3) A nationwide survey of working adults indicates that only 50% of them are satisfied with...
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3) A nationwide survey of working adults indicates that only
50% of them are satisfied with...
In a survey of 700 US adults with children, 270 of them reported that they saved...
In a survey of 700 US adults with children, 270 of them reported that they saved money for their children’s college education. Can you conclude that more than one third (33%) of US adults with children have saved money for college? Use a 0.05 significance level. a. State the null (H0) and alternative (H1) hypotheses. b. Give the test statistics and the p-value for this significance test. c. Make a decision on whether or not to reject the null hypothesis....
Suppose across all cities in Florida, the proportion of people young adults (ages 20 to 30)...
Suppose across all cities in Florida, the proportion of people young adults (ages 20 to 30) who have attended a roller derby match is 0.28. Tyree believes that the proportion of young adults in his city who have attended a roller derby match is greater than the state population proportion. He decides to test this idea using a one-sample z-test for a population proportion using a significance level of α-0. 1 0 He conducts a survey of 225 randomly selected...
A survey of 570 working adults in the United States asked each participant how many hours...
A survey of 570 working adults in the United States asked each participant how many hours they worked last week. The survey showed an average of 41.525 hours worked per week, with a standard deviation of 14.828 hours. (Note that this is the average for people that work in the United States, including part-time employees. Among full time employees, the average work week is closer to 47 hours. Link if you're interested: http://www.gallup.com/poll/175286/hour-workweek-actually-longer-seven-hours.aspx) Is there enough evidence to show that...
A survey of 1,680 randomly selected adults showed that 549 of them have heard of a...
A survey of 1,680 randomly selected adults showed that 549 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 37% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts a through e Sample proportion: 0.326786 Test statistic Critical z: P-Value z:-3.6687 ± 2.5758 0.0002 a. Is the test...
A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The...
A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The technology display below ret from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the bin distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p+0.92 Z-Value P-value Sample p 95% CI N Sample X 0.000 4.01 (0.930869,0.956847) 1 1967 2,084 0.943858 a....
A poll of 2,142 randomly selected adults showed that 92% of them own cell phones. The...
A poll of 2,142 randomly selected adults showed that 92% of them own cell phones. The technology display below results from a test of the claim that 91% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e). Test of p=0.91 vs p≠0.91 Sample X N Sample p 95% CI Z-Value P-Value 1 1970 2,142 0.919701 (0.908193,0.931210) 1.57 0.117 a. Is the...
A survey of 1 comma 567 randomly selected adults showed that 570 of them have heard...
A survey of 1 comma 567 randomly selected adults showed that 570 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 38% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (d. Sample proportion: 0.363752 Test statistic, z: negative 1.3251 Critical z: plus or minus1.9600 P-Value:...
In a survey of 3005 adults aged 57 through 85 years, it was found that 81.7%...
In a survey of 3005 adults aged 57 through 85 years, it was found that 81.7% of them used at least one prescription medication (based on data from "Use of Prescription and Over-the-Counter Medications and Dietary Supplements Among Older Adults in the United States," by Qato et al., Journal of the American Medical Association, Vol. 300, No. 24). Use a 0.01 significance level to test the claim that more than 3/4 of adults use at least one prescription medication. Enter...
According to a survey, the average American person reads for 3 hours per week. To test...
According to a survey, the average American person reads for 3 hours per week. To test if the amount of reading in Chicago is more than the national average, a researcher decides to do a hypothesis test, at a 5% significance level. She surveys 17 Chicagoans randomly and asks them about their amount of TV each week, on average. From the data, the sample mean time is 3.4 hours per week, and the sample standard deviation (s) is 0.8 hours....
3. Do you see any striking difference between problem 1 and 2? At a certain university,...
3. Do you see any striking difference between problem 1 and 2? At a certain university, the mean income of parents of the entering class is reported to by $91,600. The president of another university feels that the parents’ income for her entering class is more than $91,600. She surveys 100 randomly selected families and finds the mean income to be $96,321. Given σ = $9555. With α = 0.05, is she correct? State the null and alternative hypotheses which...
Suppose across all cities in Florida, the proportion of people young adults (ages 20 to 30) who have attended a roller derby match is 0.28. Tyree believes that the proportion of young adults in his city who have attended a roller derby match is greater than the state population proportion. He decides to test this idea using a one-sample z-test for a population proportion using a significance level of α-0. 1 0 He conducts a survey of 225 randomly selected...
A survey of 570 working adults in the United States asked each participant how many hours they worked last week. The survey showed an average of 41.525 hours worked per week, with a standard deviation of 14.828 hours. (Note that this is the average for people that work in the United States, including part-time employees. Among full time employees, the average work week is closer to 47 hours. Link if you're interested: http://www.gallup.com/poll/175286/hour-workweek-actually-longer-seven-hours.aspx) Is there enough evidence to show that...
A survey of 1,680 randomly selected adults showed that 549 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 37% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts a through e Sample proportion: 0.326786 Test statistic Critical z: P-Value z:-3.6687 ± 2.5758 0.0002 a. Is the test...
A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The technology display below ret from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the bin distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p+0.92 Z-Value P-value Sample p 95% CI N Sample X 0.000 4.01 (0.930869,0.956847) 1 1967 2,084 0.943858 a....
In a survey of 3005 adults aged 57 through 85 years, it was found that 81.7% of them used at least one prescription medication (based on data from "Use of Prescription and Over-the-Counter Medications and Dietary Supplements Among Older Adults in the United States," by Qato et al., Journal of the American Medical Association, Vol. 300, No. 24). Use a 0.01 significance level to test the claim that more than 3/4 of adults use at least one prescription medication. Enter...
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