For large U.S. companies, what percentage of their total income comes from foreign sales? A random sample of technology companies (IBM, Hewlett-Packard, Intel, and others) gave the following information.†
| Technology companies, % foreign revenue: x1; n1 = 16 | |||||||
| 62.8 | 55.7 | 47.0 | 59.6 | 55.3 | 41.0 | 65.1 | 51.1 |
| 53.4 | 50.8 | 48.5 | 44.6 | 49.4 | 61.2 | 39.3 | 41.8 |
Another independent random sample of basic consumer product companies (Goodyear, Sarah Lee, H.J. Heinz, Toys 'R' Us) gave the following information.
| Basic consumer product companies,% foreign revenue: x2; n2 = 17 | |||||||||
| 28.0 | 30.5 | 34.2 | 50.3 | 11.1 | 28.8 | 40.0 | 44.9 | ||
| 40.7 | 60.1 | 23.1 | 21.3 | 42.8 | 18.0 | 36.9 | 28.0 | ||
| 32.5 | |||||||||
Assume that the distributions of percentage foreign revenue are mound-shaped and symmetric for these two company types.
(a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to two decimal places.)
| x1 = | % |
| s1 = | % |
| x2 = | % |
| s2 = | % |
(b) Let μ1 be the population mean for
x1 and let μ2 be the
population mean for x2. Find a 98% confidence
interval for μ1 − μ2.
(Round your answers to two decimal places.)
| lower limit | % | |
| upper limit | % |
In: Math
An experimental surgical procedure is being studied as an alternative to the old method. Both methods are considered safe. Five surgeons preform the operation on two patients matched by age, sex, and other relevant factors, with the results shown. The time to complete the surgery (in minutes) is recorded. At the 5% significance level, is the new way faster?
| Old Way | New Way | Di | Di - D-bar | (Di - D-bar)2 | |
| Surgeon 1 | 36 | 29 | |||
| Surgeon 2 | 55 | 42 | |||
| Surgeon 3 | 28 | 30 | |||
| Surgeon 4 | 40 | 32 | |||
| Surgeon 5 | 62 | 56 | |||
| XXXXXXXX | XXXXXXX | sums |
*Do not use p-values
*Use and show the 5 step method
D-bar=
Sd=
Step 1: H0=
HA=
Step 2: alpha =
Step 3: Test Statistic:
Step 4: Decision Rule:
Step 5: Calculation and Decision
Reject or do not reject H0? Why?
In: Math
The population average cholesterol content of a certain brand of egg is 215 milligrams, and the standard deviation is 15 milligrams. Assume the variable is normally distributed.
If we are told the average for 25 eggs is less than 220 mg, what is the probability that the average is less than 210?
Round the answers to four decimal places.
In: Math
1). A credit card company says that their clients have a mean credit card balance of less than $3000. A random sample of 14 clients showed an average balance of $2900 with sample standard deviation of $350. At ? = 0.10, conduct all seven steps of the hypothesis test to test the claim. Assume the population is normally distributed.
(2). A researcher claims that at least 46% of U.S. adults think that the IRS is not aggressive enough in pursuing people who cheat on their taxes. In a random sample of 600 U.S. adults, 43% say that the IRS is not aggressive enough in pursuing people who cheat on their taxes. At ? = 0.01, is there enough evidence to reject the researcher’s claim? Conduct the hypothesis test.
(3). A consumer group claims that the mean annual consumption of high fructose corn syrup by a person in the U.S. is 48.8 pounds. A random sample of 120 people in the U.S. has a mean annual high fructose corn syrup consumption of 49.5 pounds. Assume the population standard deviation is 3.6 pounds. At ? = 0.05, conduct all seven steps of the hypothesis test to test the claim (no P-value).
In: Math
Independent random samples of professional football and basketball players gave the following information.
Heights (in ft) of pro football players: x1; n1 = 45
| 6.33 | 6.52 | 6.50 | 6.25 | 6.50 | 6.33 | 6.25 | 6.17 | 6.42 | 6.33 |
| 6.42 | 6.58 | 6.08 | 6.58 | 6.50 | 6.42 | 6.25 | 6.67 | 5.91 | 6.00 |
| 5.83 | 6.00 | 5.83 | 5.08 | 6.75 | 5.83 | 6.17 | 5.75 | 6.00 | 5.75 |
| 6.50 | 5.83 | 5.91 | 5.67 | 6.00 | 6.08 | 6.17 | 6.58 | 6.50 | 6.25 |
| 6.33 | 5.25 | 6.66 | 6.50 | 5.82 |
Heights (in ft) of pro basketball players: x2; n2 = 40
| 6.08 | 6.56 | 6.25 | 6.58 | 6.25 | 5.92 | 7.00 | 6.41 | 6.75 | 6.25 |
| 6.00 | 6.92 | 6.85 | 6.58 | 6.41 | 6.67 | 6.67 | 5.75 | 6.25 | 6.25 |
| 6.50 | 6.00 | 6.92 | 6.25 | 6.42 | 6.58 | 6.58 | 6.08 | 6.75 | 6.50 |
| 6.83 | 6.08 | 6.92 | 6.00 | 6.33 | 6.50 | 6.58 | 6.83 | 6.50 | 6.58 |
(a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to three decimal places.)
| x1 = | |
| s1 = | |
| x2 = | |
| s2 = |
(b) Let μ1 be the population mean for
x1 and let μ2 be the
population mean for x2. Find a 90% confidence
interval for μ1 – μ2.
(Round your answers to three decimal places.)
| lower limit | |
| upper limit |
In: Math
Statistics Anxiety
1.) Your Experiences: Share an experience you’ve had with statistics/math anxiety or test anxiety?
2.) Your Approach: What will be your approach to overcome anxiety statistics or math course?
In: Math
R Simulation:Write an R code that does the following:
(a) Generate n samples x from a random variable X that has a uniform density on [0,3].
(b) Now generate samples of Y using the equation: y = α x + β
(c) For starters, set α = 1, β = 1.
In: Math
In a recent year, according to the Bureau of Labor Statistics, the median number of years that wage and salary workers had been with their current employer (called employee tenure) was 3.5 years. Information on employee tenure has been gathered since the early 1950's using the Current Population Survey (CPS), a monthly survey of 50,000 households that provides information on employment, unemployment, earnings, demographics, and other characteristics of the U. S. population. With respect to employee tenure, the questions measure how long workers had been with their current employer, not how long they plan to stary with their employer.
Employee Tenure of 20 workers
4.1, 2.3, 3.5, 4.6, 3.1, 1.2, 3.9, 2.1, 1.0, 4.5, 3.2, 3.4, 4.1,
3.1, 2.8, 1.4, 3.4, 4.9, 5.7, 2.6
A) A congressional representative claims that the median tenure for workers from the representative's district is less than the national median tenure of 3.5 years. Thae claim is based on the representative's data shown above. Assume that the employees were randomly selected.
1) How would you test the representative's claim?
2) Can you use a parametric test, or do you need a nonparametric test? Why?
3) State the null and alternative hypothesis.
4) Test the claim using alpha = 0.05. What can you conclude? Show your work, the process that you used, and the result.
Employee tenure for a sample of male workers
3.3, 3.9, 4.1, 3.3, 4.4, 3.3, 3.1, 4.1, 2.7, 4.9, 0.9, 4.6
Employee tenure for a sample of female workers
3.7, 4.2, 2.7, 3.6, 3.3, 1.1, 4.4, 4.4, 2.6, 1.5, 4.5, 2.0
B) A congressional representative claims that the median tenure for male workers is greater that the median tenure for female workers. The claim is based on the data shown above.
5) How would you est the representative's claim?
6) Can you use a parametric test, or do you need to use a nonparametric test?
7) State the null hypothesis and the alternative hypothesis.
8) Test the claim using alpha = 0.05. What can you conclude. Show your work, the process that you used, and the result.
In: Math
This question is based on a Poisson discrete probability distribution. The distribution is important in biology and medicine, and can be dealt with in the same way as any other discrete distribution. Red blood cell deficiency may be determined by examining a specimen of blood under the microscope. The data in Table B gives a hypothetical distribution of numbers of red blood cells in a certain small fixed volume of blood from normal patients. Theoretically, there is no upper limit to the value of a POISSON distribution. In reality, you can force only so many red blood cells into a given volume. Copy the data from Table B into columns of the EXCEL worksheet, name the columns, and view the table.]
| 0 | 0.00000 |
| 1 | 0.00000 |
| 2 | 0.00000 |
| 3 | 0.00001 |
| 4 | 0.00002 |
| 5 | 0.00010 |
| 6 | 0.00031 |
| 7 | 0.00085 |
| 8 | 0.00204 |
| 9 | 0.00435 |
| 10 | 0.00839 |
| 11 | 0.01468 |
| 12 | 0.02355 |
| 13 | 0.03488 |
| 14 | 0.04797 |
| 15 | 0.06157 |
| 16 | 0.07410 |
| 17 | 0.08392 |
| 18 | 0.08977 |
| 19 | 0.09097 |
| 20 | 0.08758 |
| 21 | 0.08030 |
| 22 | 0.07027 |
| 23 | 0.05883 |
| 24 | 0.04720 |
| 25 | 0.03635 |
| 26 | 0.02692 |
| 27 | 0.01920 |
| 28 | 0.01320 |
| 29 | 0.00876 |
| 30 | 0.00562 |
| 31 | 0.00349 |
| 32 | 0.00210 |
| 33 | 0.00123 |
| 34 | 0.00069 |
| 35 | 0.00038 |
| 36 | 0.00020 |
| 37 | 0.00011 |
| 38 | 0.00005 |
| 39 | 0.00003 |
| 40 | 0.00001 |
| 41 | 0.00001 |
| 42 | 0.00000 |
| 43 | 0.00000 |
| 44 | 0.00000 |
| 45 | 0.00000 |
| 46 | 0.00000 |
| 47 | 0.00000 |
| 48 | 0.00000 |
| 49 | 0.00000 |
| 50 | 0.00000 |
| 51 | 0.00000 |
| 52 | 0.00000 |
| 53 | 0.00000 |
| 54 | 0.00000 |
| 55 | 0.00000 |
| 56 | 0.00000 |
| 57 | 0.00000 |
| 58 | 0.00000 |
| 59 | 0.00000 |
| 60 | 0.00000 |
| 61 | 0.00000 |
| 62 | 0.00000 |
| 63 | 0.00000 |
| 64 | 0.00000 |
| 65 | 0.00000 |
| 66 | 0.00000 |
| 67 | 0.00000 |
| 68 | 0.00000 |
| 69 | 0.00000 |
| 70 | 0.00000 |
| 71 | 0.00000 |
| 72 | 0.00000 |
| 73 | 0.00000 |
| 74 | 0.00000 |
| 75 | 0.00000 |
| 76 | 0.00000 |
| 77 | 0.00000 |
| 78 | 0.00000 |
| 79 | 0.00000 |
| 80 | 0.00000 |
| 81 | 0.00000 |
| 82 | 0.00000 |
| 83 | 0.00000 |
| 84 | 0.00000 |
| 85 | 0.00000 |
| 86 | 0.00000 |
| 87 | 0.00000 |
| 88 | 0.00000 |
| 89 | 0.00000 |
| 90 | 0.00000 |
| 91 | 0.00000 |
| 92 | 0.00000 |
| 93 | 0.00000 |
| 94 | 0.00000 |
| 95 | 0.00000 |
| 96 | 0.00000 |
| 97 | 0.00000 |
| 98 | 0.00000 |
| 99 | 0.00000 |
| 100 | 0.00000 |
8. What is the probability that a blood sample from this distribution will have exactly 20 red blood cells?
9. What is the probability that a blood sample from a normal person will have between 19 and 26 red blood cells? HINT: See questions 3 and 4.
10. What is the probability that a blood sample from a normal person would have fewer than 10 red blood cells?
11. What is the probability that a blood sample from a normal person will have at least 15 red blood cells? HINT: Since there is no theoretical upper limit to the Poisson distribution, the correct way to answer this question is to calculate 1 – probability of fewer than 15 red blood cells. ASSIGNMENT 3 20 INTRODUCTORY STATISTICS LABORATORY
12. A person with a red blood cell count in the lower 2.5 percent of the distribution might be considered as deficient. What is the red blood cell count below which 2.5 percent of the distribution lies? HINT: You need to determine a value X so that if you sum all the probabilities for counts up to and including that value they will sum to at least 0.025. The sum of probabilities of all counts up to but excluding X should be less than 0.025. You can proceed in the following way. Look at the table to guess how many probabilities (P[X = 0] + P[X = 1] + . . ) should be added to give a sum of approximately 0.025. Calculate sums of probabilities for your guess of X. Continue your guessing of X until you get a sum ≥ 0.025 while the sum for X-1 < 0.025.
13. What is the mean red blood cell count in this distribution?
14. What is the variance of red blood cell count in this distribution? HINT: See question 7, and remember it is a Poisson distribution.
15. Is the following statement true (1) or false (0) for this distribution? In a Poisson distribution, the variance is equal to the mean (within rounding error). Record 1 if true, 0 if false.
In: Math
6. For each of the scenarios below, identify the sampling blunder, speculate about the influence of the bias, and then make a recommendation for ridding the study of the biasing influence. a. A researcher wanted to know how people in the local community felt about the use of high-stakes testing in the public schools. The researcher spent the afternoon at Wal-Mart and randomly approached 100 shoppers to ask their opinion (they all agreed to cooperate). Random selection was accomplished with the use of a random number table (the numbers determined which shopper to target, such as the 16th to exit, then the 30th to exit, then the ninth to exit, etc.). b. A researcher wanted to know how students at a university felt about mandatory fees for all students to support a child care center for students with children. The researcher set up a table near the dormitory where many different types of students came and went. Those who stopped at the table and seemed friendly were asked to complete the questionnaire. c. To study differences in occupational aspirations between Catholic high school students and public high school students, a researcher randomly sampled (using school rosters and a random number table) 200 students from the largest Catholic high school and the largest public high school. d. To learn more about teachers' feelings about their personal safety while at school, a questionnaire was printed in a nationwide subscription journal of interest to many teachers. Teachers were asked to complete the questionnaire and mail it (postage paid) to the journal headquarters for tabulation. e. To study the factors that lead teachers in general to quit the profession, a group of teachers threatening to quit was extensively interviewed. The researcher obtained the group after placing an announcement about the study on the teachers' bulletin board at a large elementary school.
In: Math
Explain ANOVA (Analysis of Variance). How is F test for differences among more than two means used in this case to test a hypothesis? Give an example.
In: Math
For the following data, approximate the sample mean, sample variance and sample standard deviation weekly grocery bill.
Bill (in dollars) Frequency
135-139 11
140-144 20
145-149 19
150-154 7
155-159 5
In: Math
find a research article that uses hypotheses. Identify the hypotheses as null or alternative. Some articles will contain more than one hypothesis. Then, look at the discussion part of the article and see if the p-values were significant and discuss how the article presented the acceptance or rejection of the hypothesis. Include the pdf of the article with your first discussion posting.
In: Math
4. To determine the effectiveness of a proposed public relations campaign, the senior vice president for customer relations for an automobile manufacturer asked seven consumers how much they liked the company (on a scale from 0 [do not like] to 50 [like very much]) before and after viewing the primary television advertisement of the campaign. Use the following data to test whether the consumers’ ratings of the company increased, on average, after viewing the television advertisement:
|
RATING |
||
|
CONSUMER |
Before |
After |
|
1 |
42 |
49 |
|
2 |
21 |
15 |
|
3 |
31 |
35 |
|
4 |
7 |
13 |
|
5 |
29 |
41 |
|
6 |
35 |
39 |
|
7 |
14 |
22 |
a. State the null and alternative hypotheses associated with the
test.
b. If α = 0.05, what is the critical value of the associated test
statistic?
c. What is the calculated value of the associated test
statistic?
d. State your decision about the null hypothesis by comparing the
critical and calculated values of the test statistic (Parts b and
c).
e. Comment on the effectiveness of the primary television
advertisement of the campaign.
In: Math
A mixture of pulverized fuel ash and Portland cement used for grouting should have a compressive strength of more than 1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met, that is, the true mean compressive strength is more than 1300. Suppose compressive strength for specimens of this mixture is normally distributed with standard deviation of 60. A sample of 10 randomly chosen specimens has a sample mean compressive strength of 1331.26.
a) What are the appropriate null and alternative hypotheses?
b) Carry out the test from part a) at 5% level of significance stating clearly the conclusion in the context of the question.
c) What is the probability of making a Type I error in part b) and describe it in the context of the question?
d) Compute an appropriate one-sided 95% confidence bound for the true mean compressive strength and explain why you chose this bound in the context of the question.
THE ANSWERS FOR THIS PROBLEM ARE AS FOLLOWS:
a) Null: ?=1300, Alternative: ?>1300?
b) z=1.6475, z0.05 = 1.645, z>z0.05 at 5% and mean>1300, so we reject the null.
c) ? = 0.05, Reject null when true. Conclude ?>1300 when it is not.
d) lower: 1300.04 and 1300 is below. ?>1300.
Please explain the steps for how to solve this problem. Thank you!
In: Math