**Estimate a $90\%$ confidence interval of the percentage of orders that are not accurate.****Restaurant $B$ has the confidence interval $0.127<p<0.191$ with $90\%$ confidence level, compare it with results from part(a).****Conclude your results from both restaurants.**

The aim of this question is to study college-level **statistics** concepts of incorporating **confidence levels** into the **mean** and **deviation** estimates for robust business statements and **decision-making.**

The **confidence intervals** are a very crucial and integral part of basic **statistics.** Most of the market research builds its foundation on this fundamental concept. These **intervals** estimate the estimated value out of a **sample distribution** with some associate level of **confidence.** The relation between **confidence intervals** and the **confidence levels** (defined as a percentage) is drawn from experience and is available in tabulated form.

The use of **confidence levels** and **confidence intervals** helps us analytically approximate or estimate the **mean and standard deviation** from the given **sample distribution.**

## Expert Answer

**Part (a):**

The following steps will be used to find the **confidence interval:**

**Step 1**: Find the sample proportion $p$ of **non-accurate orders** $x$ to the total number of **accurate orders** $n$ from the given data.

\[ p = \dfrac{\text{number of non-accurate orders}}{\text{number of accurate orders}} \]

\[ p = \dfrac{x}{n} = \dfrac{51}{298} \]

\[ p = 0.17114 \]

**Step 2:** Find the **z-value** against the given **confidence level** from the following table:

As the confidence level for this problem is $90\%$, the **z-value** from the Table $1$ is given as:

\[ z = 1.645 \]

**Step 3**: Find the **confidence interval** by using the following formula:

\[ \text{Confidence Interval} = p \pm z \cdot \sqrt{\frac{p(1-p)}{n}} \]

By substituting the values, we get:

\[\text{Confidence Interval } = 0.17114 \pm (1.645) \cdot \sqrt{\frac{(0.17114) (1-0.17114)}{298}}\]

\[\text{Confidence Interval } = 0.17114 \pm 0.03589\]

The calculated values show that we can say with $90\%$ confidence that the **percentage** of **non-accurate orders** lies in the interval $0.135\ to\ 0.207$.

**Part (b):**

For **restaurant** $A$:

\[0.135 < p < 0.207\]

For **restaurant** $B$:

\[0.127 < p < 0.191\]

It can clearly be seen that the two **confidence intervals** are **overlapping**, as shown in Figure 1 below.

**Part (c):**

Since both the **confidence intervals** are **overlapping,** we can conclude that both restaurants have a **similar range** of **non-accurate orders. **

## Numerical Results

The **confidence interval** of Restaurant $A$ lies in the interval of $0.135-0.207$. The **confidence intervals** of both **Restaurant** $A$ and $B$ have a similar range of **non-accurate orders.**

## Example

Find the **confidence interval** of a food chain restaurant feedback with a** sample proportion $p=0.1323$** and a **confidence level** of $95\%$. The number of **positive feedback** $n=325$ and **negative feedback** $x=43$.

We can find the **z-value** from Table 1 as the **confidence level** is $95\%$.

\[ z = 1.96 \]

We can find the confidence interval using the formula given as:

\[ \text{Confidence Interval} = p \pm z \cdot \sqrt{\frac{p(1-p)}{n}} \]

Substituting the values, we get:

\[ \text{Confidence Interval} = 0.1323 \pm (1.96) \cdot \sqrt{\frac{0.1323(1 – 0.1323)}{325}} \]

\[ \text{Confidence Interval} = 0.1323 \pm 0.0368 \]

The **confidence interval** for the **restaurant’s feedback** is calculated to be $0.0955<p<0.1691$.

*Images/Mathematical drawings are created with Geogebra.*