PORTAGE LEARNING MATH 110 Final Exam - 100% CORRECT ANSWERS 2022

MATH 110 Final Exam - 100% CORRECT ANSWERS 2022 Final Exam- Requires Respondus LockDown Browser  Due No due date  Points 100  Questions 10  Time Limit 120 Minutes  Requires Respondus LockDown Browser This quiz is currently locked. Attempt History Attempt Time Score LATEST Attempt 1 117 minutes 95 out of 100 Score for this quiz: 95 out of 100 Submitted Aug 4 at 5:20pm This attempt took 117 minutes. Question 1 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) The following pie chart shows the percentages of total items sold in a month in a certain fast food restaurant.A total of 4700 fast food items were sold during the month. a.) How many were fish? b.) How many were french fries? Your Answer: For this problem: a. For Fish there were: 4700 ( .28)=1316, hovewer for Franch Fries they were: 4700 (.40)=1880 a.) Fish : 4700(.28) = 1316 b.) French Fries: 4700(.40) = 1880 Question 2 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Consider the following data: 51 56 73 48 55 Find the 40th percentile of this data. Your Answer: For this problem:n= 20, since there are 20 numbers. The numbers in order are: Since we are looking for 40th percentile I must calculate: Finally, 40 th percentile is the 8th abservation, where the 8th abservation is number 46. There are a total of twenty numbers, so n= 20. In order to find the percentiles, we must put the numbers in ascending order: For the 40th percentile: Therefore, the 40th percentile index for this data set is the 8th observation. In the list above, the 8th observation is 46. Question 3 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) In a tri-state conference, 40% attendees are from California, 10% from Oregon, and 50% from Washington. As it turns out 8 % of the attendees from California, 11% of the attendees from Oregon, and 13% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from California? Your Answer: For this problem: P(TrainICA)=.08 P(TrainIOR)=.11 P(Train)IWA)=.13 Wheras: P(CA)=.40; P(OR)=.10 and P(WA)=.50. The question is What is the probability that the person is from CA( California)?P(Train│C)=.08.. P(Train│O)=.11.. P(Train│W)=.13.. P(C)=.40,P(O)=.10,P(W)=.50. We want to find P(C│Train), so use: Question 4 5 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) Find each of the following probabilities: a. Find P(Z ≤ .17). b. Find P(Z ≥ -.34) . c. Find P(-1.14 ≤ Z ≤ 0.55). Your Answer: For this problem: While using a normal table for all of the problem mentined: a. The answer is: 0.5675 b. The answer is: c. The answer is:a. P(Z ≤ .17)=.56749. b. P(Z ≥ -.34)=1- .36693= .63307. c. P(-1.14 ≤ Z ≤ 0.55)= .70884- .12714=.5817 . As noted in earlier exams, you should use the values as they are given in the charts and tables provided. Question 5 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) Suppose that you are attempting to estimate the annual income of 1200 families. In order to use the infinite standard deviation formula, what sample size, n, should you use? Your Answer: For this problem: I must have : Therefore the smaple size, less/eaqual than 60 In order to use infinite standard deviation formula, we should have:n≤0.05(1200) n≤60 So, the sample size should be less than 60. Question 6 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) T Table (Links to an external site.) A shipment of 350 new blood pressure monitors have arrived. Tests are done on 60 of the new monitors and it is found that 8 of the 60 give incorrect blood pressure readings. Find the 80% confidence interval for the proportion of all the monitors that give incorrect readings. Answer the following questions: 1. Multiple choice: Which equation would you use to solve this problem? A. B. C. D.E. 2. List the values you would insert into that equation. 3. State the final answer to the problem Your Answer: 1. The answer is: E 2. I must calculate samples that are defective: , meaning P=.1333, while estimatin: p=.1333; n=60 (monitors tested);z=1.28, based on 80% confidence and N=350 3. As asked to state the final asnwer to the problem is: While calculating the problem using formula (E) the answer is: .1333 .0512, therefore the defective proportion is between (.0821 and .1845) We have a finite population, so we will use Case 2: E. The proportion of the sample that are defective is 8/60 = .1333 so we set P=.1333. As we mentioned previously, we estimate p by P. So, p=.1333. A total of 60 monitors were tested, so n=60. Based on a confidence limit of 80 %, we find in table 6.1 that z=1.28. The total number of monitors is 350, so set N=350. Now, we can substitute all of these values into our equation: .1333± .0512 So the proportion of the total that are defective is between .0821 and .1845. Question 7 10 / 10 ptsYou may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) 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. Your Answer: For this problem: As realized this is left tailed test, meaning P(Zz)=.02. From this perspective while utilizing standart table: Z=-2.05 To calculate the problem: The formula needs to be used: Therefore: the z score is lees than (-2.05) the null hypothesis needs to be rejected. H0: μ=85 milligrams per day. H1: μ85 milligrams per day. This is a left-tailed test, so we must find a z that satisfies P(Zz)=.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. Question 8 10 / 10 ptsYou may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) Suppose we have independent random samples of size n1 = 604 and n2 = 490. The proportions of success in the two samples are p1= .43 and p2 = .51. Find the 95% 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 Your Answer:For this problem: 1. The answer is: B 2. While focusing on the table z- score based on 95% confidence: z=1.96; n1=604; n2=490;p1=.43;p2=.51 3. While using the formula above (B): From table 6.1, we see that 95% confidence corresponds to z=1.96. Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2: B. So, the interval is (-.139,-.021). Question 9 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) Critical_values_chart (Links to an external site.) 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. Your Answer: For this problem: n=5, wheras: =1+2+5+7+9/5=4.8; =10.8 CVCC=.87834, an 95 % confidence. Finally (r): So we have a negative liner relation becouse IrI=.964 which is greater than .87834 r= - .9654 Sx = 3.3 Sy = 3.03. Note that for n=5 and 95% we get a value from the chart of .87834. The absolute of r is |r|=.9654, which is above .87834. So a negative linear relation exists. Question 10 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) Critical_values_chart (Links to an external site.) Chi-square table (Links to an external site.)F_Distribution_Table_ (Links to an external site.) 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 330 drivers who have just finished driving 4 hours or less and they give the test to 215 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 250 80 Drove 8 hours or more 140 75 Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the 1 % level of significance. Your Answer: H0:Driving hours and alternes are independent H1: driving hours and alternes are not inmdependent While calculating, based on Che sqare table the critical value is: 6.635 E: for4 hours less and past :236.15/ total 330 E : for 4 h les and fail: 93.85/total 215 E for 8h more and pased: 153.85 E for 8h more and fail 61.15 Frpm this [perspective X2=7.24 This value is greater than the value shown above 6.635, meaning the hypothesis will be rejected. H0: Driving hours and alertness are independent events. H1: 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. Quiz Score: 95 out of 100 PreviousNext