1I) Ine inree coniiuence iniervaiS Delow were consrucIeu nomm ue saine saInpie. Ule ulthem was computed at a confidence level of 90%, another at a confidence level of 95%and another at a confidence level of 98%.Which is the confidence level at 98%?B) cannot be determinedD) 31.3 u<40.7A) 29.3<42.7C) 30.3

As per given information, confidence intervals given in options are for 90%, 95% and 98%. We have to…

Answered: 1I) Ine inree coniiuence iniervaiS…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.8: Probability

Problem 31E

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"Trydint" bubble-gum company claims that 7 out of 10 people prefer their gum to "Eklypse". Test their claim at the 99 confidence level.The null and alternative hypothesis in symbols would be one of these: H0:p=0.7H1:p≠0.7 H0:p≥0.7H1:p<0.7 H0:p≤0.7H1:p>0.7 H0:μ≤0.7H1:μ>0.7 H0:μ=0.7H1:μ≠0.7 H0:μ≥0.7H1:μ<0.7 The null hypothesis in words would be one of these: The average of people that prefer Trydint gum is 0.7. The proportion of all people that prefer Trydint gum is less than 0.7. The average of people that prefer Trydint gum is not 0.7. The proportion of all people that prefer Trydint gum is 0.7 The proportion of people in a sample that prefer Trydint gum is not 0.7 The proportion of all people that prefer Trydint gum is greater than 0.7. The proportion of people in a sample that prefers Trydint gum is 0.7. Based on a sample of 330 people, 210 said they prefer "Trydint" gum to "Eklypse".The point estimate is: _______ (to 3 decimals)The 99 % confidence interval is: ____…
Based on a random sample of 50 students, the 90 percent confidence interval for the mean amount of moneystudents spend on lunch at a certain high school is found to be ($3.45, $4.15). Which of the following statements istrue?(A) 90% of the time, the mean amount of money that all students spend on lunch at this high school will bebetween $3.45 and $4.15.(B) 90% of all students spend between $3.45 and $4.15 on lunch at this high school.(C) 90% of all random samples of 50 students obtained at this high school would result in a sample meanamount of money students spend on lunch between $3.45 and $4.15.(D) 90% of all random samples of 50 students obtained at this high school would result in a 90% confidenceinterval that contains the true mean amount of money students spend on lunch.(E) Approximately 45 of the 50 students in the random sample will spend between $3.45 and $4.15 on lunchat this high school.
"Trydint" bubble-gum company claims that 7 out of 10 people prefer their gum to "Eklypse". Test their claim that it is different at the 99 confidence level.The null and alternative hypothesis in symbols would be: H0:μ=0.7H0:μ=0.7H1:μ>0.7H1:μ>0.7 H0:p=0.7H0:p=0.7H1:p>0.7H1:p>0.7 H0:p=0.7H0:p=0.7H1:p≠0.7H1:p≠0.7 H0:μ=0.7H0:μ=0.7H1:μ<0.7H1:μ<0.7 H0:p=0.7H0:p=0.7H1:p<0.7H1:p<0.7 H0:μ=0.7H0:μ=0.7H1:μ≠0.7H1:μ≠0.7 The null hypothesis in words would be: The average of people that prefer Trydint gum is 0.7. The proportion of people in a sample that prefers Trydint gum is 0.7. The proportion of people in a sample that prefer Trydint gum is not 0.7 The average of people that prefer Trydint gum is not 0.7. The proportion of all people that prefer Trydint gum is 0.7 The proportion of all people that prefer Trydint gum is less than 0.7. The proportion of all people that prefer Trydint gum is greater than 0.7. Based on a sample of 200 people, 114 said they prefer…
a)In a random sample of 31 professional actors, it was found that 24 were extroverts. Find a 98% confidence interval for p. (0.647, 0.902) (0.063, 0.9490 (0.599, 0.949) (0.647, 1.485) b) In a random sample of 88 professional actors, it was found that 20 were extroverts. Let p represent the proportion of all actors who are extroverts. Find a point estimate for e. 0.773 c. 20 b. 0.227 a. 68 d. 0.114 (0.600, 1.485)
If n=10n=10, ¯xx¯=37, and s=8s=8, construct a confidence interval at a 99% confidence level. Give your answers to one decimal place. -------- < μ< ---------- Technology
(a) What is the minimum number of students that will need to be observed to get a margin of error of 0.08 with 99% confidence? (Use a table or technology.) students (b) Dr. Victory's school superintendent, Dr. Triumph, finds out about the study and tells him that in past studies, approximately 60% of middle school students have felt bullied. Will having this knowledge change how many students will be needed for Dr. Victory's students? Yes or no? YesNo (c) Now that Dr. Triumph has told Dr. Victory that in previous studies approximately 60% of students have felt bullied, what is the minimum number of students that will need to be observed to get a margin of error of 0.08 with 99% confidence? (Use a table or technology.) students
In describing the findings of a 2010 Public ReligionResearch Institute poll, the researchers reported a 3%margin of error at a 95% confidence level. If instead theyhad used a 99% confidence level, their margin of errorwould have been:A) still 3%, because the same sample is used.B) more than 3%, because they reported greaterconfidence.C) less than 3%, because they reported greaterconfidence.D) more than 3%, because this sample is too small.E) less than 3%, because this sample is too small.
In a random sample of 69 professional actors, it was found that 38 were extroverts. Let p reoreaent the proportion of all actors who ate extroverts. Find a poibt estimate for p. Find a 95% confidence interval for p. Do you think the cinditions np>5 and nq> 5 are satisfied in rhis peoblem? In a random sample of 514 judges, it was foubd that 281 were introverts. Let p represent the proportion of all judges who are introverts. Find a point estimate for p. Find 99% confidence interval for p.
Also compute a 95% confidence interval, Cohen's D, and r^2

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