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PORTAGE LEARNING MATH 110 FINAL EXAM
INTRODUCTION TO STATISTICS | QUESTIONS
AND ANSWERS
Question 1 Not yet graded / 10 pts You may find the following files helpful throughout the exam: 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: a. fish 4700(0.28)= b. French fries 4700(0.4)= a.) Fish : 4700(.28) = 1316 b.) French Fries: 4700(.40) = 1880 Question 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.
c. P(-.59 ≤ Z ≤ - .36)= .35942- .27760= .08182. Question 5 Suppose that you are attempting to estimate the annual income of 1700 families. In order to use the infinite standard deviation formula, what sample size, n, should you use? In order to use infinite standard deviation formula, we should have: n≤0.05(1700) n≤ So, the sample size should be less than 85. Question 6 A shipment of 450 new blood pressure monitors have arrived. Tests are done on 75 of the new monitors and it is found that 15 of the 75 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:
- Multiple choice: Which equation would you use to solve this problem? A. B. C.
D.
E.
- List the values you would insert into that equation.
- State the final answer to the problem We have a finite population, so we will use Case 2: E. The proportion of the sample that are defective is 15/75 = .2 so we set P=.2. As we mentioned previously, we estimate p by P. So, p=.2. A total of 75 monitors were tested, so n=75. Based on a confidence limit of 80 %, we find in table 6.1 that z=1.28. The total number of monitors is 450, so set N=450. Now, we can substitute all of these values into our equation: .2±. So the proportion of the total that are defective is between .146 and .254. Question 7 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 9 milligrams per day. A doctor is concerned that her pregnant patients are not getting enough vitamin C. So, she collects data on 40 of her patients and finds that the mean vitamin intake of these 40 patients is 83 milligrams per day. Based on a level of significance of α = .015, test the hypothesis. H 0 : μ=85 milligrams per day. H 1 : μ<85 milligrams per day.
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 (-.1927,-.0473). Question 9 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. r= .9910 Sx = 4.2 Sy = 5.7. Note that for n=5 and 95% we get a value from the chart of .87834. The absolute of r is |r|=.9910, which is above .87834. So a positive linear relation exists. Question 10 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. 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.
