Type II error is been made.

Explanation of Solution

It is given in the question that they have been asked to look at a battery company that claims its batteries last an average of $100$ hours under normal use. So, we have,

  $\begin{array}{l}{H}_0:\mu =100\\ H_a:\mu <100\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that this sample does not show that the average life of the batteries is significantly less than $100$ hours. Thus, as we have fail to reject the null hypothesis and as it is false so the type of error conducted is Type II error in this.