Amazon.com is testing the use of drones to deliver packages for same-day delivery. In order to quote narrow time windows, the variability in delivery times must be sufficiently small. Consider a sample of 25 drone deliveries with sample variance of s^2=0.77. Amazon.com is testing the use of drones to deliver packages for same-day delivery. In order to quote narrow time windows, the variability in delivery times must be sufficiently small. Consider a sample of 25 drone deliveries with a sample variance of \( s^2 = 0.77 \).a. Construct a 90% confidence interval estimate of the population variance for the drone delivery time (to 2 decimals). Use Table 11.1.\[ \boxed{\ \ \ \ \ } \leq \sigma^2 \leq \boxed{\ \ \ \ \ } \]b. Construct a 90% confidence interval estimate of the population standard deviation (to 2 decimals). Use Table 11.1.\[ \boxed{\ \ \ \ \ } \leq \sigma \leq \boxed{\ \ \ \ \ } \](Note: Instructions reference Table 11.1, which is presumably a chi-square distribution table necessary for constructing the confidence intervals.) **Table 11.1: Selected Values from the Chi-Square Distribution Table**The table provides critical values of the Chi-Square distribution for different degrees of freedom and areas in the upper tail. The Chi-Square distribution is a probability distribution often used in hypothesis testing and data analysis, such as the chi-square test for independence.**Graph Explanation:**- The graph at the top displays a chi-square distribution curve.- The horizontal axis represents the values of \( x^2 \).- The shaded area under the curve, to the right of \( x^2 \), indicates the "Area or probability" in the upper tail.**Table Data:**- **Columns:** - The first column lists the "Degrees of Freedom," ranging from 1 to 100. - The subsequent columns show critical values corresponding to different areas in the upper tail (\(.99\), \(.975\), \(.95\), \(.90\), \(.10\), \(.05\), \(.025\), \(.01\)).- **Rows:** - Each row corresponds to a specific degree of freedom. - The values represent the critical chi-square values for the specified areas in the upper tail.**Example from the Table:**For 5 degrees of freedom:- The chi-square value for an area of \(0.05\) in the upper tail is \(11.070\).- For an area of \(0.01\), the value is \(15.086\).**Note:**- A more extensive table is available online, as mentioned in Table 3 of Appendix B.

Answered: Image /qna-images/answer/24fb89a5-6381-47b5-bd6c-7d6494a6e2c8.jpg

Amazon.com is testing the use of drones to deliver packages for same-day delivery. In order to quote narrow time windows, the variability in delivery times must be sufficiently small. Consider a sample of 25 drone deliveries with sample variance of s² = 0.77. a. Construct a 90% confidence interval estimate of the population variance for the drone delivery time (to 2 decimals). Use Table 11.1. b. Construct a 90% confidence interval estimate of the population standard deviation (to 2 decimals). Use Table 11.1. sos

Amazon.com is testing the use of drones to deliver packages for same-day delivery. In order to quote narrow time windows, the variability in delivery times must be sufficiently small. Consider a sample of 25 drone deliveries with sample variance of s² = 0.77. a. Construct a 90% confidence interval estimate of the population variance for the drone delivery time (to 2 decimals). Use Table 11.1. b. Construct a 90% confidence interval estimate of the population standard deviation (to 2 decimals). Use Table 11.1. sos

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Unfortunately, arsenic occurs naturally in some ground water. A mean arsenic level of µ = 8 parts per billion (ppb) is considered safe for agricultural use. A well in Los Banos is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of - = 7.3 ppb arsenic. It is known that o = information indicate that the mean level of arsenic in this well is less than 8 ppb? Use the classical 1.9ppb for this type of data. Does this approach. Use a = 0.01 What is the criterion (step 3) for this problem? 2.33 or more 1.645 or more -2.33 or less -1.645 or less
In order to evaluate a spectrophotometric method for the determination of titanium concentration, the method was uased on 64 alloy samples containing different certified amounts of titanium. For each alloy, eight different concentrations of titanium were made. Additionally, two spectrophotometer machines were used in the lab to control for possible variation between machines. Each machine received a random allocation of each concentration of alloy. Below you will find an ANOVA analysis for difference in certified mean concentrations detected using machines as a blocking fator. (You will have to calculate some of the values in the tables marked with ). Concentration 1.93 Machine 0.16 Residuals 14.27 Which one of the following is correct given the ANOVA table (using a 5% level of significance)? A. The method was able to detect the difference in the concentrations of titanium alloys; additionally, there was a detectable difference in the mean concentration between machines.…
The Sunny Dale Bank monitors the time required to serve customers at the drive-through window because it is an important quality factor in competing with other banks in the city. After analyzing the data gathered in an extensive study of the window operation, bank management determined that the mean time to process a customer at the peak demand period is 5 minutes, with a standard deviation of 1.5 minutes. Management wants to monitor the mean time to process a customer by periodically using a sample size of six customers. Assume that the process variability is in statistical control. Design an x-chart that has a type I error of 5 percent. That is, set the control limits so that there is a 2.5 percent chance a sample result will fall below the LCL and a 2.5 percent chance that a sample result will fall above the UCL. After several weeks of sampling, two successive samples came in at 3.70 and 3.68 minutes, respectively. Is the customer service process in statistical control?
Organic Grains LLC uses statistical process control toensure that its health-conscious, low-fat, multigrain sandwichloaves have the proper weight. Based on a previously stable andin-control process, the control limits of the x - and R -charts areUCLx = 6.56. LCLx = 5.84, UCLR = 1.141, LCLR = 0. Overthe past few days, they have taken five random samples of fourloaves each and have found the following:
A standard method for the determination of glucose in serum is reported to have a standard deviation of 0.37 mg/dL. If s = 0.37 is a good estimate of o, how many replicate determinations should be made in order for the mean for the analysis of a sample to be within (This problem requires values in your textbook's specific appendices, which you can access through the OWLV2 MindTap Reader. You should not use the OWLV2 References' Tables to answer this question as the values will not match.) Remember that you always need to round your final answer up to a whole number of replicates. a. 0.3 mg/dL of the true mean 99% of the time? b. 0.3 mg/dL of the true mean 95% of the time? c. 0.2 mg/dL of the true mean 90% of the time?
please help asap i will upvote!!
Provide one sample problem of Analysis of Variance (ANOVA)
A peditrician claimed that the average time for a drug to take effect is 15 minutes with standard deviation of eight minutes. In sample of 49 trials the average time was 18 minutes. Test the claim of a pediatrician that the average time is not equal to 15 minutes. Assume the the average time for a drug to take effect is approximately normal distributed
use the date form the second picture to answer question 3 please help me with question 3
Aresearcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 14 blue crabs from Location A contain a mean of 194 milligrams of fish and a standard deviation of 39 milligrams. The stomach contents of a sample of 8 blue crabs from Location B contain a mean of 187 milligrams of fish and a standard deviation of 44 milligrams. Al a 0 10, can you support the researcher's claim? Assume the population variances are equal. Complete parts (a) through (d) below. (a) Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: H-220 O B. Ho H-2 0 (b) Find the standardized test statistic for u - t=0.381 (Round to three decimal places as needed) Enter your answer in the answer box and then click Check Answer parts remaining Clear All Check Answer leste ppt_08_03.pptx Open file Show all P Type here to search 235 PM 2/21/2021 ts fho sk hz home end insert delet esc & @…
In our ANOVA calculation, we want our within-group variation to be higher than the between-group variation. True or False?