STAT TECH IN BUSINESS & ECON AC
18th Edition
ISBN: 9781264731657
Author: Lind
Publisher: MCG
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Chapter 15, Problem 3E
To determine
Check whether people can conclude that 10% of cars traveling on the turnpike have two or more occupants.
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The average miles per gallon for a sample of 40 cars of model SX last year was 32.1, with a population standard deviation of 3.8. A sample of 40 cars from this year’s model SX has an average of 35.2 mpg, with a population standard deviation of 5.4.
Find a 99 percent confidence interval for the difference in average mpg for this car brand (this year’s model minus last year’s).Find a 99 percent confidence interval for the difference in average mpg for last year’s model minus this year’s. What does the negative difference mean?
A special interest group reports a tiny margin of error (plus or minus 0.04 percent) for its online survey based on 50,000 responses. Is the margin of error legitimate? (Assume that the group’s math is correct.)
Suppose that 73 percent of a sample of 1,000 U.S. college students drive a used car as opposed to a new car or no car at all.
Find an 80 percent confidence interval for the percentage of all U.S. college students who drive a used car.What sample size would cut this margin of error in half?
Chapter 15 Solutions
STAT TECH IN BUSINESS & ECON AC
Ch. 15 - A recent insurance industry report indicated that...Ch. 15 - The following hypotheses are given. H0:.70H1:.70 A...Ch. 15 - The following hypotheses are given. H0:=.40H1:.40...Ch. 15 - Prob. 3ECh. 15 - Prob. 4ECh. 15 - Prob. 5ECh. 15 - Prob. 6ECh. 15 - Prob. 2SRCh. 15 - Prob. 7ECh. 15 - The null and alternate hypotheses are: H0:1=2H1:12...
Ch. 15 - Prob. 9ECh. 15 - GfK Research North America conducted identical...Ch. 15 - Prob. 11ECh. 15 - Prob. 12ECh. 15 - Prob. 3SRCh. 15 - Prob. 13ECh. 15 - Prob. 14ECh. 15 - Prob. 15ECh. 15 - Prob. 16ECh. 15 - Prob. 17ECh. 15 - Classic Golf Inc. manages five courses in the...Ch. 15 - Prob. 19ECh. 15 - Prob. 20ECh. 15 - Prob. 4SRCh. 15 - Prob. 21ECh. 15 - Prob. 22ECh. 15 - Prob. 23ECh. 15 - Prob. 24ECh. 15 - Prob. 25ECh. 15 - Advertising expenses are a significant component...Ch. 15 - Prob. 5SRCh. 15 - Prob. 27ECh. 15 - Prob. 28ECh. 15 - Prob. 29ECh. 15 - Prob. 30ECh. 15 - Prob. 31CECh. 15 - Prob. 32CECh. 15 - Prob. 33CECh. 15 - Prob. 34CECh. 15 - Prob. 35CECh. 15 - Research in the gaming industry showed that 10% of...Ch. 15 - Prob. 37CECh. 15 - Prob. 38CECh. 15 - Prob. 39CECh. 15 - Prob. 40CECh. 15 - Prob. 41CECh. 15 - Prob. 42CECh. 15 - The Consumer Confidence Survey is a monthly review...Ch. 15 - Prob. 44CECh. 15 - Prob. 45CECh. 15 - Prob. 46CECh. 15 - Vehicles heading west on Front Street may turn...Ch. 15 - Prob. 48CECh. 15 - Prob. 49CECh. 15 - Prob. 50CECh. 15 - Prob. 51CECh. 15 - Prob. 52CECh. 15 - Prob. 53CECh. 15 - Prob. 54CECh. 15 - Prob. 55CECh. 15 - Prob. 56CECh. 15 - Prob. 57CECh. 15 - Prob. 58CECh. 15 - Prob. 59CECh. 15 - Prob. 60CECh. 15 - Did you ever purchase a bag of MMs candies and...Ch. 15 - Prob. 62DACh. 15 - Prob. 63DACh. 15 - Prob. 64DA
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- You want to compare the average number of tines on the antlers of male deer in two nearby metro parks. A sample of 30 deer from the first park shows an average of 5 tines with a population standard deviation of 3. A sample of 35 deer from the second park shows an average of 6 tines with a population standard deviation of 3.2. Find a 95 percent confidence interval for the difference in average number of tines for all male deer in the two metro parks (second park minus first park).Do the parks’ deer populations differ in average size of deer antlers?arrow_forwardSuppose that you want to increase the confidence level of a particular confidence interval from 80 percent to 95 percent without changing the width of the confidence interval. Can you do it?arrow_forwardA random sample of 1,117 U.S. college students finds that 729 go home at least once each term. Find a 98 percent confidence interval for the proportion of all U.S. college students who go home at least once each term.arrow_forward
- Suppose that you make two confidence intervals with the same data set — one with a 95 percent confidence level and the other with a 99.7 percent confidence level. Which interval is wider?Is a wide confidence interval a good thing?arrow_forwardIs it true that a 95 percent confidence interval means you’re 95 percent confident that the sample statistic is in the interval?arrow_forwardTines can range from 2 to upwards of 50 or more on a male deer. You want to estimate the average number of tines on the antlers of male deer in a nearby metro park. A sample of 30 deer has an average of 5 tines, with a population standard deviation of 3. Find a 95 percent confidence interval for the average number of tines for all male deer in this metro park.Find a 98 percent confidence interval for the average number of tines for all male deer in this metro park.arrow_forward
- Based on a sample of 100 participants, the average weight loss the first month under a new (competing) weight-loss plan is 11.4 pounds with a population standard deviation of 5.1 pounds. The average weight loss for the first month for 100 people on the old (standard) weight-loss plan is 12.8 pounds, with population standard deviation of 4.8 pounds. Find a 90 percent confidence interval for the difference in weight loss for the two plans( old minus new) Whats the margin of error for your calculated confidence interval?arrow_forwardA 95 percent confidence interval for the average miles per gallon for all cars of a certain type is 32.1, plus or minus 1.8. The interval is based on a sample of 40 randomly selected cars. What units represent the margin of error?Suppose that you want to decrease the margin of error, but you want to keep 95 percent confidence. What should you do?arrow_forward3. (i) Below is the R code for performing a X2 test on a 2×3 matrix of categorical variables called TestMatrix: chisq.test(Test Matrix) (a) Assuming we have a significant result for this procedure, provide the R code (including any required packages) for an appropriate post hoc test. (b) If we were to apply this technique to a 2 × 2 case, how would we adapt the code in order to perform the correct test? (ii) What procedure can we use if we want to test for association when we have ordinal variables? What code do we use in R to do this? What package does this command belong to? (iii) The following code contains the initial steps for a scenario where we are looking to investigate the relationship between age and whether someone owns a car by using frequencies. There are two issues with the code - please state these. Row3<-c(75,15) Row4<-c(50,-10) MortgageMatrix<-matrix(c(Row1, Row4), byrow=T, nrow=2, MortgageMatrix dimnames=list(c("Yes", "No"), c("40 or older","<40")))…arrow_forward
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