### Comparing Average Annual Miles Driven Per Vehicle in Chicago to the National AverageThe average annual miles driven per vehicle in the United States is 11.1 thousand miles, with a standard deviation (σ) of approximately 600 miles. Suppose that a random sample of 41 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.9 thousand miles. Does this indicate that the average miles driven per vehicle in Chicago is different from (higher or lower than) the national average? We will use a 0.05 level of significance to determine this.**What are we testing in this problem?**- ○ single proportion- ● single mean#### (a) What is the level of significance?[Input Box]#### State the null and alternate hypotheses.- ○ \( H_0: p = 11.1; H_1: p < 11.1 \)- ○ \( H_0: \mu = 11.1; H_1: \mu > 11.1 \)- ○ \( H_0: p = 11.1; H_1: p > 11.1 \)- ○ \( H_0: \mu = 11.1; H_1: \mu \neq 11.1 \)- ○ \( H_0: p = 11.1; H_1: p \neq 11.1 \)- ● \( H_0: \mu = 11.1; H_1: \mu < 11.1 \)#### (b) What sampling distribution will you use? What assumptions are you making?- ○ The Student's t, since we assume that \( x \) has a normal distribution with known \( \sigma \).- ● The standard normal, since we assume that \( x \) has a normal distribution with known \( \sigma \).- ○ The Student's t, since we assume that \( x \) has a normal distribution with unknown \( \sigma \).- ○ The standard normal, since we assume that \( x \) has a normal distribution with unknown \( \sigma \).#### What is the value of the sample test statistic? (Round your answer to two decimal places.)[Input Box]#### (c) Find the p-value. (Round your answer to four decimal places.)[Input Box] **Topic: Understanding Sampling Distributions and P-Value Areas**### Graphical Representation of P-Value AreasIn statistical analysis, it's crucial to understand the concept of the sampling distribution and how the P-value is depicted within that distribution. Below, we illustrate this concept through graphical representations.#### Graph 1This graph shows a standard normal distribution curve with shaded areas on the tails, representing the P-value area. The curve peaks at 0, and the shaded areas lie symmetrically on both sides, beyond approximately the 1.96 mark, denoted as ± 1.96 on the x-axis. This implies a two-tailed test at a significance level where outliers are highlighted.#### Graph 2The second graph demonstrates a standard normal distribution with a single shaded area starting from 0 extending towards the right. This represents a one-tailed test where the region of interest is on the right side of the distribution, looking for significantly higher values beyond 1.#### Graph 3The third graph presents a standard normal distribution curve with the shaded area extending from the lower tail (left side) all the way to the peak at 0. This signifies a test for smaller-than-expected results, highlighting just the left tail of the distribution.#### Graph 4The fourth graph depicts a standard normal distribution with the shaded area covering from the peak (0) to the left tail. This suggests a one-tailed test focusing on significantly smaller values or detecting lower trends within the data.### Inferential Details#### Hypothesis Test Decision Analysis**(d) Decision Based on Significance Level**Based on your observations from parts (a) to (c), decide whether to reject or fail to reject the null hypothesis at the α = 0.05 level of significance:- **Option 1**: At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.- **Option 2**: At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.- **Option 3**: At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.- **Option 4**: At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.#### Application Context Interpretation**(e) Context

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Description: ### Comparing Average Annual Miles Driven Per Vehicle in Chicago to the National AverageThe average annual miles driven per vehicle in the United States is 11.1 thousand miles, with a standard deviation (σ) of approximately 600 miles. Suppose that a

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The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with o = 600 miles. Suppose that a random sample of 41 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.9 thousand miles. Does this indicate that the average miles driven per vehicle in Chicago is different from (higher or lower than) the national average? Us a 0.05 level of significance. What are we testing in this problem? single proportion single mean (a) What is the level of significance? State the null and alternate hypotheses. Но: р 11.1; H1: р < 11.1 Но: и 11.1; H1: μ > 11.1 Но: р 3D 11.1; H1: р > 11.1 Но: и 11.1; H1: µ + 11.1 %3D Ho: p = 11.1; Hi: р # 11.1 Но: и %3D 11.1; Hi: и < 11.1 O O O O O

The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with o = 600 miles. Suppose that a random sample of 41 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.9 thousand miles. Does this indicate that the average miles driven per vehicle in Chicago is different from (higher or lower than) the national average? Us a 0.05 level of significance. What are we testing in this problem? single proportion single mean (a) What is the level of significance? State the null and alternate hypotheses. Но: р 11.1; H1: р < 11.1 Но: и 11.1; H1: μ > 11.1 Но: р 3D 11.1; H1: р > 11.1 Но: и 11.1; H1: µ + 11.1 %3D Ho: p = 11.1; Hi: р # 11.1 Но: и %3D 11.1; Hi: и < 11.1 O O O O O

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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