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Mod 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)=
b.
- 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.
- 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.
The z-score is 3.75, so the data point 79 is 3.75 standard deviations above the mean.
- 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
- 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)=
- 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?
- 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?
- 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?
- 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
- 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.
- 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=.
- 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?
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.
- 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.
- 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
- 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.
- 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
- 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.
- 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
- 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.
- 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
- 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
- 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.
- 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.
- 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.
