DOC PREVIEW
TAMU STAT 303 - ex3asp07

This preview shows page 1 out of 4 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

STAT303 Sec 508-510Spring 2007Exam #3Form AInstructor: Julie Hagen CarrollOctober 10, 20071. Don’t even open this until you are told to do so.2. Please PRINT your name in the blanks provided.3. There are 20 multiple-choice questions on this exam, each worth 5 points. There is partial credit. Please mark youranswers clearly. Multiple marks will be counted wrong.4. You will have 60 minutes to finish this exam.5. If you have questions, please write out what you are thinking on the back of the page so that we can discuss it afterI return it to you.6. If you are caught cheating or helping someone to cheat on this exam, you both will receive a grade of zero on theexam. You must work alone.7. This exam is worth the same as a regular exam (this may differ from section to section.8. Good luck!1STAT303 sec 507 Exam #3, Makeup Spring 20071. Suppose that it is commonly assumed that the meanflight from College Station to Houston is 25 minutes.However, you believe that the true average is greaterthan this. You randomly choose 30 flights to take andrecord their times. The mean of your sample is 27.What would be a Type I error?A. Concluding that the mean is greater than 25 whenit really is 25 minutesB. Concluding that the mean is not greater than 25when the true mean is 24C. Concluding that the mean is not greater than 25when the true mean is 26D. Concluding that the true mean is 24 when the truemean is 25E. Two of the above are true.2. The following are confidence intervals for 1 - 2 com-puted from the same data:90% CI = (0.02, 0.09)95% CI = (0.01, 0.12)99% CI = (-0.03, 0.14)Based on the intervals above, if we were to test H0:π1= π2vs. HA: π16= π2, what would be the corre-sponding p-value?A. p-value > 0.10B. 0.10 > p-value > 0.05C. 0.05 > p-value> 0.01D. p-value < 0.01E. You need a test statistic value to determine thep-value.3. An insurance company is conducting a study compar-ing the average number of accidents for females andmales. The company wants to show on average femaleshave less accidents than males to justify lower rates forfemales. What is the appropriate hypothesis?A. H0: µfemale= µmalevs. HA: µfemale6= µmaleB. H0: πfemale= πmalevs. HA: πfemale6= πmaleC. H0: µfemale= µmalevs. HA: µfemale> µmaleD. H0: µfemale= µmalevs. HA: µfemale< µmaleE. H0: πfemale= πmalevs. HA: πfemale< πmale4. A bank wonders whether omitting the annual creditcard fee for customers who charge at least $5000 in ayear would increase the amount charged on its creditcard. The bank makes this offer to a simple randomsample of 500 existing credit card customers. The bankthen compares the amount charged this year with theamount charged last year for each of these customers.What type of test should be used to analyze this study?A. A two-sample test of proportionsB. A one-sample t-testC. A two-sample t-test since the standard deviationis unknownD. A pooled t-testE. A paired t-test5. Suppose we want to test whether the proportion of pa-tients who come down with a cold during their hospi-tal stay is the same for patients taking Echinacea everyday and patients on a placebo drug. One herb companywants to prove that it lowers the rate at which patientscatch a cold, so we set up the hypotheses: H0: π1= π2and HA: π1> π2, where π1is the proportion of peopletaking the placebo who get a cold during their hospitalstay and π2is the proportion of people taking Echi-nacea who get a cold. The resulting p-value is 0.2171.What does that mean in context of the problem?A. The probability that Echinacea doesn’t keep youfrom catching a cold is 0.2171.B. The probability that we find a difference in propor-tions at least this small assuming that Echinaceadoesn’t keep you from catching colds is 0.2171.C. Under repeated sampling, we would find that pa-tients taking Echinacea every day had the samerate of sickness as patients on a placebo 21.71%of the time, assuming Echinacea actually doesn’tkeep you from catching a cold.D. Under repeated sampling, we would find that pa-tients taking Echinacea every day had at least thismuch lower rate of sickness about 21.71% of thetime, assuming that Echinacea doesn’t keep youfrom catching a cold.E. Two of the above are true.6. Which of the following best describes the relationshipbetween a (1 − α) ∗ 100% confidence interval for µ1−µ2and a 2-sided test of hypotheses for µ1= µ2somevalue?A. There is no relationship between confidence inter-vals and hypothesis tests.B. If µ1or µ2fall within the confidence interval, wewould reject the null.C. If µ1or µ2fall within the confidence interval, wewould fail to reject the null.D. If the confidence interval contains 0, we would re-ject the null.E. If the confidence interval contains 0, we would failto reject the null.7. The purpose of pairing in an experiment is toA. make the samples independent.B. increase the degrees of freedom of the t-test so thetest has more power.C. match the observations so that there is less chanceof making an error.D. filter out the variability between the subjects.E. None of the above are correct.2STAT303 sec 507 Exam #3, Makeup Spring 20078. Suppose you tested the null hypothesis H0: µ1= µ2against the alternative HA: µ16= µ2and got an averagedifference in means of 0.5682 with a corresponding p-value of 0.0432. If you were to create a 95% confidenceinterval for the difference between the two means usingthe same data, which of the following would be true?A. The confidence interval would include 0, since0.0432 < 0.5682.B. It would be impossible to tell whether the confi-dence interval would include positive numbers ornegative numbers, since we don’t know the valueof the test statistic.C. Under repeated sampling, 95% of the time the con-fidence interval for the difference in means wouldinclude 0.5682.D. The confidence interval would not include 0.E. Two of the above are true.9. Let µ denote the mean gas mileage of all cars whenadditive is used. When additive is not used, cars have amean gas mileage of 18.25. Does using additive improvegas mileage? Test the hypotheses H0: µ = 18.25 vs.HA: µ > 18.25 at α = 0.05. A car manufacturertook a sample of 10 cars and found a sample mean of18.92 with a standard deviation of 7.47. What do youconclude about using additive?A. We have evidence to say that using additive im-proves gas mileage since our p-value is less thanα.B. Since our p-value is less than α, we do not haveevidence to say that using additive improves gasmileage.C. With such a large


View Full Document

TAMU STAT 303 - ex3asp07

Download ex3asp07
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view ex3asp07 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view ex3asp07 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?