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 willaverage a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 34 participants in thetreatment group lowered their cholesterol levels by a mean of 22.2 points with a standard deviation of 3.4 points. The 42 participants in the control group lowered theircholesterol levels by a mean of 21.2 points with a standard deviation of 1.8 points. Assume that the population variances are not equal and test the company's claim atthe 0.10 level. Let the treatment group be Population 1 and let the control group be Population 2.Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below.Ho:₁ = ₂Ha: M•M₂

The objective of this question is to test the pharmaceutical company's claim that its new…

Answered: A pharmaceutical company needs to know…

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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Kala recently switched her primary doctor to one specializing in caring for elderly patients. On her new doctor's website, it says that the mean systolic blood pressure among elderly females is 115 millimeters of mercury (mmHg). Kala believes the value is actually higher. She bases her belief on a recently reported study of 21 randomly selected, elderly females. The sample mean systolic blood pressure was 124 mmHg, and the sample standard deviation was 20 mmHg. E Assume that the systolic blood pressures of elderly females are approximately normally distributed. Based on the study, at the 0.10 level of significance, can it be concluded that μ, the population mean systolic blood pressure among elderly females, is greater than 115 mmHg? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) D (a) State the null hypothesis H, and the alternative hypothesis H₁ LL O P Р H:0…
Energy drinks come in different-sized packages: pouches, small bottles, large bottles, twin-packs, 6-packs, and so on. How is the price related to the amount of beverage? Data collected on a variety of packages revealed a mean size of 140.17 ounces with a standard deviation of 78.23 ounces. The packages had an average price of $3.66 with a standard deviation of $1.50, and the correlation between size and price was r= 0.91. A scatterplot of these data suggested that the assumptions needed for regression were reasonable. a) Interpret the value of r in context. b) Compute the slope of the regression line for predicting the price of an energy drink. Include the proper units in your answer. c) Write the equation for the least-squares regression line for these data. d) If the price was in Euros instead of US dollars, what would be the correlation between size and price? Explain. e) For this model the standard deviation of the residuals was s = 0.26. Explain what that means in context.
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the degrees of freedom within samples?
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square between samples?
You wish to evaluate the differences in educational quality between two training centres (A and B). You take a random sample of 98 individuals trained at centre A, and an independent random sample of 41 individuals trained at centre B. The individuals in both samples take the same, standardised test. The mean test score of individuals in sample A is 85.1 and the sample standard deviation is 12.2. The mean test score in sample B is 67.4 and the sample standard deviation is 11.0. Find the lower limit of a 98% confidence interval for the difference between the population means. Form the confidence interval around a positive point estimate for the difference between the means. In order to obtain a positive point estimate, make sure you subtract the lower value from the higher value when calculating the difference between the sample means. (Provide your answer as a number rounded to 2 decimal places.)
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square within samples?
Your mother claims that the baby boomer generation is thriftier than the millennials and that boomers save more money with coupons than millennials. In order to test this claim, 17 millennials and 12 boomers agreed to track how much money they saved in a month by using coupons at the grocery store. At the end of the month, the mean savings for the millenials was $18.09 with a standard deviation of $3.48, wheras the boomers saved an average of $26.97 with a standard deviation of $14.60. Perform an appropriate hypothesis test to test the null hypothesis that the true mean savings for boomers and millennials are equal versus the alternative that they are different. What is the test statistic and p-value of this test? Test statistic = -2.066; p-value = 0.031 Test statistic = 2.066; p-value = 0.061 Test statistic = -2.066; p-value = 0.939 Test statistic = 2.066; p-value = 0.969 û
Erin Richards is a junior at Central High School, and she has talked to her guidance counselor about her chances of being admitted to Tech after her graduation. The guidance counselor has told her that Tech generally accepts only those applicants who graduate in the top 10% of their high school class. The average grade point average of the last four senior classes has been 2.67, with a standard deviation of 0.58. What GPA will Erin have to achieve to be in the top 10% of her class?
One year Perry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 2.57. Also, Rita had the lowest ERA of any female pitcher at the school with an ERA of 3.26. For the males, the mean ERA was 4.304 and the standard deviation was 0.894. For the females, the mean ERA was 4.158 and the standard deviation was 0.519. Find their respective z-scores. Which player had the better year relative to their peers, Perry or Rita? (Note: In general, the lower the ERA, the better the pitcher.) Perry had an ERA with a z-score of Rita had an ERA with a z-score of (Round to two decimal places as needed.) Textbook Get more help - $ 1 G Search or type URL % 31 6 & 7 * 8 + ( 9 0 Clear all + Check answer delete
The amount of Derek's monthly phone bill is normally distributed with a mean of $74 and a standard deviation of $8. What percentage of his phone bills are between $58 and $90? Use the empirical rule to solve the problem
Suppose you've just been hired as a statistics professor, but you've never taught undergrads before. While you're doing your best, you're not sure you're fully up to par with all the other excellent statistics professors out there. You decide to look at your 41 students' grades on their midterm exam. They averaged grades of 81 on this exam, with a standard deviation of 10. You reason that if they're higher than normal, you might be making things too easy for them. Likewise, if they're lower than normal, you might be making things too hard. Seeking a Goldilocks effect, you compare this with the national average of statistics midterm exam scores - this is 79. BEING SPECIFIC, what kind of statistical analysis should you perform, and WHY? Three parts of your answer will factor into the grade on this answer: the type of statistical test, one specific aspect of this test (hint- see the earlier questions), and your explanation as to why.
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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