Question

1. A random sample of 429 college students was interviewed about a number of matters. For...

1. A random sample of 429 college students was interviewed about a number of matters. For each of the below, write a sentence summarizing the results (i.e “I am 95% confident that….”) a. They reported that they had spent an average of $378.23 on textbooks during the previous semester. If the sample standard deviation for these data is $15.78, construct an estimate of the population mean at the 95% confidence level.

b. They also reported that they had visited the health clinic an average of 1.5 times a semester. If the sample standard deviation is 0.3, construct an estimate of the population mean at the 95% confidence level.

c. On average, the sample had missed 2.8 days of classes per semester because of illness. If the sample standard deviation is 1.0, construct an estimate of the population mean at the 95% confidence level.

d. On average, the sample had missed 3.5 days of classes per semester for reasons other than illness. If the sample standard deviation is 1.5, construct an estimate of the population mean at the 95% confidence level.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The Research Center interviewed a random sample of 1,546 adult Anericans to determine the average time...
The Research Center interviewed a random sample of 1,546 adult Anericans to determine the average time they spent sleeping per night. The sample mean was found to be 7.2 hours of sleep, with a sample standard deviation of 1.4 hours. Construct a 95% confidence interval for the true mean. Interpret for confidence interval in words.
A survey is given to students to study the cost of textbooks this semester. A random...
A survey is given to students to study the cost of textbooks this semester. A random sample of 12 students had an average cost of $284.90 and a standard deviation of $96.10. Find a 95% confidence interval for the average textbook cost for all students assuming the cost of textbooks are normal.
In a survey, 91 college students were interviewed about how much time they spent on school...
In a survey, 91 college students were interviewed about how much time they spent on school e-mail each week. The sample average was 3.34 hours with a sample standard deviation of 7.728 hours. Construct a 90% confidence interval for the population mean. Also, find the margin of error. Round your answers to 3 decimal places. Группа выборов ответов (1.994,4.686) margin of error=1.346 (1.994,4.686) margin of error=2.692 (1.731,4.949) margin of error=1.609 (1.731,4.949) margin of error=3.218 (1.994,4.686) margin of error=3.340
A random sample of 24 observations is used to estimate the population mean. The sample mean...
A random sample of 24 observations is used to estimate the population mean. The sample mean and the sample standard deviation are calculated as 104.6 and 28.8, respectively. Assume that the population is normally distributed. Construct the 90% confidence interval for the population mean. Then, construct the 95% confidence interval for the population mean. Finally, use your answers to discuss the impact of the confidence level on the width of the interval.
Construct a 95% confidence interval for the population mean. A sample of 34 college students had...
Construct a 95% confidence interval for the population mean. A sample of 34 college students had mean annual earnings of $4520 with a standard deviation of $677
How much do students pay, on the average, for textbooks during the first semester of college?...
How much do students pay, on the average, for textbooks during the first semester of college? From a random sample of 400 students the mean cost was found to be $357.75, and the sample standard deviation was $37.89. Assuming that the population is normally distributed, find the margin of error of a 91% confidence interval for the population mean.
A random sample of 25 college students attending the RWU Spring Concert showed that they had...
A random sample of 25 college students attending the RWU Spring Concert showed that they had an average amount of $15.00 cash in their pockets. This same sample data had a sample standard deviation of $2.50. a. (3 pts.) What is the point estimate for the population mean µ, amount of cash in the pockets of a RWU student? b. (4 pts.) Find the 95% confidence interval for the population mean µ, amount of cash in the pockets of a...
1. Consider a normally distributed population with a standard deviation of 64. If a random sample...
1. Consider a normally distributed population with a standard deviation of 64. If a random sample of 90 from the population produces a sample mean of 250, construct a 95% confidence interval for the true mean of the population. 2.A psychologist is studying learning in rats (to run a maze). She has a sample of 40 rats and their mean time to run the maze if 5 minutes with a standard deviation of 1. Estimate the average time to run...
1. The 95% confidence interval states that 95% of the sample means of a specified sample...
1. The 95% confidence interval states that 95% of the sample means of a specified sample size selected from a population will lie within plus and minus 1.96 standard deviations of the hypothesized population mean. T or F 2. Which of the following is NOT necessary to determine how large a sample to select from a population? Multiple Choice The level of confidence in estimating the population parameter The size of the population The maximum allowable error in estimating the...
4. Find the margin of error E. In a random sample of 151 college students, 84...
4. Find the margin of error E. In a random sample of 151 college students, 84 had part-time jobs. Find the margin of error E for the 95% confidence interval used to estimate the population proportion. Round your answer to four decimal places. Answer 0.0792 5. Find the minimum sample size required to estimate the population proportion p: Margin of error: 0.10; confidence level: 95%; from a prior study, is known to be 66%. 6. Find the minimum sample size...