Question

In a study of the accuracy of fast food drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90 %

Confidence intervals
ANSWERED
asked 2021-03-09
In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: \(\displaystyle{0.127}{<}{p}{<}{0.191}\)</span>. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___<p<___ ​(Round to three decimal places as​ needed.) (This is used directly in response to B).

Answers (1)

2021-03-10

Step 1 It is given that 51 out of 298 orders are not accurate. The sample proportion is \(\displaystyle\frac{{51}}{{298}}={0.17}\). The sample size is 298. For the confidence interval 90%, the two-tailed z value at 0.05 level of significance is 1.645. The 90% confidence interval for the population proportion is calculated as follows: \(\displaystyle{C}{I}=\hat{{{p}}}\pm{z}_{{\frac{\alpha}{{2}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(\displaystyle={0.171}\pm{\left({1.645}\right)}\sqrt{{{\frac{{{0.17}{\left({1}-{0.17}\right)}}}{{{298}}}}}}\)
\(\displaystyle={0.171}\pm{0.036}\)
\(\displaystyle={\left({0.135},{0.207}\right)}\) Thus, the percentage of orders that is not accurate of restaurant A lies between 0.135 and 0.207. Step 2 The percentage of orders that are not accurate of restaurant B lies between 0.127 and 0.191. From the intervals it can be observed that neither restaurant appears to have a significant different percentage of orders that are not accurate because these two confidence intervals are overlapped.

0
 
Best answer

expert advice

Need a better answer?

Relevant Questions

asked 2021-05-14
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
\(\begin{array}{|c|c|}\hline 11.8 & 7.7 & 6.5 & 6 .8& 9.7 & 6.8 & 7.3 \\ \hline 7.9 & 9.7 & 8.7 & 8.1 & 8.5 & 6.3 & 7.0 \\ \hline 7.3 & 7.4 & 5.3 & 9.0 & 8.1 & 11.3 & 6.3 \\ \hline 7.2 & 7.7 & 7.8 & 11.6 & 10.7 & 7.0 \\ \hline \end{array}\)
a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. \([Hint.\ ?x_{j}=219.5.]\) (Round your answer to three decimal places.)
MPa
State which estimator you used.
\(x\)
\(p?\)
\(\frac{s}{x}\)
\(s\)
\(\tilde{\chi}\)
b) Calculate a point estimate of the strength value that separates the weakest \(50\%\) of all such beams from the strongest \(50\%\).
MPa
State which estimator you used.
\(s\)
\(x\)
\(p?\)
\(\tilde{\chi}\)
\(\frac{s}{x}\)
c) Calculate a point estimate of the population standard deviation ?. \([Hint:\ ?x_{i}2 = 1859.53.]\) (Round your answer to three decimal places.)
MPa
Interpret this point estimate.
This estimate describes the linearity of the data.
This estimate describes the bias of the data.
This estimate describes the spread of the data.
This estimate describes the center of the data.
Which estimator did you use?
\(\tilde{\chi}\)
\(x\)
\(s\)
\(\frac{s}{x}\)
\(p?\)
d) Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa. [Hint: Think of an observation as a "success" if it exceeds 10.] (Round your answer to three decimal places.)
e) Calculate a point estimate of the population coefficient of variation \(\frac{?}{?}\). (Round your answer to four decimal places.)
State which estimator you used.
\(p?\)
\(\tilde{\chi}\)
\(s\)
\(\frac{s}{x}\)
\(x\)
asked 2021-05-14
When σ is unknown and the sample size is \(\displaystyle{n}\geq{30}\), there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(\displaystyle{n}\geq{30}\), use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
asked 2021-06-02
Let p be the population proportion for the situation. (a) Find point estimates of p and q, (b) construct 90% and 95% confidence intervals for p, and (c) interpret the results of part (b) and compare the widths of the confidence intervals. In a survey of 2223 U.S. adults, 1334 say an occupation as an athlete is prestigious.
asked 2021-06-16
Let p be the population proportion for the situation. (a) Find point estimates of p and q, (b) construct 90% and 95% confidence intervals for p, and (c) interpret the results of part (b) and compare the widths of the confidence intervals. In a survey of 2202 U.S. adults, 1167 think antibiotics are effective against viral infections.
...