### Statistical Analysis of Car Rental MileageA consumer advocacy group is conducting a large study on car rental practices. They aim to perform a statistical test on the mean monthly mileage ($\mu$) of cars rented in the U.S. this year. The consumer group suspects that the mean monthly mileage this year is less than last year's mean of 2850 miles.They plan to perform the test by selecting a random sample of monthly mileages. The mean of this sample is calculated to be 2715 miles, with a standard deviation of 750 miles.**Based on this information, complete the following:**#### (a) Hypotheses Formulation1. **Null Hypothesis ($H_0$):** - The mean monthly mileage of cars rented this year is equal to last year’s mean.2. **Alternative Hypothesis ($H_1$):** - The mean monthly mileage of cars rented this year is less than last year’s mean.#### (b) Type I Error- If the group decides not to reject the null hypothesis, they risk a Type I error. A Type I error occurs when a true null hypothesis is incorrectly rejected.#### (c) Type II Error- Suppose the true mean monthly mileage of cars rented this year is 2700 miles. Fill in the blanks to describe a Type II error: - A Type II error would be failing to reject the hypothesis that $\mu$ is less than 2850 when, in fact, $\mu$ is 2700. This exercise provides a clear understanding of hypothesis testing, potential errors, and the importance of statistical analysis in real-world applications.

Here, the claim is the mean mileage (µ) is less than 2850 miles. So, the alternative hypothesis will…

Answered: A consumer advocacy group is doing a…

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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The site traffic for a business's website was monitored for 10 days, showing daily traffic of 14,782; 14,801; 15,264; 15,907; 14,794; 15,079; 15,439; 15,464; 15,580; 14,774. Calculate the mean of the dataset.
According to a report published by the USDA 4 years ago, a typical American consumes an average of 6.66 pounds of potato chips per year.A researcher at the USDA would like to determine if the average amount of potato chips consumed each year by a typical American has changed since the original report was published. The researcher collected data from a random sample of 31 Americans and found that the mean amount of potato chips consumed each year by the sample was 6.37 pounds with a standard deviation of 0.603 pounds.Using α=0.05α=0.05, test the hypothesis that the mean amount of potato chips consumed each year by a typical American is different than 6.66 pounds. Use the p-value method. Determine the test statistic for the hypothesis test. Round the solution to four decimal places.Determine the p-value (range) for the hypothesis test. p-value < 0.001 0.001 < p-value < 0.01 0.01 < p-value < 0.02 0.02 < p-value < 0.05 0.05 < p-value < 0.10 0.10 < p-value…
A particular fruit’s weights are normally distributed, with a mean of 700 grams and a stand deviation of 18 grams. The heaviest 6% of fruits weigh more than how many grams?
What level of measurement do the variance and standard deviation assume?
When sandy made landfall in October of 2012, it became the second most costly hurricane in the US. As of march 2014, the estimated cost was over $68 billion. much of the damage was caused by flooding. the rainfall in areas affected by sandy follow a normal distribution. the average rainfall during the storm was 7.3 inches with a standard deviation of 2.44 inches (www.accuweather.com). Areas in Maryland and New jersey saw more than the other affected regions. a) what is the probability that a randomly selected region has less than 5 inches of rain? b) what is the probability that a randomly selected region had between 10 and 12 inches of rain? c) what is the probability that a randomly selected region had more than 20 inches of rain?
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Most water treatment facilities monitor the quality of their drinking water on an hourly basis. One variable measured is pH, which measures the degree of alkalinity or acidity in the water. A pH below 7.0 is acidic, one above 7.0 is alkaline, and a pH of 7.0 is neutral. One water-treatment plant has a target pH of 8.5. (Most try to maintain a slightly alkaline level) The mean and standard deviation of 1 hour’s test results based on 49 water samples at this plant are: = 8.42; s = 0.16 Does this sample provide sufficient evidence that the mean pH level in the water differs from 8.5? do in excel if needed

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