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[LATEST UPDATE] MATH 110 Module 1-10 Exams with Answer Key - Portage Learning | 100% GUARANTEED PASS. Module 1 Exam 1 1.Define each of the following: a) Observation b) Element c) Variable Observation- all the information collected for each element in a study Element- in a data set, the individual and unique entry about which data has been collected, analyzed and presented in the same manner Variable- a particular, measurable attribute that the researcher believes is needed to describe the element in their study. 2.Explain outliers An outlier is a value which is out of place compared to the other values. It may be too large or too small compared to the other values 3.Look at the following data and see if you can identify any outliers: 53 786 789 821 794 805 63 777 814 2333 783 811 795 788 780 Outliers: 53 63 2333

a) How many were burgers? b) How many were fish? a) Burgers, 2900(0.12)=

b) Fish, 2900(0.28)=

Mod 2 Exam

1. During an hour at a fast food restaurant, the following types of sandwiches are ordered: Turkey Turkey CheeseburgerHamburger Fish Chicken Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Chicken Fish Turkey Fish Hamburger Fish Cheeseburger Fish Cheeseburger Hamburger Fish Fish Cheeseburger Hamburger Fish Turkey Turkey Chicken Fish Chicken Cheeseburger FishTurkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken Turkey Fish Hamburger Chicken Fish a) Make a frequency distribution for this data. b) Make a relative frequency distribution for this data. Include relative percentages on this table. a. b.
2. Consider the following data: 430 389 414 401 466 421 399 387 450 407 392 410 440 417 471

Find the 40th percentile of this data.

3. Consider the following data: {29, 20, 24, 18, 32, 21} a) Find the sample mean of this data. b) Find the range of this data. c) Find the sample standard deviation of this data. d) Find the coefficient of variation. a.

4. Suppose that you have a set of data that has a mean of 49 and a standard deviation of 8. a) Is the point 57 above, below, or the same as the mean. How many standard deviations is 57 from the mean. b) Is the point 33 above, below, or the same as the mean. How many standard deviations is 33 from the mean. c) Is the point 31 above, below, or the same as the mean. How many standard deviations is 31 from the mean. d) Is the point 79 above, below, or the same as the mean. How many standard deviations is 79 from the mean. a) The data point 57 is above the mean. Now use the z-score to determine how many standard deviations 57 is above the mean. We are told that the mean is 49 and the standard deviation is 8. So, the z-score is given by: The z-score is 1, so the data point 57 is 1 standard deviation above the mean. b) The data point 33 is below the mean. Now use the z-score to determine how many standard deviations 33 is below the mean. We are told that the mean is 49 and the standard deviation is 8. So, the z-score is given by: The z-score is -2, so the data point 33 is 2 standard deviations below the mean (the negative sign indicates that the point is below the mean). c) The data point 31 is below the mean. Now use the z-score to determine how many standard deviations 31 is below the mean. We are told that the mean is 49 and the standard deviation is 8. So, the z-score is given by: The z-score is -2.25, so the data point 31 is 2.25 standard deviations below the mean (the negative sign indicates that the point is below the mean). d) The data point 79 is above the mean. Now use the z-score to determine how many standard deviations 79 is above the mean. We are told that the mean is 49 and the standard deviation is 8. So, the z-score is given by: The z-score is 3.75, so the data point 79 is 3.75 standard deviations above the mean.

5. Consider the following set of data: {20, 5, 12, 29, 18, 21, 10, 15} a) Find the median. b) Find the mode of this set. a) In order to find the median, we must first put the numbers in ascending order: 5, 10, 12, 15, 18, 20, 21, 29. Notice that there are two “middle” numbers, 15 and 18. The median is the average of these two numbers. Median = (15+18)/2 = 16.5. b) No number occurs more than once, so there is “no mode”.

Exam 3

1. Find the answer to each of the following by first reducing the fractions as much as possible: a) P(412,3)= b) C(587,585)=
2. Suppose you are going to make a password that consists of 5 characters chosen from {1,2,4,9,d,i,k,m,n,w,z}. How many different passwords can you make if you cannot use any character more than once in each password?

3. Suppose A and B are two events with probabilities: P(A)=.35, P(Bc^ )=.45, P(A∩B)=.25. Find the following: a) P(A𝖴B). b) P(Ac^ ). c) P(B). a. For P(A 𝖴 B). Use P(A 𝖴 B)=P(A)+P(B)-P(A∩B). But for this equation, we need P(B) which we can find by using P(B)=1-P(Bc^ ). So, P(B)=1-.45= .55. P(A 𝖴 B)=.35+.55-.25=. b. For P(Ac^ ). Use P(A)=1-P(Ac^ ) which may be rearranged to (Ac^ )=1-P(A). P(Ac^ )=1-.35=.65. c. For P(B). Use (B)=1-P(Bc^ ). P(B)=1-.45=.55.
4. Suppose A and B are two events with probabilities: P(Ac^ )=.50, P(B)=.65, P(A∩B)=.30. a) What is (A│B)? b) What is (B│A)?

5. In a manufacturing plant, three machines A, B, and C produce 30 %, 20 %, and 50 %, respectively, of the total parts production. The company's quality control department determined that 3 % of the parts produced by machine A, 2.5 % of the parts produced by machine B, and 4 % of the parts produced by machine C are defective. If a part is selected at random and found to be defective, what is the probability that it was produced by machine B?
6. The probability that a certain type of battery in a smoke alarm will last 3 years or more is .85. The probability that a battery will last 7 years or more is .25. Suppose that the battery is 3 years old and is still working, what is the probability that the battery will last at least 7 years?
7. Suppose that 7 out of 17 people are to be chosen to go on a mission trip. In how many ways can these 7 be chosen if the order in which they are chosen is not important.

Exam 4

1. In a large shipment of clocks, it has been discovered that 21 % of the clocks are defective. Suppose that you choose 7 clocks at random. What is the probability that 2 or less of the clocks are defective.
2. Find each of the following probabilities: (use standard normal distribution table to get z-score) a. Find P(Z ≤ 1.27). b. Find P(Z ≥ -.73). c. Find P(-.09 ≤ Z ≤ .86). a. P(Z ≤ 1.27) =0. b.P(Z ≥ -0.73= 1- P(Z ≤ -0.73)=1- 0.23270=0. c. P(-0.09 ≤ Z ≤0 .86)= P(Z≤0 .86)- P(Z≤ - 0.09) 0.80511-0.46414=.
3. A company manufactures a large number of rods. The lengths of the rods are normally distributed with a mean length of 4.0 inches and a standard deviation of .75 inches. If you choose a rod at random, what is the probability that the rod you chose will be: a) Less than 3.0 inches? b) Greater than 3.7 inches? c) Between 3.5 inches and 4.3 inches?

a. We must find the z-score for x=3.0: So, we want P(Z ≤ -1.33). From the table, we find. P(Z ≤ - 1.33)=.09176. b. We must find the z-score for x=3.7: So, we want P(Z ≥-.4). Since this is greater than, we must use: P(Z ≥-.4)=1.0-P(Z ≤-.4)=1- .34458= .65542. c. We must find the z-score for x=3.5: and the z-score for x=4.3: So, we want P(-.67 ≤ Z ≤ .4) P(-.67 ≤ Z ≤ .4)=P(Z≤ .4)-P( Z ≤ -.67). P(-.67 ≤ Z ≤ .4)=.65542-.25143=.40399.

4. A life insurance salesperson expects to sell between zero and five insurance policies per day. The probability of these is given as follows: Policies Sold Per Day Probability, f(x)

Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data.

5. An archer is shooting arrows at a target. She hits the target 68% of the time. If she takes 15 shots at the target, what is the probability that she will hit the target exactly 12 times?

Exam 5

1. Suppose that you take a sample of size 20 from a population that is not normally distributed. Can the sampling distribution of x̄ be approximated by a normal probability distribution? No because the sample has to be at least 30 to use sampling distribution of x̄ or be normally distributed.
2. Suppose that you are attempting to estimate the annual income of 2000 families. In order to use the infinite standard deviation formula, what sample size, n, should you use? Your Answer: n N ≤ 0. n 2000 ≤ 0. n ≤ 0.05(2000)= Sample size must be less than 100
3. Suppose that in a large hospital system, that the average (mean) time that it takes for a nurse to take the temperature and blood pressure of a patient is 150 seconds with a standard deviation of 35 seconds. What is the probability that 30 nurses selected at random will have a mean time of 155 seconds or less to take the temperature and blood pressure of a patient? We calculate the standard deviation of the sample distribution: Calculate the z-score: So, we want to find P(Z < .782) on the standard normal probability distribution table. Recall that P(Z < .78) = .78230. Therefore, there is a 0.78230 probability that a simple random sample of 30 nurses will have a mean time of 155 seconds or less.

4. Suppose that in a very large city 9.8 % of the people have more than two jobs. Suppose that you take a random sample of 70 people in that city, what is the probability that 9 % or more of the 70 have more than two jobs? Now we find the z-score: We want P(Z>-0.23). From the standard normal table, we find: P(Z>-.23)=1- P(Z<-.23)=1-.40905=.59095. So there is a .59095 probability that the percentage of the sample that have more than two jobs is more than 9 %.

Exam 6

1. A new drug for migraine headaches has been introduced to the market. You would like to know if migraine patients prefer the old drug or the new drug. So, you sample 190 patients that have used both drugs and you find that 52% of the sampled patients prefer the new drug, 48% of the patients prefer the old drug. Find the 90% confidence limit for the proportion of all patients that prefer the new drug. Can you be 90% confident that a majority of all the patients prefer the new drug? p=.52 n = 190 Based on a confidence limit of 90 %, we find in table 6.1 that z=1. So, the 90% confidence limit is: Notice that the proportion that like the new drug may be as small as .46 which is less than the .5. So, we cannot be 90% confident that a majority of all the patients will prefer the new drug.

2. Suppose that you are a nurse and you are assigned to do checkups of people one day per week in a certain village. You have a total of 300 patients in the village. You have the option of doing the checkups in the mornings or in the afternoons. Therefore, you ask 35 patients and find that 62% prefer afternoon appointments while 38% prefer morning appointments. Find the 95% confidence limit for the proportion of all patients that prefer afternoon appointments. Since 62% prefer afternoon, we set P = .62. As we mentioned previously, we estimate p by P. So, p=.62. The total population is 300, so set N=300. A total of 35 patients were surveyed, so Based on a confidence limit of 95 %, we find in table 6.1 that z=1.96. Now, we can substitute all of these values into our equation: So the proportion of the total who prefer afternoon appointments is between .469 and .771.

Exam 7

1. a) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error. The newspaper in a certain city had a circulation of 15,000 per day in 2010. You believe that the newspaper’s circulation is more than 15,000 today. b) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error. A certain website had 3500 hits per month a year ago. You believe that the number of hits per month is less than that today. a. H 0 :^ μ^ =15,000^ circulation H 1 : μ >15,000 circulation Type I error: Reject the null hypothesis that the mean of circulations is 15,000 even though it is correct. Type II error: Do not reject the null hypothesis when the mean of circulations is greater than 15, circulations. b. H 0 :μ^ =3500^ hits H 1 :μ <3500 hits Type I error: Reject the null hypothesis even though the hits per month in a year are at least 3500. Type II error: Do not reject the null hypothesis when the mean of hits per month is less than 3500.

2. Suppose that we have a problem for which the null and alternative hypothesis are given by: H 0 : μ=322. H 1 :μ≠ 322. Is this a right-tailed test, left-tailed test, or two-tailed test. Find the z value based on a level of significance of .06. Two-tailed test P(Z<z)=0.06/2=0.03 and P(Z>z)=0.06/2=0. z=-1.88 and z=1.
3. It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each day. The standard deviation of the population is estimated to be 12 milligrams per day. A doctor is concerned that her pregnant patients are not getting enough vitamin C. So, she collects data on 45 of her patients and finds that the mean vitamin intake of these 45 patients is 81 milligrams per day. Based on a level of significance of α = .02, test the hypothesis. H 0 : μ=85 milligrams per day. H 1 : μ<85 milligrams per day. This is a left-tailed test, so we must find a z that satisfies P(Z<z)=.02. In the standard normal table, we find z. 02 = -2.05. For a left-tailed test, we will reject the null hypothesis if the z-score is less than -2.05. We find the z-score: Notice that since the z-score is less than -2.05, we reject the null hypothesis.
4. A mayor claims that the unemployment rate in his city is 4 %. Many people think that the unemployment rate is higher. So, 95 residents of the city are contacted and it is found that 8 of them are unemployed. Can the mayor’s claim be supported to a level of significance of α = .02, test the hypothesis. H 0 : p=.04. H 1 : p>.04. Since this is a right-tailed test. z=1-0.02=0.98=2.05. Since this is a right-tailed test, and the z-score is greater than 2.05, we reject the null hypothesis.

Exam 8

Suppose we have independent random samples of size n 1 = 420 and n 2 = 510. The proportions of success in the two samples are p 1 = .38 and p 2 = .43. Find the 99% confidence interval for the difference in the two population proportions. Answer the following questions:

1. Multiple choice: Which equation would you use to solve this problem? A. B. C. D.
2. List the values you would insert into that equation.
3. State the final answer to the problem From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2: B. So, the interval is (.-0.1333, - 0.03326).

2. In certain hospital, nurses are required to constantly make rounds to check in on all of the patients. The nursing supervisor would like to know if there is a difference between the number of rounds completed per shift by the nurses on the day shift compared to the nurses on the night shift. So, the nursing supervisor checks the records of 70 day shift nurses and finds that they complete an average (a mean) of 30 rounds per shift with a standard deviation of 4.6 rounds per shift. The nursing supervisor also checks the records of 84 night shift nurses and finds that they complete an average (a mean) of 25 rounds per shift with a standard deviation of 5. rounds per shift. a) Find the 90% confidence interval for estimating the difference in the population means (μ 1 - μ 2 ). b) Can you be 90% confident that there is a difference in the means of the two populations? Answer the following questions:
3. Multiple choice: Which equation would you use to solve this problem? A. B. C. D.
4. List the values you would insert into that equation.
5. State the final answer to the problem 90% confidence corresponds to z=1.645. n 1 =70, n 2 =84, s 1 =4.6, s 2 =5.7, x̄ 1 =30, x̄^2 =25^ A. b) Since the entire confidence interval is positive, we can be 90 % sure that there is a difference in the means of the two populations.

3. A head librarian supervises a number of libraries in a large county. He wants to know if full-time library workers and part-time library workers re-shelve books at the same rate. So, he checks the records of 40 full-time library workers and finds that they re-shelve an average of 185 books per hour with a standard deviation of 17. books per hour. The records of 40 part-time library show that they re-shelve an average of 190 books per hour with a standard deviation of 9.2 books per hour. Using a level of significance of α=.10, is there enough evidence to indicate a difference in the mean number of books re-shelved by full-time workers compared to part-time workers? Answer the following questions:
4. Multiple choice: Which equation would you use to solve this problem? A. B. C. D.
5. List the values you would insert into that equation.
6. State the final answer to the problem H 0 : μ 1 - μ 2 = 0 H 1 : μ 1 - μ 2 ≠0. Since this is a two-tailed test, we must find the z that satisfies: P(Z<z)=.1/2=.05 and P(Z > z)=.1/2=.05. In the standard normal table, z=-1.645 and z=1.645. We will reject the null hypothesis if the z-score is less than -1.645 or the z-score is greater than 1.645. We now find the z-score: Since the z-score is between -1.645 and 1.645, we do not reject the null hypothesis.

4. Consider the following dependent random samples Observations 1 2 3 4 5 6 x-values 8.1 7.6 8.3 8.4 7.9 7. y-values 8.4 8.4 8.5 8.9 8.1 7. a) Determine^ the^ difference^ between^ each^ set^ of^ points,^ xi -^ yi b) Do^ hypothesis testing^ to^ see^ if^ μd <^0 at^ the^ α^ =^ .025. Since we are testing whether or not μd < 0, then our null and alternate hypothesis will be set as follows: H 0 : μd = 0 H 1 : μd < 0 n=6. This is a left-tailed test. Note that for t. 025 = -2.571 for 6-1 = 5. We find the mean in the usual way: The sample standard deviation is given by: Then using the mean, d = -.4333, and the standard deviation, sd= .2422, that we found above: Since t < t. 025 , we reject the null hypothesis.
5. A new energy drink is supposed to improve a person’s time in the one mile run. The times, in seconds, of eight runners with and without the drink are given below: Runner 1 2 3 4 5 6 7 8 x-time (before) 254 276 276 265 271 273 268 281 y-time (after) 265 269 277 279 266 273 275 279 Find the 95 % confidence interval for mean of the differences, μd. Answer the following questions:

1. Multiple choice: Which equation would you use to solve this problem? A. B. C. D.
2. List the values you would insert into that equation.
3. State the final answer to the problem Note that n=8. We will define , di = xi - yi. After doing the appropriate calculations, we find that d =-2. sd= 7.558. When we look at the student’s t chart for 95% confidence (the 95% is found along the bottom row of the chart) and DOF=8-1=7 (the df=7 is found in the leftmost column) we find that t=2.365. Then D.

Exam 9

1. Suppose you have 45 data points and you calculate the sample correlation coefficient and find that r = .32. Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer. Note that for n=45 and 95% we get a value from the chart of .29396. The absolute of r is |r|=.32, which is above .29396. So a positive linear relation exists.
2. Suppose you have 60 data points and you calculate the sample correlation coefficient and find that r = .20. Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer. Note that for n=60 and 95% we get a value from the chart of .25420. The absolute of r is |r|=.20, which is below .25420. So no linear relation exists.
3. Compute the sample correlation coefficient for the following data: Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer. n=5 r=-0.966 IrI=0.966 c=0.87834 x=6.4 y=12 Sx=2.702 Sy=3. The absolute value of negative r is greater than the critical value correlation coefficient. There is a negative linear relation between the variables.
4. Find the best fit line for the following data:

r = .9910 Sx = 4.2 Sy = 5.7. y= 1.345 x +1.123.

Exam 10

1 Find the value of X^2 for 17 degrees of freedom and an area of .005 in the right tail of the chi-square distribution. Look across the top of the chi-square distribution table for .005, then look down the left column for 17. These two meet at X^2 =35.718.

2. Find the value of X^2 for 10 degrees of freedom and an area of .005 in the left tail of the chi-square distribution. Since the chi-square distribution table gives the area in the right tail, we must use 1 - .005 = .995. Look across the top of the chi-square distribution table for .995, then look down the left column for 10. These two meet at X^2 =2.156.
3. Find the value of X^2 values that separate the middle 80 % from the rest of the distribution for 9 degrees of freedom. In this case, we have 1-.80=.20 outside of the middle or .20/2 = .1 in each of the tails. Notice that the area to the right of the first X^2 is .80 + .10 = .90. So we use this value and a DOF of 9 to get X^2 = 4.168. The area to the right of the second X^2 is .10. So we use this value and a DOF of 9 to get X^2 = 14.684.
4. Find the critical value of F for DOF=(4,17) and area in the right tail of. In order to solve this, we turn to the F distribution table that an area of .05. DOF=(4,17) indicates that degrees of freedom for the numerator is 4 and degrees of freedom for the denominator is 17. So, we look up these in the table and find that F=2.96.

5. The mayor of a large city claims that 30 % of the families in the city earn more than $ 100,000 per year; 52 % earn between $ 30,000 and $ 100,000 (inclusive); 18 % earn less than $ 30,000 per year. In order to test the mayor’s claim, 285 families from the city are surveyed and it is found that: 90 of the families earn more than $ 100,000 per year; 135 of the families earn between $ 30,000 and $ 100,000 per year (inclusive); 60 of the families earn less $ 30,000. Test the mayor’s claim based on 5 % significance level. We will set H 0 : The mayor’s distribution is correct. H 1 : The mayor’s distribution is not correct. This is a multinomial experiment, for multinomial experiments, we use the chi-square distribution. Calculate the degrees of freedom for three possible outcomes: DOF=3-1=2. Our level of significance is 5 % (.05). So look up DOF of 2 and .05 on the Chi-square distribution table to get 5.991. For the expected frequencies, we will use Ei = npi. So, the expected frequencies (of the n=285 in the sample) based on the mayor’s distribution: This is smaller than the critical value of 5.991. Therefore, we do not reject the null hypothesis.

6. A trucking company wants to find out if their drivers are still alert after driving long hours. So, they give a test for alertness to two groups of drivers. They give the test to 395 drivers who have just finished driving 4 hours or less and they give the test to 565 drivers who have just finished driving 8 hours or more. The results of the tests are given below. Passed Failed Drove 4 hours or less 290 105 Drove 8 hours or more 350 215 Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the 1 % level of significance. H 0 : Driving hours and alertness are independent events. H 1 : Driving hours and alertness are not independent events. We have two rows and three columns, so # of Rows =2 and # of Columns=2. The degrees of freedom are given by: DOF = (# of Rows-1)(# of Columns-1)=(2-1)(2-1)=1. Using this, along with .01 (for the 1% level of significance) we find in the chi-square table a critical value of 6.635. This value is greater than the critical value of 6.635. So, we reject the null hypothesis.