A manufacturing company packages shipments in either large or small boxes. A random sample of 40 shipmentsthat are packaged in large boxes is found to have a mean of 15 pounds and a standard deviation of 2.8 pounds. Aseparate random sample of 50 shipments that are packaged in small boxes is found to have a mean of 10 poundsand a standard deviation of 1.5 pounds. The manager would like to know if the data provide convincing evidencethat the true mean weight of all shipments that are packaged in small boxes is less than the true mean weight of allshipments that are packaged in large boxes. The manager tests Ho: Hs – = 0, Hạ: Hs – M < 0, where u = the truemean weight of all shipments that are packaged in large boxes and Us = the true mean weight of all shipments thatare packaged in small boxes. The conditions for inference have been met. What are the values of the test statisticand P-value for a t-test about a difference in means?Find the t-table here and the z-table here.O t=-10.19. The P-value is less than 0.0005.O z=-10.19. The P-value is less than 0.0002.O t= 10.19. The P-value is less than 0.0005.O z = 10.19. The P-value is less than 0.0002.

Answered: Image /qna-images/answer/89774d9c-ea16-4495-85b1-2a10e1410184.jpg

Answered: A manufacturing company packages…

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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A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 38 participants in the treatment group lowered their cholesterol levels by a mean of 23.2 points with a standard deviation of 4.4 points. The 44 participants in the control group lowered their cholesterol levels by a mean of 21.5 points with a standard deviation of 2.9 points. Assume that the population variances are not equal and test the company's claim at the 0.01 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
The average McDonald's restaurant generates $3.6 million in sales each year with a standard deviation of 0.9. Trent wants to know if the average sales generated by McDonald's restaurants in Kentucky is different than the worldwide average. He surveys 27 restaurants in Kentucky and finds the following data (in millions of dollars): 4.1, 2.8, 4.4, 4.5, 5.3, 5, 3.7, 2.9, 3.8, 4.8, 3.6, 2.3, 3.7, 2.9, 2.9, 4, 1.1, 5.2, 2.9, 5, 4, 4, 5.9, 3.2, 2.2, 4.3, 3.8 Perform a hypothesis test using a 3% level of significance. Step 1: State the null and alternative hypotheses. Ho: [? v] ? v На: ? ? v (So we will be performing a Select an answer test.) Step 2: Assuming the null hypothesis is true, determine the features of the distribution of point estimates using the Central Limit Theorem. By the Central Limit Theorem, we know that the point estimates are Select an answer v with distribution mean and distribution standard deviation Step 3: Find the p-value of the point estimate. P( ? v ? v = P( ? ♥ ?…
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According to the Bureau of Labor Statistics, the mean salary for registered nurses in Kentucky was $56,279. The distribution of salaries is assumed to be normally distributed with a standard deviation of $5,902. Someone would like to determine if registered nurses in Ohio have a greater average pay. To investigate this claim, a sample of 168 registered nurses is selected from the Ohio Board of Nursing, and each is asked their annual salary. The mean salary for this sample of 168 nurses is found to be $56,485.465. Describe the sampling distribution of the sample mean salary when samples of size 168 are selected. Mean (μ y-bar): _____ Standard Deviation (σ y-bar): _____ Shape: The distribution of y-bar is __________ (normally distributed/not normally distributed) because __________ (the population of salaries is NOT normally distributed/the sample size is large/the sample size is NOT large/the population of salaries is normally distributed) What conjecture has been made? The mean…
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1) According to Consumer Expenditures, the average consumer spent $1729 on apparel and services in 2017. In 2018, a random sample of 36 consumers had a mean of $1667.11 and a standard deviation of $351.69. Can it be said that at the 5% level of significance that the mean level of consumption differed in 2018 with respect to 2017?
A parenting magazine reports that the average amount of wireless data used by teenagers each month is 5 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 25 teenagers at her high school, and calculates a mean of 4.7 Gb per month with a standard deviation of 0.9 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.01 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. E Tables E Keypad Answer Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the average amount of wireless data used by teenagers each month is less than 5 Gb. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the average amount of…
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