Question

1. Ginger root is used by many as a dietary supplement. A manufacturer of supplements produces capsules that are advertised to contain at least 500 mg. of ground ginger root. A consumer advocacy group doubts this claim and tests the hypotheses H0: ? = 500 Ha: ? < 500 based on measuring the amount of ginger root in a SRS of 100 capsules. Suppose the results of the test fail to reject H0 when, in fact, the alternative hypothesis is true. In this case the consumer advocacy group will have

a. committed a Type I error.

b. committed a Type II error.

c. no power to detect a mean of 500.

2. A researcher reports that a test is “significant at 5%.” This test will be

a. Significant at 1%.

b. Not significant at 1%.

c. Significant at 10%.

3. Suppose the average Math SAT score for all students taking the exam this year is 480 with standard deviation 100. Assume the distribution of scores is normal. The senator of a particular state notices that the mean score for students in his state who took the Math SAT is 500. His state recently adopted a new mathematics curriculum and he wonders if the improved scores are evidence that the new curriculum has been successful. Since over 10,000 students in his state took the Math SAT, he can show that the P-value for testing whether the mean score in his state is more than the national average of 480 is less than 0.0001. We may correctly conclude that

a. there is strong statistical evidence that the new curriculum has improved Math SAT scores in his state.

b. although the results are statistically significant, they are not practically significant, since an increase of 20 points is fairly small.

c. these results are not good evidence that the new curriculum has improved Math SAT scores.

4. I want to construct a 92% confidence interval. The correct z* to use is

a. 1.75

b. 1.41

c. 1.645

5. The teacher of a class of 40 high school seniors is curious whether the mean Math SAT score µ for the population of all 40 students in his class is greater than 500 or not. To investigate this, he decides to test the hypotheses

H_{0}: µ = 500

Ha: µ & gt; 500

at level 0.05. To do so, he computes that average Math SAT score of all the students in his class and constructs a 95% confidence interval for the population mean. The mean Math SAT score of all the students was 502 and, assuming the standard deviation of the scores is α = 100, he finds the 95% confidence interval is 502 ± 31. He may conclude

a. H0 cannot be rejected at level α = 0.05 because 500 is within confidence interval.

b. H0 cannot be rejected at level α = 0.05, but this must be determined by carrying out the hypothesis test rather than using the confidence interval.

c. We can be certain that H0 is not true.

6. I wish to find a 95% confidence interval for the mean number of times men change channels with a remote control during a commercial. Based on a preliminary study, I estimate s = 15. How many commercials’ worth of data do I need to have a margin of error no more than 3?

a. 10

b. 97

c. 96

7. Experiments on learning in animals sometimes measure how long a laboratory rat takes to find its way through a maze. Suppose for one particular maze, the mean time is known to be 20 seconds with a standard deviation of = 2 seconds. Suppose also that times for laboratory rats are normally distributed. A researcher decides to test whether rats exposed to cigarette smoke take longer on average to complete the maze. She exposes 25 rats to cigarette smoke for 15 minutes and then records how long each takes to complete the maze. The mean time for these rats is 20.6 seconds. Are these results significant at the = 0.05 level? Assume the researcher’s rats can be considered a SRS from the population of all laboratory rats.

a. Yes.

b. No.

c. The question cannot be answered since the results are not practically significant.

8. The times for untrained rats to run a standard maze has a N (65, 15) distribution where the times are measured in seconds. The researchers hope to show that training improves the times. The alternative hypothesis is

a. Ha: µ > 65.

b. Ha: > 65.

c. Ha: µ < 65.

9. Does taking garlic tablets twice a day provide significant health benefits? To investigate this issue, a researcher conducted a study of 50 adult subjects who took garlic tablets twice a day for a period of six months. At the end of the study, 100 variables related to the health of the subjects were measured on each subject and the means com-pared to known means for these variables in the population of all adults. Four of these variables were significantly better (in the sense of statistical significance) at the 5% level for the group taking the garlic tablets as compared to the population as a whole, and one variable was significantly better at the 1% level for the group taking the garlic tablets as compared to the population as a whole. It would be correct to conclude

a. there is good statistical evidence that taking garlic tablets twice a day provides some health benefits.

b. there is good statistical evidence that taking garlic tablets twice a day provides benefits for the variable that was significant at the 1% level. We should be somewhat cautious about making claims for the variables that were significant at the 5% level.

c. None of the above.

10. To assess the accuracy of a kitchen scale a standard weight known to weigh 1 gram is weighed a total of n times and the mean, , of the weighings is computed. Suppose the scale readings are normally distributed with unknown mean, µ , and standard deviation = 0.01 g. How large should n be so that a 90% confidence interval for µ has a margin of error of ± 0.0001?

a. 165

b. 27061

c. 38416

11.Suppose that you are a student worker in the statistics department and agree to be paid by the Random Pay system. Each week the Chair flips a coin. If the coin comes up heads, your pay for the week is $80; if it comes up tails, your pay for the week is $40. You work for the department for 100 weeks (at which point you have learned enough probability to know the system is not to your advantage). The probability that , your average earnings in the first two weeks, is greater than $65 is

a. 0.2500.

b. 0.3333.

c. 0.5000.

12. A set of ten cards consists of five red cards and five black cards. The cards are shuffled thoroughly and I am given the first four cards. I count the number of red cards X in these four cards. The random variable X has which of the following probability distributions?

a. The binomial distribution with parameters n = 10 and p = 0.5.

b. The binomial distribution with parameters n = 4 and p = 0.5.

c. None of the above.

13. In a pre-election poll, 400 of the 500 probable voters polled favored the incumbent. In this poll, the sample proportion, , of those favoring the challenger is

a. 0.80

b. 0.20

c. 0.50

14. Incomes in a certain town are strongly right skewed with mean $36000 and standard deviation $7000. A random sample of 10 households is taken. What is the probability the average of the sample is more than $38000?

a. 0.3875

b. 0.1831.

c. Cannot say.

15. The scores of individual students on the American College Testing (ACT) Program composite college entrance examination have a normal distribution with mean that varies slightly from year to year and standard deviation 6.0. You plan to take an SRS of size n of the students who took the ACT exam this year and compute the mean score of the students in your sample. You will use this to estimate the mean score of all students this year. In order for the standard deviation of to be no more than 0.1, how large should n be?

a. At least 60.

b. At least 3600.

c. This cannot be determined because we do not know the true mean of the population.

16. We want to take a sample of 100 items out of a large batch for quality control purposes. Based on past history, the proportion of defective items is 4%. Can we use the normal approximation to the binomial distribution to find the probability of finding more than 5 defective items in the sample of 100?

a. Yes, because n is large.

b. No

c. We do not have enough information.

17. The SAT scores of entering freshmen at University X have a N (1200, 90) distribution and the SAT scores of entering freshmen at University Y have a N (1215,110) distribution. A random sample of 100 freshmen is sampled from each University, with the sample mean of the 100 scores from University X and the sample mean of the 100 scores from University Y. The probability that is greater than µ Y, the population mean for University Y, is

a. 0.0475.

b. 0.0869.

c. 0.4325.

Quizzes prepared by Dr. Patricia Humphrey, Georgia Southern University

18.Suppose that you are a student worker in the Statistics Department and they agree to pay you using the Random Pay system. Each week the Chair flips a coin. If it comes up heads, your pay for the week is $80, and if it comes up tails, your pay for the week is $40. You work for the department for two weeks. Let be the average of the pay you receive for the first and second week. The sampling distribution of is : $40 $60 $80

a. given by the following probability distribution.

Probability: 0.25 0.50 0.25

b. approximately normal with mean $60 and standard deviation 14.14.

c. exactly normal with mean $60 and standard deviation 14.14.

19. From previous polls, it is believed that 66% of likely voters prefer the incumbent. A new poll of 500 likely voters will be conducted. In the new poll, if the proportion favoring the incumbent has not changed, what is the mean and standard deviation of the number preferring the incumbent?

a. µ = 330, s = 10.59

b. µ = 0.66, s = 0.021

c. µ = 330, s = 18.17

20. From previous polls, it is believed that 66% of likely voters prefer the incumbent. A new poll of 500 likely voters will be conducted. In the new poll, if the proportion favoring the incumbent has not changed, what is the probability that more than 68% will favor the incumbent?

a. 32%.

b. 82.7%

c. 17.1%

21.A local farmer is interested in comparing the yields of two varieties of tomatoes. In an experimental field, she selects 40 locations and assigns 20 plants from each variety at random to the locations. She determines the average per plant (in pounds). She computes a 95% confidence interval for the difference in mean yields between the two varieties using the two-sample t procedures with the resulting interval (2.13, 6.41). For testing using the two-sample t procedures we can say that

a. the P-value could be greater than 0.05.

b. the P-value must be less than 0.05.

c. no information about the P-value can be obtained without the test statistic.

22.As the degrees of freedom become larger, the difference between the t and z distributions becomes

a. Narrower

b. Stays the same

c. Wider

23.Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled through weed and soil treatment. An experiment was conducted to compare peach-tree seedling growth with soil and weeds treated with one of two herbicides. In a field containing 10 seedlings, five were randomly selected throughout the field and assigned to receive Herbicide A. The remainder received Herbicide B. Soil and weeds for each seedling were treated with the appropriate herbicide, and at the end of the study period the height in (cm) was recorded for each seedling. The following results were obtained:

Herbicide A 87 80 80 76 73

Herbicide B 78 77 74 68 62

A 90% confidence interval for the difference in mean heights for the two herbicides is (0.2, 14.6). Which statement is correct?

a. The P -value for a test of the null hypothesis of equal means and the alternative of different means would be greater than 10% since the interval doesn’t include 0.

b. A 95% confidence could not include zero either, since we would be even moreconfident of a difference in the groups.

c. Both (a) and (b) are incorrect.

24. I did an eggsperiment with a fellow instructor one day. His calculus class was studying volumes of solids by rotating curves around the x-axis. We modeled the volume of an egg as an ellipsoid, and measured eggs with calipers, using the calculus formula. Each egg was also measured for volume using a water displacement method. We wanted to know if the two methods agreed or not. The data were

A 95% confidence interval for the average difference (calculus – water) is

a. (-1.69, 5.04)

b. (-7.54, 10.90)

c. (-2.29, 5.64)

25.Suppose that a random sample of 41 state college students is asked to measure the length of their right foot in centimeters. A 95% confidence interval for the mean foot length for students at this university turns out to be (21.709,25.091). Which of the following is true?

a. The sample mean was 23.4 cm.

b. The margin of error is 3.382

c. If the confidence level is changed to 90% we will get a wider interval.

Quizzes prepared by Dr. Patricia Humphrey, Georgia Southern University

26.A local farmer is interested in comparing the yields of two varieties of tomatoes. In an experimental field, he selects 20 locations and assigns 10 plants from each variety at random to the locations. He determines the yield per plant (in pounds). The mean yield for plants of variety 1 was = 16.3 pounds with a standard deviation = 3 pounds. The mean yield for plants of variety 2 was =18.4 pounds with a standard deviation = 4 pounds. The standard error of the difference in sample means is

a. 2.10 pounds.

b. 2.50 pounds.

c. 1.58 pounds.

27. The manager of an automobile dealership is considering a new bonus plan to increase sales. Currently, the mean sales rate per salesperson is five automobiles per month. The correct set of hypotheses to test the effect of the bonus plan is

a. H_{0}: µ < 5 H_{a}: µ ≤ 5

b. H_{0}: µ ≤ 5 H_{a}: µ > 5

c. H_{0}: µ > 5 H_{a}: µ ≤ 5

28.A bank is investigating ways to entice customers to charge more on their credit cards. (Banks earn a fee from the merchant on each purchase, and hope to collect interest from the customers as well). A bank selects a random group of customers who are told their “cash back” will increase from 1% to 2% for all charges above a certain dollar amount each month. Of the 500 customers who were told the increase applied to charges above $1000 each month, the average increase in spending was $527 with standard deviation $225. Of the 500 customers who were told the increase applied to charges above $2000 each month, the average increase in spending was $439 with standard deviation $189. When testing whether or not the increases in spending are different, the test is significant at

a. 5%

b. 1%

c. 0.5%

29.A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained an SRS of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be = 6 hours with a standard deviation s1 = 3 hours. The researcher also obtained an independent SRS of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be = 4 hours with a standard deviation s2 = 2 hours. Let µ 1 and µ 2 represent the mean amount of time spent in extracurricular activities per week by the populations of all high school students in the suburban and city school districts, respectively. If the researcher used the more accurate software approximation to the degrees of freedom, he would have used which of the following for the number of degrees of freedom for the two-sample t procedures?

a. 39.

b. 59.

c. 98.

30. We wish to see if the dial temperature for a certain model oven is properly calibrated. Four ovens of a certain model are selected at random. The dial on each is set to 300° F and after one hour, the actual temperature of each is measured. The temperatures measured are 305°, 310°, 300°, and 305°. Assuming that the actual temperatures for this model when the dial is set to 300° are normally distributed with mean µ , we test whether the oven is properly calibrated by testing the hypotheses

H 0: µ = 300, H a: µ ≠ 300.

Based on the data, the P -value for this test is

a. between 0.10 and 0.05.

b. between 0.05 and 0.025.

c. between 0.025 and 0.01.

31.An inspector inspects large truckloads of potatoes to determine the proportion with major defects prior to using the potatoes to be made into potato chips. She intends to compute a 95% confidence interval for p. To do so, she selects an SRS of 50 potatoes from the over 2000 potatoes on the truck. Suppose that only 2 of the potatoes sampled are found to have major defects. Which of the following assumptions for inference about a proportion using a confidence interval are violated?

a. The population is at least 10 times as large as the sample.

b. n is so large that both the count of successes n and the count of failures n (1 – ) are 10 or more.

c. There appear to be no violations.

32. A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective. An SRS of 100 parts from supplier 2 finds 10 defective. Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. A 98% confidence interval for p1 – p2, the difference in the two proportions is

a. -.05 ? 0.033.

b. -.05 ? 0.068.

c. – .05 ? 0.074.

33. I want to estimate the proportion of individuals in my area who think the public school system needs major overhauling. If I believe the proportion will be about 35%, how many individuals will I need to sample if I want a 95% margin of error to be no more than 3%?

a. At least 30.

b. At least 1068.

c. At least 972.

34.A poll finds that 54% of the 600 people polled favor the incumbent. Shortly after the poll is taken, it is disclosed that he had an extramarital affair. A new poll finds that 50% of the 1030 polled now favor the incumbent. The standard error for a confidence interval for the candidate’s latest support level is

a. 0.016

b. 0.020

c. 0.025

35. A student believes that 20% of all students think pepperoni is their favorite pizza. He performs a test of hypothesis, H0: p = 0.2, having taken a sample of 200 students and finding that 52 think pepperoni is their favorite. He finds a p-value of 0.0338, so rejects the null at ? = 0.05. He then computes a 95% confidence interval for the true proportion and finds it is (0.199, 0.321). He is confused. 20% is in the interval! What is the difference?

a. He made a mistake in one of his calculations.

b. The two use different values of p in computing the standard deviation.

c. He should have used the plus four confidence interval.

36. A poll finds that 54% of the 600 people polled favor the incumbent. Shortly after the poll is taken, it is disclosed that he had an extramarital affair. A new poll finds that 50% of the 1030 polled now favor the incumbent. We want to know if his support has decreased. The test statistic is

a. z = 1.56

b. z = -2.57

c. z = -1.55

37. I want to know which of two manufacturing methods will be better. I create 10 prototypes using the first process, and 10 using the second. There were 3 defectives in the first batch and 5 in the second. Find a 95% confidence interval for the difference in the proportion of defectives.

a. (-0.62, 0.22)

b. (-0.56, 0.22)

c. (-0.493, 0.160)

38. A sample of 75 students found that 55 of them had cell phones. The margin of error for a 95% confidence interval estimate for the proportion of all students with cell phones is

a. 0.084

b. (0.633, 0.833)

c. 0.100

39.A poll finds that 54% of the 600 people polled favor the incumbent. Shortly after the poll is taken, it is disclosed that he had an extramarital affair. A new poll finds that 50% of the 1030 polled now favor the incumbent. We want to know if his support has decreased. In computing a test of hypothesis with , what is the estimate of the overall proportion, ?

a. 52%

b. 52.5%

c. 51.5%

40. 100 rats whose mothers were exposed to high levels of tobacco smoke during pregnancy were put through a simple maze. The maze required the rats to make a choice between going left or going right at the outset. 80 of the rats went right when running the maze for the first time. Assume that the 100 rats can be considered an SRS from the population of all rats born to mothers exposed to high levels of tobacco smoke during pregnancy (note that this assumption may or may not be reasonable, but researchers often assume lab rats are representative of large populations since they are often bred to have uniform characteristics). Let p be the proportion of rats in this population that would go right when running the maze for the first time. A 90% confidence interval for p is

a. 0.8 ± .040.

b. 0.8 ± .066.

c. 0.8 ± .078.

Next to these stages, there are four types of loyalty. The first type is no loyalty: this can be seen as the typical action supporter that is giving money to fundraiser friends/family. There is no connection between the action supporter and War Child, non-loyal customers add a small amount to the financial account of the organization. The second type is inertia loyalty: this type can be a potential Friend, the person is giving a donation because they always have done it. This person is not loyal to one charity because there is no deeper connection and is easy with switching from charity to charity. This person is reachable to get a deeper connection with to become a potential Friend. It is needed to give the potential Friend attention and please the person. The third type is latent loyalty: this type is based on situational influences such as time, social life, physical factors, reasons why to support charity and the mood of the person. Latent loyalty means that there are repeated purchases. The last type is premium loyalty: this is the greatest type an organization can get. This type indicates the loyal Friend who is proud to discover more and tell about War Child to everyone. (Griffin, 2010) Using the stages and types of loyalty, War Child will get a better understanding about their potential Friends. Increased loyalty can give benefits to more respects of the company. Increased loyalty will be cost saving because there will be reduced marketing costs, more positive word-of-mouth, satisfied loyal customers whereby the failure costs decrease. (Griffin, 2010) 1.2.4 Customer relationship management Kotler (2014, p. 9) stated in the book ‘’Principles of marketing’’ that customer relationship management (CRM) can be mentioned as delivering superior customer value and satisfaction by building and maintaining profitable customer relationships. (Kotler, 2014) Using a suitable approach for the action supporters will lead to charitable giving. To build and maintain profitable customer relationships, customer relationship management need to be implemented. It is very important to deliver customer value and make the action supporters satisfied. By doing this,>

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