Instructor S Name STP 231 Test 2A Fall 2011

Instructor’s name STP 231 Test 2A Fall 2011

STP 231 Statistics for Life Science Majors

Instructor’s Name Print Name______

Test #2 Fall 2011

Honor Statement:

By signing below you confirm that you have neither given nor received any unauthorized assistance on this exam. This includes any use of a graphing calculator beyond those uses specifically authorized by the Mathematics Department and your instructor. Furthermore, you agree not to discuss this exam with anyone until the exam testing period is over. In addition, your calculator’s memory and menus may be checked at any time and cleared by any testing center proctor or Mathematics Department instructor.

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Instructions: a.)The exam is worth a total of 102 points; please make sure your exam has all pages before you begin. Problems 1-9 are worth 8 points and problems 10-14 are worth 6 points. Make sure that you have all 11 pages to your exam. If you do not have 11 pages, then contact the student worker at the check out desk. b.) Show all work in detail or your answer will not receive any credit. Include appropriate units on all questions that apply. Write neatly and box all answers. c.) Please ask the student worker at the check out desk if you need scratch paper. d.) No calculators or computers that do symbolic algebra, like the Casio FX-2, TI-89, or TI- 92, may be used. e.) The formulas and tables are shown on the last page. You may take these pages off to help you during the exam f.) Part 1 of the exam is for free response. Show your work or explain the process for each problem to receive credit. g.) Part 2 is the multiple choice part of the exam. A table is given on page three for you to write in the letter for the correct answer. You may take the first page off to help you write the correct letter in for each answer. h.)You may have one 4 by 6 note card. It must not contain any worked out problems. It must be attached to the back of your test when the exam is turned in.

For Part I – Free Response. Show all work (as described on page 1).

1. A children’s shoe manufacturer is interested in estimating the population mean length of fourth grader’s feet. A random sample of 20 fourth graders’ feet yielded a 23.095 centimeters. Assume that s = 1.280 centimeters. Construct a 95 percent confidence interval for  the population mean length of all fourth graders’ feet.

2. Human body temperature. Summary statistics for female versus male body temperatures in Fahrenheit degrees for two independent samples of adult Americans are given below:

Sample Gender Sample size Sample mean body temperature Sample standard deviations 1 Females 65 98.394 0.743 2 Males 65 98.105 0.699

SE  0.1265 If the degrees of freedom is 64, and the standard error Y1Y2 then find the 95% confidence interval for the difference between the population means 1  2

FA11 T2 Form A © 2011 Arizona State University, Department of Mathematics and Statistics 3 Instructor’s name STP 231 Test 2A Fall 2011 For Part II – Multiple Choice: Circle your answer choice on the exam AND fill in the answer with the letter of the answer that you believe is the correct answer.

Problem Letter of Problem Letter of Problem Letter of Problem Letter of Number Answer Number Answer Number Answer Number Answer 3. 7. 10. 13.

4. 8. 11. 14.

5. 9. 12.

6. XXXXX XXXX

Circle the correct letter 3. Find the z0.09

A. -1.28 B. -1.34 C. 1.34 D. 1.28 E. None of the these

Use the following information for problems 4-6 to find the requested information According to research data based on weather, age, and course, the time that it takes a person to run a 5K is normally distributed with a mean of 35.0 minutes and a standard deviation of 3.16 minutes.

4. What would be the percentage of people that would run the 5K between 32 and 37 minutes?

A. 45.9 % B. 56.5 % C. 32.0 % D. 76.6% E. None of these

5. If a group of 4 individuals are chosen at random, what percentage of the group mean times would be less than 33 minutes?

A. 36.7 % B. 63.3 % C. 26.4 % D. 10.2% E. None of these

6. According to research data based on weather, age, and course, the time that it takes a person to run a 5K is normally distributed with a mean of 35.0 minutes and a standard deviation of 3.16 minutes. If 96.56% of runners finish within 31 and 38 minutes. What is the probability that for a random sample of 9 individuals, that all individuals finish within 31 to 38 minutes?

A. 0.2202 B. 0.0251 C. 0.7298 D. 0.9656 E. None of these

7. Cholesterol levels in men 18 to 24 years of age are normally distributed with a mean of 178 and a standard deviation of 41. What cholesterol level corresponds to the 90th percentile, the level at which treatments may be necessary?

A. 230.48 B. 125.52 C. 211.45 D. 268.00 E. None of the these

Find the necessary sample size. 8. The effects of various diets on the weight gain of young turkeys over a third week period were studied. How many turkeys should be studied if the standard deviation of previous experiments was 90 grams and the standard error of the group mean should be no more than 30 grams?

A. 3 B. 4 C. 6 D. 9 E. None of these

Find the requested information 9. A researcher was interested in comparing the GPAs of students at two different colleges. Indepen- dent normally distributed random samples of 8 students from college A and 13 students from col-

lege B yielded the following GPAs. y 1 = 3.1125, y 2 = 3.4385, s1 = 0.4357, s2 = 0.5485, n1 = 13, SE n2 = 8. Find the standard error for y1- y 2

A. 0.32 B. 0.052 C. 0.10 D. 0.23 E. None of these

Determine the correct answer 10. The mean annual income for adult women in one city is $32,520 and the standard deviation of the incomes is $5,190. The distribution of incomes is skewed to the right. For samples of size 50, which of the following statements best describes the sampling distribution of the mean?

A. Y is uniformly distributed. B. Y is approximately normally distributed or is normally distributed. C. The distribution of Y is skewed to the right. D. Nothing can be said about the distribution of Y .

11. Which statements are true about Normal Probability Plots?

1. Normal probability plots are used to determine if the data is normally distributed. 2. Normal probability plots are used to determine if the data contains outliers. 3. The data points are ranked from smallest to largest and are then plotted versus their normalized scores. 1 1 4. Transformations on the variable such as y, log Y , , can be performed to transform non- Y Y normal distributions to normal distributions

A. 1, 2, 3 B. 1, 2, 3, 4 C. 2, 3, 4 D. 1, 3, 4 E. 1, 2, 4

12. Which of the following statements are true about sampling distributions?

1. The sample size affects the shape of the sampling distribution. 2. The mean of all of the samples in the sampling distribution is not affected by the sample size. 3. As the sample size increases, the sample means are closer to the population mean. 4. The calculation for the Z-score is the same for a variable Y as it is for the sampling distribution for Y .

5. Y is the only parameter that can be estimated by a sampling distribution

A). 1, 3, 4 B) 1, 2, 3 C) 1, 2, 4, 5 D) 1, 2, 4, 5 E) 1, 3, 4, 5

13. Which of the following statements regarding t-curves is/are true? 1. The t-curve with 10 degrees of freedom is flatter and wider than the standard normal curve. 2. The total area under a t-curve with 10 degrees of freedom is greater than the area under the standard normal curve. 3. The t-curve with 10 degrees of freedom more closely resembles the standard normal curve than the t-curve with 20 degrees of freedom.

A) 2, 3 B) 3 C) 1 D) 2 E) 1, 3 F) None of these

14. Determine which of the following statements about confidence intervals are true. From a sample size of n = 106 adults, the body temperature was measured. If the 95% confidence interval for the body temperature is 98.08o F < m < 98.32 o F .

1. We are 95% confident that the average body temperature of the all of the individuals in the population is between 98.08˚F and 98.32˚F 2. We are 95% confident that the average body temperature of the 106 individuals in the sample is between 98.08˚F and 98.32˚F 3. 95 % confidence interval is wider than a 99 % confidence interval

A) 2, 3 B) 3 C) 1 D) 2 E) 1, 3 F) None of these

Formula Sheet

Descriptive Measures Binomial Probability y y-m n! y = z = nCj = n s (n- j )! j ! 2 y ( y- y ) m= s = i N n Pr{Y= j }= nCj� pj (1 p ) n- j

Probability Rules P(E c )  1 P(E)

P(E1 or E2 )  P(E1 )  P(E2 )  P(E1 and E2 )

P(E1 and E2 ) For any probabilities P( E1 E2 )  P(E1 )

If P(E1 ) and P(E2 ) are independent, then P(E1 and E2 )  P(E1 )  P(E2 )

Sampling Distribution of the Sample Mean my = m s s = y n

y - m z = s n

Confidence Interval for One Population Mean s s Guessed SD y弊 t with df = n -1 SE = Desired SE = a / 2 ny n n

s2 s 2 Standard Error of y- y SE = 1+ 2 1 2 ( y1- y 2 ) n1 n 2 2  s2 s2   1 2      n1 n2  df    2 2 Degrees of Freedom for 1  2  s2   s2   1   2   n   n   1    2  n1 1 n2 1

2 2 s1 s2 Nonpooled confidence interval for 1  2 : y1  y2   t / 2  n1 n2