Interpret a P-value: The manufacturer of a medication designed to lower blood pressureclaims that the mean systolic blood pressure for people taking their medication is less than135. To test this claim, blood pressure is measured for a sample of 500 people who are takingthe medication. The P-value for testingHo: μ = 135 versusH₁: µ< 135 isP = 0.001.a. The manufacturer concludes that because the P-value is very small, we can be fairlycertain that the mean pressure is less than 135, but we cannot conclude that it is a lotsmaller. Is this conclusion justified?

Consider the variable X, the systolic blood pressure of a person. Suppose the statement is given…

Answered: Interpret a P-value: The manufacturer…

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 doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 4 points with 99% confidence assuming s=16.4 based on earlier studies? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required? LOADING... Click the icon to view a partial table of critical values. A 99% confidence level requires nothing subjects. (Round up to the nearest subject.) A 90% confidence level requires nothing subjects. (Round up to the nearest subject.) How does the decrease in confidence affect the sample size required? A. Decreasing the confidence level decreases the sample size needed. B. Decreasing the confidence level increases the sample size needed. C. The sample size is the same for all levels of confidence.
One group of accounting students took a distance learning class, while another group took the same course in a traditional classroom. At α = .10, is there a significant difference in the mean scores listed below? Exam Scores for Accounting Students Statistic Distance Classroom Mean scores x⎯⎯1x¯1 = 9.0 x⎯⎯2x¯2 = 10.2 Sample std. dev. s1 = 2.1 s2 = 2.4 Number of students n1 = 19 n2 = 19 Assume μ1 is the mean score of distance learning and μ2 is the mean score of classroom learning. The appropriate hypotheses is H0: μ1 – μ2 = 0 vs. H1: μ1 – μ2 ≠ 0 (b) Specify the decision rule. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.) Reject the null hypothesis if tcalc < _______. Or tcalc > ______________. (c) Find the test statistic tcalc. (Round your answer to 4 decimal places. A negative value should be indicated by a minus sign.) tcalc ______________
To gauge their fear of going to a dentist, a random sample of adults completed the Modified Dental Anxiety Scale questionnaire (BMC Oral Health, Vol. 9, 2009). Scores on the scale range from zero (no anxiety) to 25 (extreme anxiety). The mean score was 11 and the standard deviation was 4. Assume that the distribution of all scores on the Modified Dental Anxiety Scale is approximately normal with μ = 11 and o=4. Complete parts a through c below. a. Suppose you score a 10 on the Modified Dental Anxiety Scale. Find the z-value for this score. Z= (Round to two decimal places as needed.) b. Find the probability that someone scores between 10 and 19 on the Modified Dental Anxiety Scale. The probability is (Round to four decimal places as needed.) c. Find the probability that someone scores above 20 on the Modified Dental Anxiety Scale. The probability is (Round to four decimal places as needed.) B **** 4
In the experiment, you repeated measurements 13 times for a resistor. After data processing, you determined that the standard deviation of the measurements is 4.3 N. What is the standard error of the mean in unit of N?
You are testing against based on an SRS of 20 observations from a Normal population. The t statistic is t= -2.25. Using Table B, the P-value falls between and .
What is the F value? Is the null hypothesis rejected or not rejected? Draw Conclusion.
To gauge their fear of going to a dentist, a random sample of adults completed the Modified Dental Anxiety Scale questionnaire (BMC Oral Health, Vol. 9, 2009). Scores on the scale range from zero (no anxiety) to 25 (extreme anxiety). The mean score was 11 and the standard deviation was 4. Assume that the distribution of all scores on the Modified Dental Anxiety Scale is approximately normal with µ = 11 and o = 4. Complete parts a through c below. a. Suppose you score a 10 on the Modified Dental Anxiety Scale. Find the z-value for this score. (Round to two decimal places as needed.) 1)
A population has a mean of µ = 20. A sample is selected from this population and a treatment is administered to each individual in the sample. The scores from this sample are summarized as follows: n = 16 ,X = 28 ,s2 = 64. Do the sample data support the conclusion that the treatment has a significant effect? Test with α = .05.
A standardized math test of fifth grade students produces a mean ?μ = 100 with a ?σ = 18. A sample of n = 36 students participated in "boost" training for their math skills. The sample M = 104 with S = 24. Did this "boost" training significantly improve students' math performance? Use ?α = .01.
For a sample with a standard deviation of s = 8, a score of X = 42 corresponds to z = –0.25 what is the mean
In a class of twenty-five 15-year old girls the average height is M = 62.60 inches. The national average for girls at this age is reported as follows: µ = 63.80 inches with a standard deviation, s = 2.66 inches. 1. What is the effect size Cohen’s d and what does it tell us about the effect? 2. Compute the z-statistic.
Suppose you had obtained a test statistic of 2.45 (this is incorrect). What would the P-value be? Draw a conclusion based on a = 0.01. Select one: O a. 0.01 < P-value < 0.02, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O b. 0.005 < P-value < 0.01, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O c. 0.01 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O d. 0.005 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet…

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