**Step-by-Step Decision Process on Hypothesis Testing****Part 3 of 5**Progress Bar: Approximately 60% complete.**Part 4 of 5**1. **Make the Decision** - Instruction: Choose one option to either accept or reject the null hypothesis. - Options: A dropdown menu is available for selection. - Icons: A checkmark (X) and a refresh symbol are present for confirming or resetting choices.**Part 4 of 5**Progress Bar: Approximately 80% complete.**Part 5 of 5**1. **Summarize the Results** - Instruction: Choose an option that describes the conclusion based on the data. - Sentence Structure: "There is (Choose one) that the average number of sick days is greater than 5." - Options: A dropdown menu is available for selection. - Icons: A checkmark (X) and a refresh symbol are present for confirming or resetting choices.This step-by-step guide provides a structured approach for making statistical decisions regarding hypotheses, specifically focusing on determining whether to accept or reject the null hypothesis based on the data. **Sick Days in Bed**A researcher wishes to determine if the average number of sick days a worker takes per year is greater than 5. A random sample of 37 workers at a large department store had a mean of 5.4. The standard deviation of the population is 1.4. Is there enough evidence to support the researcher's claim at $\alpha = 0.01$? Assume the variable is normally distributed. Use the P-value method with tables.**Instructions for Hypothesis Testing****Part 0/5****Part 1 of 5: Formulate Hypotheses**- **State the hypotheses and identify the claim:** - $ H_0 $ (Null Hypothesis): $[ \text{Choose one} ]$ - $ H_1 $ (Alternative Hypothesis): $[ \text{Choose one} ]$ - This hypothesis test is a $[ \text{Choose one} ]$ test.**Part 1/5: Progress Indicator****Part 2 of 5: Calculate the Test Value**- **Compute the test value:** - Always round z-score values to at least two decimal places. - $ z = [ \text{Input Box} ]$**Part 2/5: Progress Indicator****Part 3 of 5: Determine the P-value**- **Find the P-value:** - Round the answer to at least four decimal places. - $ P\text{-value} = [ \text{Input Box} ]$**Graphical Elements:**- The image contains selection boxes for choosing symbols and values related to hypothesis expressions and mathematical calculations, such as comparison operators and statistical symbols.

Given that Sample size n = 37 Sample mean = 5.4 Population SD = 1.4 Level of significance = 0.01

Answered: Sick Days in Bed A researcher wishes to…

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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Assume that you have been provided the following information for a sample of cars The covariance between weight and miles per gallon is equal to -4.9 pound-miles. The standard deviation of the cars' mileages is 7.75. The standard deviation of the cars' weights is 0.94. Estimate the correlation coefficient to the first decimal place.
Consider a set of data in which the sample mean is 30.230.2 and the sample standard deviation is 7.87.8. Calculate the z-score given that x=31.4x=31.4. Round your answer to two decimal places.
A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1539 and the standard deviation was 315. The test scores of four students selected at random are 1940, 1290, 2240, and 1420. Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for 1940 is. (Round to two decimal places as needed.) The Z-score for 1290 is. (Round to two decimal places as needed.) The Z-score for 2240 is. (Round to two decimal places as needed.) The Z-score for 1420 is. (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. OA. The unusual value(s) is/are. CD (Use a comma to separate answers as needed.) OB. None of the values are unusual.
P3. Show that the mle of unknown standard deviation when the mean is known is different from the sample standard deviation when the random sample is taken from a normal population characterized by two parameters, mean and variance. Which estimator is unbiased? Justify your answer. Show that a sample mean is a minimum variance unbiased estimator of the mean of a normal population with known variance.
Exam scores in a large class are normally distributed with a mean of 55 and standard deviation of 10. The teacher decides to boost the scores by adding 20 points to everyone’s score. Elizabeth’s original score on the exam was 75, so her new score is 95. The z-score corresponding to her new score (relative to the other new scores) will be:
Exam scores in a large class are normally distributed with a mean of 15.7 and a standard deviation of 7.43. Rana's score on the exam was 9, and Sara's score was 25. The absolute value for the difference between the two ladies z-score is O a. 4.5760 O b. 2.4226 O c. 1.2517 O d. 2.1534
If we meet these conditions, the sampling distribution of the mean will have a normal shape and ... The mean of the sampling distribution will be u, (i.e., the same as the population mean) and the standard deviation of the sampling distribution will be (that's the population standard deviation divided by the square root of the sample size). In the AP Stats Guy video, he talks about the number of text messages his students send during class. Suppose the average number of text messages his students send during class is u = 30 text messages. If we take samples of say, n = 36 students at a time, we would expect the mean of the sampling distribution we create to be the same as the population mean, 30 text messages. If we can further say that standard deviation of the number of text messages is 12 text messages, by how much would we expect the sample means to vary? (hint, use the formula above) text messages
The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players’ salaries and 35 football players’ salaries is selected. The mean salary is determined for both samples. Let represent the difference in the mean salaries for baseball and football players. Which of the following represents the shape of the sampling distribution for ? skewed right since the populations are both right skewed skewed right since the differences in salaries cannot be negative approximately Normal since both sample sizes are greater than 30 approximately Normal since the sum of the sample sizes is greater than 30
The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 15 days.In what range would you expect to find the middle 95% of most pregnancies?Between and .If you were to draw samples of size 55 from this population, in what range would you expect to find the middle 95% of most averages for the lengths of pregnancies in the sample?Between and .
The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players' salaries and 35 football players' salaries is selected. The mean salary is determined for both samples. Let -T, represent the difference in the mean salaries for baseball and football players. Which of the following represents the standard deviation of the sampling distribution for - -- ? O 0.41 million O 0,57 million Save and Exit Next Submit Mark this and return hp & 4 6. 7 8. 9. y h k
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