Chapter PVI, Problem 27RE

(a)

To determine

To explain can you use these data to estimate the number of loaves sold on the busiest $10\%$ of all days.

Answer to Problem 27RE

No, we cannotuse these data to estimate the number of loaves sold on the busiest $10\%$ of all days.

Explanation of Solution

It is given in the question that Clarksburg Bakery is trying to predict how many loaves of bread to bake. It is also given,

  $\begin{array}{l}\text{Mean}=103\\ \text{Median}=100\\ \text{SD}=9\\ \text{Min}=95\\ \text{Max}=140\\ {\text{Q}}_1=97\\ {\text{Q}}_2=105.5\end{array}$

The histogram is given for the data and from this histogram, we note that $115$ loaves or more are not normal and this assume the last $100$ days are typical. No, we cannot use these data to estimate the number of loaves sold on the busiest $10\%$ of all days.

(b)

To determine

To explain why you can use these data to construct a $95\%$ confidence interval for the mean number of loaves sold per day.

Explanation of Solution

It is given in the question that Clarksburg Bakery is trying to predict how many loaves of bread to bake. It is also given,

  $\begin{array}{l}\text{Mean}=103\\ \text{Median}=100\\ \text{SD}=9\\ \text{Min}=95\\ \text{Max}=140\\ {\text{Q}}_1=97\\ {\text{Q}}_2=105.5\end{array}$

Thus, we can use these data to construct a $95\%$ confidence interval for the mean number of loaves sold per day because the sample size is large and the central limit theorem says that sample mean will be approximately normal.

(c)

To determine

To calculate a $95\%$ confidence interval and carefully interpret what that confidence interval means.

Answer to Problem 27RE

We are $95\%$ confident that on a average the baker will sell between $101.2\text{ and 104}\text{.8}$ loaves of bread a day.

Explanation of Solution

It is given in the question that Clarksburg Bakery is trying to predict how many loaves of bread to bake. It is also given,

  $\begin{array}{l}\text{Mean}=103\\ \text{Median}=100\\ \text{SD}=9\\ \text{Min}=95\\ \text{Max}=140\\ {\text{Q}}_1=97\\ {\text{Q}}_2=105.5\end{array}$

Thus, $\alpha =0.05$ so by using the calculator $\text{TI}-89$ , the confidence interval is calculated as:

  $(101.2,104.8)$ .

So, we are $95\%$ confident that on a average the baker will sell between $101.2\text{ and 104}\text{.8}$ loaves of bread a day.

(d)

To determine

To find out how many days’ data could they have used.

Answer to Problem 27RE

  $25$ days data they could have used.

Explanation of Solution

It is given in the question that Clarksburg Bakery is trying to predict how many loaves of bread to bake. It is also given,

  $\begin{array}{l}\text{Mean}=103\\ \text{Median}=100\\ \text{SD}=9\\ \text{Min}=95\\ \text{Max}=140\\ {\text{Q}}_1=97\\ {\text{Q}}_2=105.5\end{array}$

Now, if the bakery would have been satisfied with a confidence interval whose margin of error was twice as wide then, the number of days data they could use is calculated as:

So, from above we get the margin of error $1.79$ . For this we have,

  $\begin{array}{l}ME=z*\times \frac{\sigma }{\sqrt{n}}\\ \Rightarrow 2\times 1.79=1.96\times \frac{9}{\sqrt{n}}\\ \Rightarrow \frac{3.58}{1.96}=\frac{9}{\sqrt{n}}\\ \Rightarrow \sqrt{n}=\frac{9\times 1.96}{3.58}\\ \Rightarrow n=25\end{array}$

Thus, $25$ daysdata they could have used.

(e)

To determine

To explain does your confidence interval provide strong evidence that this estimate was incorrect.

Answer to Problem 27RE

Yes, our confidence interval provide strong evidence that this estimate was incorrect.

Explanation of Solution

It is given in the question that Clarksburg Bakery is trying to predict how many loaves of bread to bake. It is also given,

  $\begin{array}{l}\text{Mean}=103\\ \text{Median}=100\\ \text{SD}=9\\ \text{Min}=95\\ \text{Max}=140\\ {\text{Q}}_1=97\\ {\text{Q}}_2=105.5\end{array}$

And the confidence interval is calculated as: $(101.2,104.8)$ . Since $100$ loaves per day is too low and the entire confidence interval is above that so, yes we can say that our confidence interval provide strong evidence that this estimate was incorrect.