Stats Final guide Essay

Student A -> z = (83 – 65) / √144 = 18 / 12 = 1.5

d. It should appear in the tail signifying a negative relationship, with .05 in that tail.

18. Suppose that the scores of architects on a particular creativity test are normally distributed. Using a normal curve table, what percentage of architects have Z scores:

Fry Brothers heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of

1. A researcher is interested in whether students who attend private high schools have higher average SAT Scores than students in the general population. A random sample of 90 students at a private high school is tested and and a mean SAT score of 1030 is obtained. The average score for public high school student is 1000 (σ= 200).

Assume that a set of test scores is normally distributed with a mean of 100 and a standard deviation of 20. Use the 68-95-99.7 rule to find the following quantities:

14. On a standard measure of hearing ability, the mean is 300 and the standard deviation is 20. Give the Z scores for persons who score (a) 340, (b) 310, and (c) 260. Give the raw scores for persons whose Z scores on this test are (d) 2.4, (e) 1.5, (f) 0, and (g)-4.5.

Assume 50 random samples of the same sample size are taken from a population, and a 90% confidence interval is constructed from each sample. How many of the intervals would you expect to contain the true population mean?

31. A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks with a standard deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value.

The probability of getting the Z value would be 1.000 minus .999 which equals .001.

We are looking at a very large group of tests. Forty five tests were selected from this group. The mean of the sampled tests is 58 and the standard deviation is 13. Find the 95% confidence interval for the mean grade for the whole class.

Final Exam Review Questions Solutions Guide You will probably want to PRINT THIS so you can carefully check your answers. Be sure to ask your instructor if you have questions about any of the solutions given below. 1. Explain the difference between a population and a sample. In which of these is it important to distinguish between the two in order to use the correct formula? mean; median; mode; range; quartiles; variance; standard deviation. Solution: A sample is a subset of a population. A population consists of every member of a particular group of interest. The variance and the standard deviation require that we know whether we have a sample or a population. 2. The following numbers represent the weights in pounds of six 7year old

What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least $112,000? d. What is the likelihood of selecting a sample with a mean of more than $100,000? e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000.

From refering normal distribution table we get that at 95% probability we get that value of "z" is 0.3289

A population has a mean of 75 and a standard deviation of 8. A random sample of 800 is selected. The expected value of is