It is known that the average body temperature of healthy individuals is is normally distributed with a mean, μ = 37 ^oC and standard deviation, σ = 0.6 ^oC.  

We found the average temperature of a random sample of 24 SIBT students in a Statistics tutorial.

1. What is the probability that the average body temperature of this sample of students will be less than 37.4 ^oC? (4 dp) Answer   
2. What is the probability that average body temperature of these students will be more than 37.1 ^oC?  (4 dp)

The math department is trying to determine if their tests have a gender bias. They look at scores at old exams which were normally distributed and take a random sample of students. They found that the scores of the 16 male students had a normal distribution with a mean of 73.4 and a standard deviation of 6.8. The scores of the 17 females students had a normal distribution with a mean of 77.4 and a standard deviation of 6.2. Using a significance level of 0.05, test the claim the tests have a gender bias.

The math department is trying to determine if their tests have a gender bias. They look at scores at old exams which were normally distributed and take a random sample of students. They found that the scores of the 16 male students had a normal distribution with a mean of 73.4 and a standard deviation of 7.1. The scores of the 17 females students had a normal distribution with a mean of 77.4 and a standard deviation of 6.2. Using a significance level of 0.05, test the claim the tests have no gender bias.

True or false? 1. Watching videos that present misperceptions followed by an explanation of the right answers tend to improve student learning. 2. Students tend to learn more when they find the explanation of misconceptions confusing 3. If a video presents a new idea, then students tend to revise their own misconceptions 4. If a video presents misconceptions first, then students tend to report confusion: and they don't learn anything. 5. Videos alone do not promote student learning

Suppose the University of Oklahoma decides to alter its tuition schedule by separating its students based on how many credit hours they have accumulated. Students with fewer than 15 credit hours get a 13% reduction in tuition while students with 45-90 and more than 90 credit hours face an increase in tuition of 22 and 71%, respectively. Fully explain whether this pricing strategy is rooted in a sound understanding of the price elasticity of demand, or not.

Ti-84

A research group claims that less than 25% of students at one medical school plan to go into general practice. It is found that among a random sample of 120 of the school's students, 20% of them plan to go into general practice. At the 0.10 significance level, do the data provide sufficient evidence to conclude that the percentage of all students at this school who plan to go into general practice is less than 25%? Use the confidence interval approach.

A class survey in a large class for first-year college students asked, "About how many hours do you study in a typical week?". The mean response of the 427 students was x¯¯¯ = 12 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 7 hours in the population of all first-year students at this university. What is the 99% confidence interval (±0.001) for the population mean?

Confidence interval is from ____to_____ hours.

A researcher is trying to predict test scores from student’s level of stress. What is the standardized regression equation for the following data taken from 10 students? The average level of stress for students is 9.50 with a standard deviation of 2.76. The average test score for students is 12.50 with a standard deviation of 5.13. The correlation stress and test scores is 0.88.

a) ZY’ = 0.54 (ZX) b) ZY’ = 1.32 (ZX) c) ZY’ = .88 (ZX) d) ZY’ = 1.86 (ZX)

Suppose a fitness center has two weight-loss programs. Fifteen students complete Program A, and fifteen students complete Program B. Afterward, the mean and standard deviation of weight loss for each sample are computed (summarized below). What is the difference between the mean weight losses, among all students in the population? Answer with 95% confidence.

Prog A - Mean 10.5 St dev 5.6

Prog B - Mean 13.1 St dev 5.2

You would like to study the height of students at your university. Suppose the average for all university students is 68 inches with a SD of 20 inches, and that you take a sample of 17 students from your university.

a) What is the probability that the sample has a mean of 64 or more inches? probability = .204793 (is this answer correct or no? and I need help with part b too.)

b) What is the probability that the sample has a mean between 63 and 68 inches?