16) A sample of 125 tobacco smokers who recently completed a new smoking-cessationprogram were asked to rate the effectiveness of the program on a scale of I to 10, with10 corresponding to "completely effective" and 1 corresponding to "completelyineffective". The average rating was 4.1 and the standard deviation was 4.0.E37Construct a 95% confidence interval for the mean score.A) 3.8 < u<4.4B) 3.7 <4.5D) 0 u<4.1C) 3.4

Answered: 16) A sample of 125 tobacco smokers who…

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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1. In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with garlic processed tablet form. Cholesterol levels were measured before and after the treatment. The chang (before-after) in their levels of LDL cholesterol (in mg/dL) have a mean of 4.2 and a standard devia 19.4. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in redu LDL cholesterol?
20) A student randomly selects 10 paperbacks at a store. The mean price is $8.75 with a standard deviation of $1.50. Construct a 95% confidence interval for the population standard deviation, o. Assume the data are normally distributed.
Three experiments investigating the relation between need for cognitive closure and persuasion were performed. Part of the study involved administering a "need for closure scale" to a group of students enrolled in an introductory psychology course. The "need for closure scale" has scores ranging from 101 to 201. For the 75 students in the highest quartile of the distribution, the mean score was x = 176.90. Assume a population standard deviation of ? = 7.71. These students were all classified as high on their need for closure. Assume that the 75 students represent a random sample of all students who are classified as high on their need for closure. Find a 95% confidence interval for the population mean score ? on the "need for closure scale" for all students with a high need for closure. (Round your answers to two decimal places.)
Compute a 95% confidence interval for the true population mean given the following 10 data that represent a sample of ceramics vessels with the following rim diameters (in centimeters), the sample mean to be x̄ = 15.3 cm and the sample standard deviation is s = 3.5 cm
1) Construct a 94% confidence interval for the population mean, μ. Assume the population has a normal distribution.A sample of 20 part-time workers had mean annual earnings of $3120 with a standard deviation of $677. Round tothe nearest dollar. (Show work) Use TI -84 calculator.
Which is cheaper: eating out or dining in? The mean cost of a steak, salad, and rice bought at the grocery store is $13.04. A sample of 30 neighborhood restaurants showed a mean price of $12.15 and a standard deviation of $2 for a comparable restaurant meal. 1.1) Develop appropriate hypotheses for a test to determine whether the sample data support the conclusion that the mean cost of a restaurant meal is less than $13.04. 1.2) At alpha = 0.05, what is your conclusion? Show a detailed solution to support your conclusion. (Find p-values at: https://www.socscistatistics.com/pvalues/tdistribution.aspx)
5.37 Dental anxiety study. To gauge their fear of going to a den- tist, 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 anx- iety) 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 ap- proximately normal with µ = 11 and o = 4. a. Suppose you score a 10 on the Modified Dental Anxiety Scale. Find the z-value for your score. b. Find the probability that someone scores between 10 and 15 on the Modified Dental Anxiety Scale. c. Find the probability that someone scores above 20 on the Modified Dental Anxiety Scale.
16. Hospital Noise Levels For a sample of 24 operating Ok Hospital Noise Levels For a sample of 24 operating rooms taken in the hospital study mentioned in Exercise 19 in Section 7–1, the mean noise level was 41.6 decibels, and the standard deviation was 7.5. Find the 95% confidence interval of the true mean of the noise levels in the operating rooms.
5. a). What is a fundamental difference between the t statistic and a z-score in terms of variability? b). What value is estimated with a confidence interval using the t statistic?
One method for measuring air pollution is to measure the concentration of carbon monoxide, or CO, in the air. Suppose Nina, an environmental scientist, wishes to estimate the CO concentration in Zagreb, Croatia. She randomly selects 45 locations throughout the city measures the CO concentration at each location. Based on her 45 samples, she computes the margin of error for a 95% t-confidence interval for the mean concentration of CO in Zagreb, in µg/m³, to be 4.28. What would happen to the margin of error if Nina decreases the confidence level to 90%? Nina increases the confidence level to 99%? Nina decreases the sample size to 33 locations? Nina increases the sample size to 65 locations? Answer Bank Increase Stay the same Decrease
A. True or False? A researcher compares the BMI of two populations of individuals from different socioeconomic backgrounds and finds that the proportion of individuals considered obese and from a poor background is .37 (n = 112), and the proportion of individuals considered obese and from an affluent background is .27 (n = 87); a 95% confidence interval for the risk difference for obesity among members of the poor population compared to the affluent population is (-0.03, 0.23). B. Why is it true or false?
7 6. 10 12 14 15 t95% 2.45 2.37 2.31 2.26 2.21 2.17 2.15 1.96

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