In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 non-accurate ones.

in a study of the accuracy of fast food

  • 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:

confidence level z table

Table 1

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.

Comparison of A and B Restaurants

Figure 1

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.


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.

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