A poll of 1188 Americans showed that 46.8% of the respondents prefer to watch the news rather than read or listen to it.Use those results with a 0.10 significance level to test the claim that fewer than half of Americans prefer to watch thenews rather than read or listen to it. Use the P-value method. Use the normal distribution as an approximation to thebinomial distribution.Let p denote the population proportion of all Americans who prefer to watch the news rather than read or listen to it.Identify the null and alternative hypotheses.Ho: P = 0.5H₁: p < 0.5(Type integers or decimals. Do not round.)Identify the test statistic.z =(Round to two decimal places as needed.)

A pool of 1188 Americans showed that 46.8% of the despondents prefer to watch the news rather than…

Answered: A poll of 1188 Americans showed that…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A phone company finds that the mean data usage for its residential customers is 10GBs, and the median for the same group is 4GBs. This indicates that the company's residential customer usage data has a right-skewed distribution True or False True False
A construction company that installs drywall wanted to investigate how a person’s age affects how much dry wall they can install in a week. On one hand, it would seem that the younger workers would be able to install more, but it seems that experience would also play a role. Two random samples were taken. Assume the distributions are normal.Dry Wall Results by Age Age 18-21(sheets/week) Age 25-28(sheets/week) 92 104 109 116 93 96 81 104 115 108 109 108 84 116 121 112 94 120 116 108 a) Set up the null and alternative hypotheses to see if there is a difference in the average number of sheets of dry wall that a person can install per week based on age. b) Check the conditions. c) Calculate the test statistic. d) Find the p-value. e) Using α = 0.05, state your conclusion in the context of the problem.
Jason collects information about annual family income from each student at his college. He computes both the mean and median family income across all students and is surprised to find that the mean income is much higher than the median income. What is the most likely reason for the difference in the mean and the median? Select one: a. The sample size is too small. b. The median tends to produce lower estimates of central tendency. c. The median is not an appropriate estimate for annual family income's level of measurement. Incorrect d. The mean is more affected by a small number of extreme values.
A researcher wants to know if the news station a person watches is a factor in the amount of time (in minutes) that they watch. The table below shows data that was collected from a survey. FOX CNN Local 60 59 50 69 55 31 61 49 28 70 16 58 49 49 15 55 41 32 59 58 46 53 66 38 Assume that all distributions are normal, the three population standard deviations are all the same, and the data was collected independently and randomly. Use a level of significance of α=0.1α=0.1.H0: μ1=μ2=μ3H0: μ1=μ2=μ3Ha:Ha: At least two of the means differ from each other. For this study, we should use The test-statistic for this data = (Please show your answer to 4 decimal places.) The p-value for this sample = (Please show your answer to 4 decimal places.)
A psychologist believes that youths identified as lesbian, gay, or bisexual already know their sexual orientation before they turned 18. To verify this, she took a random sample of 654 young adults that identified as lesbian, gay, or bisexual and asked them the age when they first realized their sexual orientation. It is known that the mean age when a young adult first realized they might be lesbian, gay, or bisexual is normally distributed. R COMMANDER OUTPUT sd se (mean) mean n 14.72199 14.9814 0.57568 654 95 percent confidence interval: 13.572 15.872
Which statement accurately describes the proportions in the tails of a normal distribution? The proportion in the left-hand tail is smaller than the proportion in the right-hand tail. O Proportions in both the left-hand and right-hand tails tend to be relatively large. OProportions in both tails are the same., O Proportions in both tails are negative.
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Do students perform worse when they take an exam alone than when they take an exam in a classroom setting? Eight students were given two tests of equal difficulty. They took one test in a solitary room and they took the other in a room filled with other students. The results are shown below. Exam Scores Alone 83 77 76 84 83 76 70 76 Classroom 92 73 77 86 86 81 75 81 Assume a Normal distribution. What can be concluded at the the a = significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: H₁: Select an answer ✓ Select an answer ✓ Select an answer ✓ Select an answer ✓ b. The test statistic ? | = c. The p-value = d. The p-value is ? ✓ a e. Based on this, we should f. Thus, the final conclusion is that ... 0.05 level of significance level of Select an answer (please enter a decimal) Select an answer (Please enter a decimal) (please show your answer to 3 decimal places.) (Please show your answer to 4 decimal places.) Select an answer…
A researcher wants to know if there is a difference between the mean amount of sleep that people get for various types of employment status. The table below shows data that was collected from a survey. Full Time Worker Part Time Worker Unemployed 6 6 7 5 6 7 8 9 9 7 6 8 8 8 8 8 8 8 6 6 8 8 7 6 9 Assume that all distributions are normal, the three population standard deviations are all the same, and the data was collected independently and randomly. Use a level of significance of α=0.01α=0.01.H0: μ1=μ2=μ3H0: μ1=μ2=μ3H1:H1: At least two of the means differ from each other. For this study, we should use Select an answer 2-PropZTest 2-PropZInt 1-PropZInt 2-SampTInt 1-PropZTest T-Test χ²-Test ANOVA χ²GOF-Test TInterval 2-SampTTest The test-statistic for this data = (Please show your answer to 3 decimal places.) The p-value for this sample = (Please show your answer to 4 decimal places.) The p-value is Select an answer less than (or equal to) alpha greater…
Español A r.able indicator of a baby's health is the weight gained in the fir year of the baby's life. Assume that the population of all such weight gains for baby boys is approximately normally distributed. A study claimed that the mean of this population is 3.454 kg. As a practicing pediatrician, you want to test this claim. So, you select a random sample of baby boys, and you record the weight each gained in their first year. Follow the steps below to construct a 95% confidence interval for the population mean of all the weight gains for baby boys in their first year. Then state whether the confidence interval you construct contradicts the study's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. Take Sample Sample size: 0 Point estimate: 0 Sample standard deviation: 0 Critical value: 0 Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you…

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