Solve the problem.9) The table shows the number of households burglarized in a sample of households with dogs and ina sample of households without dogs. Assume that you plan to use a significance level of a = 0.01to test the claim that p1 < p2. Find the critical value(s) for this hypothesis test. Do the data supportthe claim that a smaller proportion of households with pet dogs are burglarized?9)Household with Dog Household without DogNumber of households in sampleNumber of households burglarized2061402011A) z = -2.575; noB) z = -2.33; noC) z = 2.33; yesD) z = -1.96; yes

Answered: Image /qna-images/answer/cb3963b5-fa47-4e5a-a8a0-5afcc84792be.jpg

Answered: Solve the problem. 9) The table shows…

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 mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey &8 employees. The number of sick days they took for the past year are as follows: 12; 6; 15; 3; 10; 8; 8; 8. Let X= the number of sick days they took for the past year. Should the personnel team believe that the mean number is about 10? Conduct a hypothesis test at the 5% level. What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.) What is the p-value? (Round your answer to four decimal places.) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to three decimal places.)
8
In a certain population, the mean price of the stock prices was 230, the median was 255, the variance was 1225, the first quartile was 240. and the third quartile was 270. Find the second quartile ? What percentage of data values is less than or equal 270? OApproximately 68%. OApproximately 50% OApproximately 75% OApproximately 25% OApproximately 95% OApproximately 100% What is the coefficient of variation? 018.78% 0532.61% 015.22% 0657.14% O14.89% Determine whether the data value 160 is an outlier or not ? OYes. ONo. OCan't conclude.
A survey claims that 9 out of 10 doctors recommend brand Wfor their patients who have children. To test this claim against the alternative that the actual proportion of doctors who recommend brand Wis less than 90%, a random sample of 100 doctors results in 85 who indicate that they recommend brand W. 1. What decision should you makeusing a significance level of 5%? 2. Summarize your conclusions in the context of the problem.
The Bureau of Labor Statistics reports that the official unemployment rate for Black people was 10.4% and 4.7% for White people in February 2015. Select all correct answers for this question. O The samples of white and black people are independent. The explanatory variable is the unemployment rate. The response variable is the unemployment rate. The response variable is race.
The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 331 homes in Oregon found that 132 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) H0: H1: Based on the hypotheses, compute the following:Test Statistic = p-value = Based on the above we choose to The correct summary would be: that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.
D? please and thank you
= Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 43 of the 49 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 92 of the 109 subjects developed rhinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test? OA. Ho: P₁ P2 H₁: P₁ = P2 O D. Ho: P₁ P2 H₁: P₁ P2 Identify the test statistic. Z= (Round to two decimal places as needed.) Identify the P-value. OB. Ho: P₁ P2 H₁: P₁ P2 ? O E. Ho: P₁ P2 H₁: P₁ P2 P-value= (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is the significance level of a =…
Express the alternative hypothesis in symbolic form. A psychologist claims that the mean amount of time (in hours) that females spend in talk therapy is more than that of men. Assume that two samples are independent. Let the female talk therapy times be the first population and the male talk therapy times be the second population.
Tutorial 1 Identify the population and the sample. 1) A surve y of 1378 Malaysian households found that 27% of the households own a computer. 2) W hen 1094 Malaysian households were surveyed; it was found that 67% of them owned two cars. 3) A survey of 2625 elementary school children found that 28% of the children could be classified as obese. Determine whether the numerical value is a parameter or a statistic. Explain your reasoning. 4) A recent sur vey by the alumni of a major university indicated that the average salary of 10,500 of its 175,000 graduates was RM95,000. 5) The average salary of all assembly - line employees at a certain car manufacturer is RM41,000.. 6) A survey of 1162 students was taken from a university with 10,000 students. Provide an appropriate response. 7) Explain the difference between a sample and a population. Determine whether the data are qualitative or quantitative. 8) the colors of automobiles on a used car lot 9) the number of complaint letters received…
The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 317 homes in Oregon found that 166 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 5% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) Ho: P Test Statistic = 0° p-value = H₁: p Based on the hypotheses, compute the following: #v oº 48 48 <-Need answer, please. <-Need answer, please OB 0° Based on the above we choose to Fail to reject the null hypothesis ✓ OB The correct summary would be: There is not enough evidence to support the claim that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.

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