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# Exam 2 Fall 05 - 1

By Dolores Mills,2014-05-27 19:12
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Exam 2 Fall 05 - 1

Stat 303: Section 505

Fall 2005

Form A

Instructor: Elizabeth Young

Instructions:

1.) Don’t EVEN open this until I tell you. (Meanwhile, read the rest of the instructions.)

2.) Be SURE to mark your Form on the scantron!

3.) Sign your name on the line on the scantron. With this signature, you agree to follow the Aggie Honor Code:

“An Aggie does not lie, cheat, or steal or tolerate those who do.”

4.) There are 20 multiple-choice questions, each worth 5 points. Please mark your answers CLEARLY with a #2 pencil. Multiple marks will be counted wrong.

5.) You have 50 minutes to finish this exam. It is worth 20% of your final grade.

6.) Good luck!

1, 8, 13, 16, and 17 have changed slightly!

1.) NC State posts the grade distributions for its courses online. Students in Stat 101 can make A’s, B’s, C’s, D’s, or F’s, and the students’ grades on a 4-point scale can be 4, 3, 2,

1, or 0. In 2003, the Stat 101 students’ grades had the following distribution:

0 1 2 3 4 X

0.01 0.05 0.30 0.44 0.20 P(X)

What are the mean and variance of X?

A.) 2.77 and 2.9

B.) 2.77 and .7371

C.) 2.0 and 1.581

D.) 2.0 and 2.5

E.) It cannot be determined from the information given.

2.) A chemical engineering student has developed a new insecticide for beetles. The old insecticide killed about 95% of beetles when used at a concentration level of 3% insecticide to 97% water. If the new insecticide also kills the same percentage of beetles when used at the same concentration level, what is the probability that 490 or more beetles will die when we spray the new pesticide on 500 beetles?

A.) 0.0016

B.) 0.0010

C.) 0.0020

D.) 0.9990

E.) 0.0931

3.) The weights of baskets of strawberries should be normally distributed with mean 18 oz and standard deviation 2 oz, but the machine that loads the baskets actually is loading them with a mean of 19 oz. You take a sample and find a p-value which is LARGER than your desired α level. Which of the following statements is TRUE?

A.) You made a Type II error: you rejected the null hypothesis when you shouldn’t have.

B.) You made a Type I error: you rejected the null hypothesis when you shouldn’t have.

C.) You didn’t make any error at all; you made the correct decision about whether to reject.

D.) If you increased your sample size, it would increase the power of the test.

E.) If you increased the probability of making a Type I error, then power would decrease.

4.) In one test for ESP (extrasensory perception), the experimenter looks at cards hidden from the subject. Each card contains a star, circle, wave, or square. As the experimenter looks at each card, the subject names the shape on the card. If the subject is randomly guessing, he should be correct 1 time out of 4 on average. We record 500 attempts for a subject. What is the probability, if our subject is just guessing, that at least 30% of his answers are correct?

A.) 0.9951

B.) 0.0136

C.) 0.0049

D.) 0.0098

E.) 0.0025

5.) Psychologists often measure the amount of time it takes rats to find their way through mazes. One psychologist believes that playing rap music to rats will increase the amount of time it takes the rats to complete the maze. The mean time for rats who have not had music played to them to complete the maze is 20 seconds; the researcher plays rap music to a sample of 50 rats and finds that their average time to complete the maze is 22 seconds. What are the psychologist’s null and alternative hypotheses?

A.) H: μ ? 22 H: μ > 22 oa

B.) H: x ? 22 H: x > 22 oa

C.) H: μ ? 20 H: μ < 20 oa

D.) H: μ ? 20 H: μ > 20 oa

E.) H: x ? 20 H: x > 20 oa

6.) The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to count the moths. When traps are checked periodically, the mean number of moths trapped is 0.6 with a standard deviation of 0.7. (Note that the number of moths trapped is strongly right-skewed.) If we take a sample of 50 traps, what is the probability we will catch an average of at least 0.63 moths in our traps?

A.) 0.3030

B.) 0.6179

C.) 0.3821

D.) 0.6970

E.) 0.4840

7.) John and Sally are preparing to take the SAT. (They don’t know each other, so their scores should be independent of one another.) On different days, they get different questions, so their scores are not always exactly the same. John’s scores are normally

distributed with a mean of 1320 and a standard deviation of 13. Sally’s scores are normally distributed with a mean of 1300 and standard deviation of 15. What is the probability that Sally scores higher on the SAT than John?

A.) 0.0781

B.) 0.3124

C.) 0.0244

D.) 0.8438

E.) 0.1562

8.) A hypothesis test found that the use of aspirin lowered heart attack rates, and the p-value found was 0.0325. (The null hypothesis was that patients on aspirin had the same heart attack rate as patients on a placebo drug, while the alternative hypothesis was that the patients on aspirin had a lower heart attack rate.) What does this p-value mean in context of the problem?

A.) If we were to repeat this study many times, the probability that we would make a Type II error over the long run would be 0.0325.

B.) The probability that we would reject the hypothesis that patients on aspirin had the same heart attack rate as patients on the placebo drug assuming that patients on aspirin actually had the same heart attack rate as patients on the placebo drug is 0.0325.

C.) The probability that this few people on aspirin or even fewer had heart attacks, assuming that people on aspirin had the same heart attack rate as patients on the placebo, is 0.0325.

D.) About 3.25% of the confidence intervals for the true proportion of patients on aspirin would include π.

E.) The probability that we would reject the null hypothesis assuming that patients on aspirin actually had a higher heart attack rate is 1 0.0325 = 0.9675.

29.) If X ~ N(3, 5) (i.e. the standard deviation is 5), what is the probability that (-1 ? X ?

2) ?

A.) -0.8000

B.) 0.2119

C.) 0.6000

D.) 0.4207

E.) 0.2088

10.) The calcium level in the blood of healthy young adults varies with mean 9.4 mg per deciliter and standard deviation 0.4 mg per deciliter. A clinic in rural Guatemala measures the blood calcium level of 160 healthy pregnant women at their first visit for prenatal care. The mean blood calcium level of the Guatemalan women is = 9.44. Is x

this an indication that the mean calcium level in pregnant women from Guatemala is different from 9.4? Test the hypotheses at alpha = 0.05:

H: μ = 9.4 o

H: μ ? 9.4 a

Find the p-value for the test, and make your conclusions.

A.) 0.9204, and we don’t have enough evidence to say that the mean calcium level for Guatemalan women is different from 9.4.

B.) 0.2076, and we don’t have enough evidence to say that the mean calcium level

for Guatemalan women is different from 9.4.

C.) 0.1000, and we can’t conclude that the mean calcium level for Guatemalan women is different from 9.4.

D.) 0.4602, and we can’t conclude that the mean calcium level for Guatemalan

women is different from 9.4.

E.) 0.1038, and we don’t have enough evidence to say that the mean calcium level for Guatemalan women is different from 9.4.

11.) A veterinarian is interested in the Vitamin C content of a variety of supplemented alfalfa hay. Suppose that the Vitamin C content in the population has standard deviation σ = 8 mg per gram. A sample of 15 cuttings has mean Vitamin C content 152 mg per gram. What is a 95% confidence interval for mean Vitamin C content? If I were to run a hypothesis test that the true mean Vitamin C content was 148 mg per gram against the alternative that the mean Vitamin C content was different from 148 mg per gram, would I reject it at α = 0.05?

A.) (144.60, 151.40). Yes.

B.) (144.60, 151.40). No.

C.) (147.95, 156.05). Yes.

D.) (147.95, 156.05). No.

E.) (142.68, 153.32). No.

12.) Students in the Greene County school district have the following racial distribution (percentages are the percentages of students of that race):

Race Asian/Pacific Hispanic Middle Eastern White / Non-Hispanic Other

Percentage 7% ? 1% 61% 2%

What is the probability that TWO randomly selected students are BOTH Hispanic?

A.) 0.29

B.) 0.58

C.) 0.0841

D.) 0.3721

E.) The events are not disjoint, so it cannot be determined.

13.) What is the best interpretation of the confidence interval in problem 11?

A.) If I were to repeat the study by taking multiple samples of 15 cuttings many times, about 95% of the time, I would capture the true mean Vitamin C content.

B.) About 95% of the confidence intervals that I create will contain the mean 152 mg per gram of Vitamin C.

C.) The probability that the interval I selected above contains the true mean Vitamin C content is 0.95.

D.) 95% of the time, the true mean μ will fall in the interval I selected above.

E.) The probability that the interval I selected above contains 152 is 0.95.

14.) Sue is trying to estimate the mean time for rats to go through a new maze that she created. The standard deviation for times to go through similar mazes is σ = 2 seconds, and she thinks that the standard deviation for her maze will be similar. She wants to estimate the time to go through her new maze to within 1 second with 95% confidence. How large a sample does she need to take?

A.) 15

B.) 16

C.) 10

D.) 26

E.) 11

15.) To determine whether the mean concentration of bottles of peroxide is actually greater than the advertised value of 0.04 ml, a researcher takes a sample of 40 bottles. In testing the null hypothesis H: μ ? 0.04 against the alternative H: μ > 0.04, he finds a oa

test statistic z = 2.53. What is the p-value for his test, and what is his conclusion?

A.) 0.9943, and he concludes that the true mean peroxide concentration is not greater than 0.04.

B.) 0.0114, and he concludes that the true mean peroxide concentration is not greater than 0.04

C.) 0.0057, and he concludes that the true mean peroxide concentration is not greater than 0.04.

D.) 0.0114, and he concludes that the true mean peroxide concentration is greater than 0.04.

E.) 0.0057, and he concludes that the true mean peroxide concentration is greater than 0.04.

2216.) Suppose that X ~ N(8, 4) and Y ~ N(10, 3). (So 4 and 3 are the standard deviations

of X and Y, respectively.) What is the distribution of X + Y?

2 A.) N(18, 7) 2 B.) N(2, 7) 2 C.) N(18, 5) 2 D.) N(18, 6) 2 E.) N(18, 8)

17.) Suppose that a weight is known to weigh 0.30 grams, but students’ measurements are not exactly accurate; in fact, they vary according to a normal distribution with mean 0.30 and standard deviation 0.041. Dr. Obsessive requires his students to measure things 10 times and report the average of their 10 measurements. What is the sampling distribution of the students’ averages of the 10 measurements?

A.) ~ N(0.30, 0.090)

B.) ~ N(0.30, 0.002)

C.) ~ N(0.30, 0.041)

D.) ~ N(0.30, 0.013)

E.) ~ N(0.30, 0.004)

18.) In Dr. Obsessive’s class (refer to question above), Julie needs to create a 99% confidence interval for the true mean weight. The average of her 10 measurements is 0.303. What is her confidence interval? Does it contain the true mean?

A.) (0.270, 0.336); Yes

B.) (0.270, 0.336); No

C.) (0.267, 0.333); Yes

D.) (0.267, 0.333); No

E.) (0.278, 0.328); Yes

19.) Which of the following is / are true?

A.) All parameters have a sampling distribution, since they vary from sample to sample.

B.) Parameters are usually unknown and we test hypotheses concerning parameters as a way to find out what they are.

C.) Random chance will cause a parameter to vary from sample to sample.

D.) Two of the above are true.

E.) None of the above is true.

20.) Birth weights for babies born at 8 months into the pregnancy vary according to a normal distribution with mean 7 lbs and standard deviation 0.4 lbs. Jack wants to know thwhether his newborn son, born at 8 months, is above the 75 percentile for babies born thththis early. What is the 75 percentile of birth weights? (Hint: the 75 percentile is Q, 3

and we found that Q for the standard normal was 0.67 on the homework.) 3

A.) 15.825 lbs.

B.) 2.956 lbs.

C.) 6.732 lbs.

D.) 7.268 lbs.

E.) 7.508 lbs.