PSY315 Week Two Practice Problems

4. Calculate the following measures of central tendency for the set of cube measurement data. Show your work or explain your procedure for each.

13. For the scores (3.0, 3.4, 2.6, 3.3, 3.5, 3.2): N = 6; Sum = 19; Mean = Sum/N = 19/6 = 3.17; Median = (2.6+3.3)/2 = 2.95; Sum of Squared Deviations = (3-3.17)2+(3.4-3.17)2+(2.6-3.17)2+(3.3-3.17)2+(3.5-3.17)2+(3.2-3.17)2 = 0.533; SD2 = SS/N = 0.533/6 = 0.089; SD = 0.089 = 0.298.

A company that makes cola drinks states that the mean caffeine content per 12-ounce bottle of cola is 45 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles of the cola has a mean caffeine content of 45.5 milligrams with a standard deviation of 6.1 milligrams. At a = 0.08, can you reject the company’s claim?

Using the formula from part a: Var[X+2W]=864, making the z-value (120-110)/29.39=.34. The probability of exceeding .34 is .3669.

18) A sample of candies have weights that vary from 2.35 grams to 4.75 grams. Use this information to find the upper and lower limits of the first class if you wish to construct a frequency distribution with 12 classes. A) 2.35-2.65 Answer: B 19) Assume that the heights of men are normally distributed. A random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 1.4 inches. Construct a 99% confidence interval for the population standard deviation, . A) (0.8, 2.1) B) (1.0, 2.6) C) (0.9, 2.5) D) (1.0, 2.4) Answer: C B) 2.35-2.55 C) 2.35-2.54 D) 2.35-2.75

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.

Problem 5: (3.5 points). A manufacturing process produces items whose weights are normally distributed, with a mean of 120 grams and standard deviation 10 grams.

A life insurance company wants to update its actuarial tables. Assume that the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 71 years and a standard deviation of 3.5 years. What proportion of the plan participants are expected to see their 75th birthday? Note: Write your answers with two places after the decimal, rounding off as appropriate.

The data can be put into a frequency distribution and it will be clear both from the numbers and the bar chart if the numbers show a symmetrical distribution or positive or negative skew. The mean for the data set is 18.9, the median is 19.0, and the mode 19.0. The range is 43.2 and the standard deviation is roughly 7.669. These numbers were calculated from the body fat column of the data since these are the numbers that will help us decide whether the claim about men having a body fat average of 20% is accurate or not. The bar chart of men's body fat at the end of the paper represents the distribution of body fat among the 252 men. The numbers were collected by going in increments. Body fat from 0 to 4.9 and 5.0 to 9.9 and so on, going in increments of 5, and all the way up to 40. The outliers in the distribution are 0 to 5 and 35 to 40 percent body fat. The overall look of the distribution looks symmetrical but due to the higher 10 to 15 column it might be able to be argued that the data is skewed to the left. For the purpose of this analysis we will continue to assume that it is symmetrical until further

21) Seventy two percent of all observations fall within 1 standard deviation of the mean if the data is normally distributed.

4) Describe how the Mean and Stand Deviations were calculated (HINT: describe the steps in each process) The mean is calculated by summarizing all the results and dividing the result among the number of participants.

Students participating in a drug education program are given a drug awareness test at the beginning of the program.  The mean score for the population of 1,231 students is 61 with a standard deviation of 12.  As program director, you're curious as to how parents of the students perform on a drug awareness test and whether or not they are significantly different from the students.  Selecting a random sample of 50 parents, you administer the test and discover a mean score of 56.

3) No, the ages are not normally distributed because with the view of the histogram, it is not bell shaped.

Students Who Care is a student volunteer program in which college students donate work time in community centers for homeless people. Professor Gill is the faculty sponsor for this student volunteer program. For several years Dr. Gill has kept a record of the total number of work hours volunteered by s student in the program each semester. For students in the program, for each semester the mean number of hours was 29.1 hours with a standard deviation of 1.7 hours. Find an interval for the number of hours volunteered in which at least 88.9% of the students in this program would fit.

From this number above, we may estimate the Mean, median and standard deviation as next: Mean: 16.66;