A survey is given to students to study the cost of text books this semester. A...
A survey is given to students to study the cost of text books this semester. A random sample of 36 students had an average cost of $392 a standard deviation of $102. Find a 90% confidence interval for mean cost of textbooks this semester for all students.
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A survey is given to students to study the cost of
text books this semester. A...
= Question Help For a random sample of 400 students, the mean cost for textbooks during...
= Question Help For a random sample of 400 students, the mean cost for textbooks during the first semester of college was found to be $374.75, and the sample standard deviation was $30.81. Assuming that the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean The margin of error for a 98% confidence interval is (Round to two decimal places as needed.) I
The student affairs office of a public university wishes to study absenteeism among students at the...
The student affairs office of a public university wishes to study absenteeism among students at the School of Business during the semester. A random sample of 100 business students reveals the following: •Absenteeism: mean = 9.7 classes & standard deviation = 8.0 classes •48 business studentswere absent more than 5 classes a.Construct a 95% confidence interval estimate of the population mean number of absences for business students during the semester. b.Construct a 90% confidence interval estimate of the population proportion...
4. A survey asked a random sample of 363 first-year students how many hours they studied...
4. A survey asked a random sample of 363 first-year students how many hours they studied during a particular week. The mean was 15.3 hours. Suppose we know that the population standard deviation is 8.5 hours. Construct a 90%, 95% and 99% confidence interval for the mean study time of all first year students at this university. Interpret the 90% confidence interval.
ame: A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students...
ame: A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students paid for their education by student loans. Find the 90% confidence of the true proportion of students who paid for their education by student loans. A survey of 1898 people found that 45% of the adults said that dandelions were the toughest weeds to control in their yards. Find the 95% confidence interval of the true proportion who said that dandelions were the toughest...
find the following confidence intervals Near the beginning of the semester, you administered a survey to...
find the following confidence intervals Near the beginning of the semester, you administered a survey to 10 students. Suppose we wanted to use your sample to calculate the mean age of college students. If your sample mean is 20.4 years and sample standard deviation is 1.27 years, then what are the following confidence intervals? (Round your answers to three decimal places.) 80% CI: LCL = years. 80% CI: UCL = years. 90% CI: LCL = years. 90% CI: UCL =...
A random sample of students was asked for the number of semester hours they are taking...
A random sample of students was asked for the number of semester hours they are taking this semester. The standard deviation was found to be 4.7 semester hours. a) how many students should be included in the sample to be 90% sure that the sample mean is within 1 semester hour of the population mean for all students at this college?
as 2009 Mansour conducted a survey to estimate the average number of the hours studied weekly...
as 2009 Mansour conducted a survey to estimate the average number of the hours studied weekly by college students. He chose random sample of 16 students and found that the average time they study weekdy was 15 hours with a standard deviation of 2 hours. (10 marks) a. What is the point estimate of the population mean? b. Which Test should Mansour use to find the mean confidence Interval (t or Z). Justify your answer? c. Using the 90% level...
The college bookstore tells prospective students that the average cost of its textbooks is $52 with...
The college bookstore tells prospective students that the average cost of its textbooks is $52 with a standard deviation of $4.50. A group of smart statistics students thinks that the average cost is higher. In order to test the bookstore’s claim against their alternative, the students will select a random sample of size 100. Assume that the mean from their random sample is $52.80. What is the Z-score? What percentage of samples would have a higher average cost for text...
1. The college bookstore tells prospective students that the average cost of its textbooks is $52...
1. The college bookstore tells prospective students that the average cost of its textbooks is $52 with a standard deviation of $4.50. A group of smart statistics students thinks that the average cost is higher. In order to test the bookstore’s claim against their alternative, the students will select a random sample of size 100. Assume that the mean from their random sample is $52.80. What is the Z-score? What percentage of samples would have a higher average cost for text...
A random sample of 225 full time students at a large two-year college is taken to...
A random sample of 225 full time students at a large two-year college is taken to estimate the amount they spent on text books this semester. The sampled students spent an average of $200 with a standard deviation of $75. Thus, a 95% confidence interval for the average amount spent by all full time students at that school would be: Group of answer choices $200 ± $0.33 $200 ± $5 $200 ± $10 $200 ± $75
= Question Help For a random sample of 400 students, the mean cost for textbooks during the first semester of college was found to be $374.75, and the sample standard deviation was $30.81. Assuming that the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean The margin of error for a 98% confidence interval is (Round to two decimal places as needed.) I
4. A survey asked a random sample of 363 first-year students how many hours they studied during a particular week. The mean was 15.3 hours. Suppose we know that the population standard deviation is 8.5 hours. Construct a 90%, 95% and 99% confidence interval for the mean study time of all first year students at this university. Interpret the 90% confidence interval.
ame: A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students paid for their education by student loans. Find the 90% confidence of the true proportion of students who paid for their education by student loans. A survey of 1898 people found that 45% of the adults said that dandelions were the toughest weeds to control in their yards. Find the 95% confidence interval of the true proportion who said that dandelions were the toughest...
as 2009 Mansour conducted a survey to estimate the average number of the hours studied weekly by college students. He chose random sample of 16 students and found that the average time they study weekdy was 15 hours with a standard deviation of 2 hours. (10 marks) a. What is the point estimate of the population mean? b. Which Test should Mansour use to find the mean confidence Interval (t or Z). Justify your answer? c. Using the 90% level...
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