Midterm 1

A random sample of 10 countries was selected for your final. Here is some information on them from the 1997 CIA World Factbook.

Country	Total Fertility Rate (i.e. Average number of Children born per woman)	Percentage Change in Gross Domestic Product, 1998 -1999 (estimated)	Female Literacy Rate (i.e. women over age 15 who can read and write)
Afghanistan	6.07	-0.5%	15%
Cambodia	5.81	-1.5%	22%
Costa Rica	2.85	-2.1%	95%
Indonesia	2.66	-2.8%	78%
Italy	1.16	+3.1%	96%
Jordan	4.94	-0.4%	49%
Nigeria	6.17	+1.0%	47%
Russia	1.35	-1.4%	97%
United Arab Emirates	3.62	+1.2%	80%
United States	2.06	+4.0%	97%
AVERAGE	3.669	0.06%	67.6%
STAND. DEV.	1.8486	2.1186%	30.27%

1. What is the correlation between Female Literacy Rate and fertility?

-.93

2. What is the regression equation for predicting fertility from the Female Literacy Rate?

fertility = -5.667(literacy) + 7.5

3. Cuba's Female Literacy Rate is 96%. What is Cuba's predicted fertility rate?

2.06 = -5.667( .96) + 7.5 (about 2.1 children)

4. What is the median Percentage Change in Gross Domestic Product?

-.45%

5. Economists use Percentage Change in Gross Domestic Product as a predictor of increased prosperity (positive change) or recession (negative change or no change). The average for the 10 countries above is a +0.06% change so some economists are using this as a signal of "good times" ahead for the global economy, others are arguing that there will be recession next year. The long-run expected change is 0% (no change).

Please test the hypothesis that there will be no change next year using information from the sample of 10 countries above. Formulate the null and the alternative, and compute a test statistic. Give me a conclusion in plain English.

Null: average = 0%

Alternative: average > 0%

use a t-test, this is a sample of 10 where the population standard deviation is unknown.

SD+ is 2.2332

should get a t = .085 with 9 degrees of freedom

No evidence for good times ahead. Do not reject the null.

An analysis of all pizza delivery times for a well-known national chain reveals the following distribution of delivery times (in minutes):

Time to Delivery	15	20	30	40	45
Percentage of Deliveries	10%	20%	40%	20%	10%

Assume that distribution of deliveries is a stable historical fact. You and 15 of your best friends order pizza to be delivered to 16 different places around Los Angeles. Treat this like a random sample.

6. What is the expected value for the average delivery time?

30 minutes

7. What is the standard deviation of the population of delivery times?

9.2195

8. What is the standard error for the average delivery time of the pizzas you and your friends ordered?

2.305

9. The pizzas ordered by you and your friends took an average of 35 minutes for delivery. The company claims that their pizza deliveries take an average of 30 minutes. Is there evidence to suggest that the company is not being truthful about its delivery time claims? State the null and the alternative hypothesis, perform a test, use a 1% level of significance as your rule, then briefly explain your conclusions from the test.

Null: average = 30

Alternative: average > 30

test is a Z. Results in a Z of 2.169 or about 2.2 The p-value is 1.39%

Close, but not enough evidence to reject the null. 1.39% is not less than 1%

The following comes from a recent article in U.S. News and World Report :

Going outside the medical mainstream

Can 42 percent of Americans be wrong?

BY AVERY COMAROW

The numbers were stunning, the clinical findings hardly less so. The Journal of the American Medical Association, in an issue devoted to alternative medicine, last week reported that 42.1 percent of Americans surveyed in 1997 used therapies outside mainstream medicine, like chiropractic, acupuncture, folk remedies, and homeopathy.

The 42.1 percent figure, reported by a team led by David Eisenberg of Beth Israel Deaconess Medical Center in Boston, was up by almost a fourth from the 33.8 percent Eisenberg found just seven years ago in a similar survey. And if the group he surveyed is typical, visits by all Americans to practitioners of alternative medicine totaled 629 million, exceeding those to conventional physicians.

For the sake of this final, assume that 33.8% figure measured in 1990 is a stable, long-run, historical fact about the usage of alternative medicine. Also assume that the newer 1997 survey had 144 respondents.

10. Test the hypothesis that the use of alternative medical therapies by Americans increased between 1990 and 1997. State the null and the alternative, perform a test, and state a p-value. Please use a 5% level of significance as your decision rule. Do you believe that more Americans are using alternative medicine in 1997?

Null: true percentage = 33.8%

Alternative: true percentage > 33.8%

test is a Z. 33.8% is a parameter.

SD = SQRT(.338 * .662) = .4730 (must get this right for full credit)

SE = 3.9419 %

Z = 2.11 approximately round to a Z=2.15, the p-value is 1.6%. I'd reject the null in favor of the alternative, more Americans are using alternative medicine in 1997.

11. Suppose you go to medical school and replicate this study (i.e. draw a new sample) in 2003. Let's assume that the 42.1% figure is now the stable, long run fact about the usage of alternative medicine by Americans. What is the chance that your sample of 100 will have at least 50% of the respondents using alternative medicine?

SD changes to SQRT(.421*.579) so SE changes to 4.937%. The resulting chance is 5.48%

12. Amstar Corporation sued Domino's Pizza, Inc. claiming that Domino's use of the name "Domino" on its pizza infringed on the Domino Sugar trademark (owned by Amstar Corp.). Both sides presented survey evidence on whether Domino's Pizza, Inc.'s use of the name "Domino" tended to create confusion among consumers.

Amstar surveyed 525 people in 10 cities in the eastern United States (two of the cities had Domino's Pizza outlets). The persons interviewed were women reached at home during the day who identified themselves as the household member responsible for grocery buying. Shown a Domino's Pizza box, 44.2% of those interviewed indicated their belief that the company that made the pizza made other products. 72% of that group (31.6% of all respondents) believed that the pizza company also make sugar.

Assume that a multistage cluster sampling method was used to draw the sample and that for our purposes, it produced a random sample of 525 people. Now that you have taken Statistics 10, do you think this is a good study? Answer yes or no and please justify your response. Please be brief, a few sentences will be sufficient. No calculations are required to answer this question.

ANSWER: No. This was taken from an actual case and it's an example of a bad study. Looking for two basic ideas: 1. selection bias (somehow only women who were at home during the day got into their sample. There is nothing wrong with the sampling method, this is pretty standard for a multistage cluster, but there seems to be something odd going on with the interview process.) 2. response bias -- somehow, women who didn't have Domino's Pizza outlets in their cities (only 2 of the 10 cities had pizza outlets) were able to express beliefs about the pizza company.

13. A survey is carried out by the planning department to determine the distribution of household size in a certain city. They draw a simple random sample (SRS) of 1000 households. But after several visits, the interviewers find people at home in only 853 of the sample households. Rather than face such a high non-response rate, the planners draw a second batch of households and use the first 147 completed interviews to bring the first sample up to 1000 households. They counted 3,087 people living in these 1000 households (the 853 + 147 households) and estimate the average household size in the city to be about 3.1 persons.

Is the average like to be (circle one):

(i) too high

(ii) too low

(iii) about right

Explain your choice. Be brief. No calculations are required to answer this question.

The average is too high. 1. By drawing a second batch and using the first 147 completed interviews, they have effectively contaminated what was probably a reasonable sample. 2. Those families which made it into the interview are probably larger families; there is a better chance of catching a larger family at home other things being equal. Another way to think about this is that it is tough to catch a person who lives at home alone at home, it is much easier to catch someone at home if there is more than one person living in that house. The result is the average will be too high.

One year, there were about 3,000 institutions of higher learning in the U.S. (including junior colleges and community colleges). As part of a continuing study of higher education, the government took a simple random sample (SRS) of 400 of these institutions. The average enrollment in the 400 sample schools was 3,700 and the SD was 6,500. The government estimates the average enrollment at all 3,000 institutions to be around 3,700; they put a give or take number of 325 around this estimate. Say whether each of the following statements is true or false, and explain. If you need more information to decide, say what you need and why.

14. An approximate 68% confidence interval for the average enrollment of all 3000 institutions runs from 3,375 to 4,025.

This is TRUE. 325 is the standard error. A 68% confidence interval is 3700 + or - the 325.

15. About 68% of the schools in the sample had enrollments in the range 3,700 ± 6,500.

This is FALSE. You are actually being asked about the sample itself here. The 6,500 is the sample standard deviation. You can't make this statement unless you know that the sample is normally distributed.

16. I guess you get 10 points just for attending. J Have a happy winter break. Best wishes for the future.