Interpret the confidence interval. Select the correct choice below and fill in the answer boxes to complete your choice.Can you assist me on how the assignment chose 1.35 as the answer for the question? I was able to calculate the confidence interval for part A, but don't understand the second part. Thank you so much in advance. **Transcription for Educational Website:**---**Problem 11.2.69**One year at a university, the algebra course director decided to experiment with a new teaching method that might reduce variability in final-exam scores by eliminating lower scores. The director randomly divided the algebra students who were registered for class at 9:40 A.M. into two groups. One of the groups, called the control group, was taught the usual algebra course; the other group, called the experimental group, was taught by the new teaching method. Both classes covered the same material, took the same unit quizzes, and took the same final exam at the same time. The final-exam scores (out of 40 possible) for the two groups are shown in the accompanying tables. Find a 90% confidence interval for the ratio of the population standard deviations of final-exam scores for students taught by the conventional method and for students taught by the new method. Assume that both populations are normally distributed. (Note: s₁ = 7.637, s₂ = 7.240, and for df = (19, 19), F₀.₀₅ = 1.85.)- Click here to view the data table.- Click here to view page 1 of the F-distribution.- Click here to view page 2 of the F-distribution.- Click here to view page 3 of the F-distribution.- Click here to view page 4 of the F-distribution.---The 90% confidence interval for \(\frac{\sigma_1}{\sigma_2}\) is \(0.74\) to \(1.44\).**(Round to two decimal places as needed.)****Interpret the confidence interval. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to two decimal places as needed.)**- A. We can be 90% confident that the population standard deviation final-exam score for students taught by the conventional method is somewhere between \[ \] and \[ \] times greater than for those taught by the new method.- B. We can be 90% confident that the population standard deviation final-exam score for students taught by the conventional method is somewhere between 1.35 times less than and 1.44 times greater than for those taught by the new method.- C. We can be 90% confident that ### Data Table#### OverviewThis table displays collected data from a control group and an experimental group. Both groups have multiple readings organized in columns, which may represent steps in a study or results from different observations.#### Control Group- **Readings**: - Row 1: 38, 38, 37, 36 - Row 2: 36, 35, 34, 33 - Row 3: 28, 27, 27, 27 - Row 4: 25, 25, 25, 25 - Row 5: 24, 23, 22, 22 - Row 6: 18, 18, 18, 17 - Row 7: 15, 15, 14, 14, 13#### Experimental Group- **Readings**: - Row 1: 39, 37, 36, 36 - Row 2: 35, 35, 33, 33 - Row 3: 32, 28, 26, 25 - Row 4: 25, 24, 24, 23 - Row 5: 19, 18, 17, 17#### AnalysisThe data suggests comparative observations which may indicate differences between the control and experimental conditions. The readings provide a basis for statistical analysis to determine if there are significant changes or patterns according to the variables being tested.#### Graphs/DiagramsThis section includes no visual graphs or diagrams. Data interpretation relies on numerical comparison and statistical analysis.

From the given information, The confidence interval σ1/σ2 is 0.74 and 1.44 There is a 90% chance that the ratio of two standard deviations is lies between 0.74 and 1.44. Thus, the correct option is, We can be 90% confident that population standard deviation final-exam score for students taught by the conventional method is somewhere between 0.74 times less than and 1.44 times greater than for those taught by the new method.