Answer the following question about this study. Use the applet above, if necessary.Find the 90% confidence interval the statistics class is looking for.(3.48, 4.38)(3.03, 4.53)(3.18, 4.68) According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of theirdrinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student iscurious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week bystudents at his college. He plans to use a 90% confidence interval. He surveys a random sample of 71 students. The sample mean is 3.93alcoholic drinks per week. The sample standard deviation is 3.78 drinks.Let's do some exploration with technology to see how changes in the confidence level affect the confidence interval. Use the applet below toanswer the following questions.-♡Central probability = 0.50-ŅAt-score = 0.6750central probability =degrees of freedom =N0.5050-♡2₂3

Sample size (n) = 71Sample mean (x̅) = 3.93Standard deviations (s) = 3.78 Confidence level (c) =…

Answered: According to the website…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.3: Measures Of Spread

Problem 1GP

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According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 95% confidence interval. He surveys a random sample of 66 students. The sample mean is 3-75 alcoholic drinks per week. The sample standard deviation is 3.68 drinks. Construct the 95% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college. Your answer should be rounded to 2 decimal places.
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According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 90% confidence interval. He surveys a random sample of 63 students. The sample mean is 3.7 alcoholic drinks per week. The sample standard deviation is 3.2 drinks. Construct the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college. 1. The t value is - 2. The standard error has a value of -- 3. The margin of error has a value of 4. Construct the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college.
How Much Alcohol Do College Students Drink? According to the website www.collegedrinkingprevention.gov, “About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall.” A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 90% confidence interval. He surveys a random sample of 71 students. The sample mean is 3.93 alcoholic drinks per week. The sample standard deviation is 3.78 drinks. Here is an applet you can use when directed to find the critical T-value.Note: the "Central Probability" is the area of the middle shaded region. 1. Find the critical T-value for this 90% confidence interval. Hint: Use the applet to find the T-value for 90% confidence with df = 71 – 1 = 70.
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The term “bleeding heart liberal” is due to the (supposed) excess sympathy that liberals feel for criminals, the poor, the sick, and other marginalized groups. At the same time, however, members of marginalized groups are also more likely to be liberal, arguably because they themselves have experienced circumstances that make them more empathetic to the plight of others. If this hypothesis is true, we might expect that self-proclaimed liberals would have higher mean scores on an index capturing the number of “bad things” that have happened to them in the last year. In the GSS, of 56 persons claiming to be “strong conservatives,” the mean score on such an index was 3.75 (sd = 4.15), while, of 46 persons claiming to be “strong liberals,” the mean score on the index was 4.4 (sd = 2.30). Is the hypothesis reasonable?
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