### Analysis of Incinerator Compliance with Environmental Standards#### IntroductionIncinerators can emit hazardous pollutants into the atmosphere. A study was conducted where stack gas samples from 30 incinerators in a major city were analyzed to determine compliance with environmental standards.#### Hypothesis Testing for Compliance**Sample Analysis:** Out of 30 samples, 18 passed the environmental test.**Research Question:** Can we conclude that fewer than half of the incinerators meet the standard?**Hypotheses:**- Null Hypothesis ($H_0$): $p \leq 0.5$- Alternate Hypothesis ($H_1$): $p > 0.5$**P-value & Conclusion:**- **P-value Input:** A box is provided for the user to input the P-value calculation with specified tolerances. - The decision rule is: - (a) Reject $H_0$ - (b) Cannot Reject $H_0$**Significance Level:** 90%#### Additional Analysis**Question:** How many samples must pass to conclude that at least 70% of incinerators meet the standards?**Hypotheses:**- Null Hypothesis ($H_0$): $p \geq 0.7$- Alternate Hypothesis ($H_1$): $p < 0.7$**Determine Pass Threshold:**- An input field is provided for calculating the number of samples ($n_{\text{passed}}$) that need to pass for the conclusion to hold.Users are prompted to test hypotheses and are given multiple-choice options for hypothesis formulations, as well as input boxes to perform calculations and determine outcomes based on the significance level.

Answered: Incinerators can be a source of…

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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Suppose IQ scores were obtained for 20 randomly selected sets of twins. The 20 pairs of measurements yield x=98.34, y=100.45, r=0.940, P-value=0.000, and y=−3.86+1.06x, where x represents the IQ score of the twin born second. Find the best-predicted value of y given that the twin born second has an IQ of 103? Use a significance level of 0.05.
A researcher claims that the incidence of a certain type of cancer is less than 5%. To test this claim, the a random sample of 4000 people are checked and 170 are determined to have the cancer. In this example, the p-value was determined to be 0.015. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%)
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Suppose IQ scores were obtained for 20 randomly selected sets of twins. The 20 pairs of measurements yield x= 99.11, y = 100.1, r=0.872, P-value = 0.000, and y = 7.99 +0.93x, where x represents the IQ score of the twin bom first. Find the best predicted value of y given that the twin born first has an IQ of 96? Use a significance level of 0.05. 1 Critical Values of the Pearson Correlation Coefficient r E Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is. Critical Values of the Pearson Correlation Coefficient r a = 0.01 NOTE: To test Ho: p=0 against H,: p#0, reject H, if the absolute value of r is greater than the critical value in the table. (Round to two decimal places as needed.) a = 0.05 4 0.950 0.990 0.878 0.959 0.811 0.917 0.754 0.875 0.834 0.798 0.765 0.735 0.708 8 0.707 9 0.666 10 0.632 11 0.602 0.576 0.553 12 0.684 13 0.661 0.641 0.623 0.606 14 0.532 0.514 15 16 17 0.497 0.482 0.468 0.456 0.444 0.590 0.575…
The percentage of drivers wearing seat belts in the Limpopo province was 68.5% in 2017 and 72% in 2018. Assume that these percentages are based on random samples of 1000 drivers in 2017 and 1050 in 2018. Suppose you must test at 5% level of significance whether the proportion of all drivers in Limpopo province who used seat belts in 2017 is lower than that for 2018. Determine (i) the alternative hypothesis to be tested: (ii) the value of test statistic: (iii) the p-value and conclusion
In a random sample of males, it was found that 25 write with their left hands and 217 do not. In a random sample of females, it was found that 58 write with their left hands and 450 do not. Use a .05 significance level to test the claim that the rate of left-handedness among males is less than that among females. Identify the test statistic.z=?
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According to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.38. Suppose a random sample of 105 traffic fatalities in a certain region results in 51 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the a = 0.01 level of significance? Because npo (1 - Po 10, the sample size is 5% of the population size, and the sample the requirements for testing the hypothesis satisfied. (Round to one decimal place as needed.) can be reasonably assumed to be random, What are the null and alternative hypotheses? is given to be random, cannot be reasonably assumed to be random, Ho: versus H,: (Type integers or decimals. Do not round.) is given to not be random, Find the test statistic, zo. Z0 = (Round to two decimal places as needed.) Find the P-value.…
A national organization that conducts research on the cost and quality of health care in the U.S. reported that, in 2012, U.S. families spent an average of $9,590 on health care expenses. Suppose you decide to test whetherthe average in 2015 is greater than the average in 2012. After conducting the appropriate statistical test, you find a P-value of 0.022. If the level of significance is 0.05, which of the following is the best interpretation of the P-value? a. The P-value of 0.022 indicates that there is a 2.2% chance that the 2015 average is greater than the average amount spent in 2012.b. The P-value of 0.022 provides weak evidence that the average in 2015 average is greater than the average amount spent in 2012.c. The P-value of 0.022 provides strong evidence that the 2015 average is greater than the average amount spent in 2012.d. The P-value of 0.022 indicates that there is a very low probability that the 2015 average is different than average amount spent in 2012.
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