Presentation Transcript

1. Free Response Question Practice

2. Chapter 10
3. 2002 #5 Sleep researchers know that some people are early birds (E), preferring to go to bed by 10 p.m. and arise by 7 a.m., while others are night owls (N), preferring to go to bed after 11 p.m. and arise after 8 a.m. A study was done to compare dream recall for early birds and night owls. One hundred people of each of the two types were selected at random and asked to record their dreams for one week. Some of the results are presented below.
4. a) The researchers believe that night owls may have better dream recall than do early birds. One parameter of interest to the researchers is the mean number of dreams recalled per week with μE representing this mean for early birds and μN representing this mean for night owls. The appropriate hypotheses would then be H0: μE – μN =0 and Ha: μE – μN < 0. State two other pairs of hypotheses that might be used to test the researchers’ belief. Be sure to define the parameter of interest in each case. pE= proportion of early birds who recall dreams pN= proportion of night owls who recall dreams H0: pE– pN = 0 vs. Ha: pE – pN< 0 OR H0: pE= pNvs. Ha: pE< pN
5. b) Use the data provided to carry out a test of the hypotheses about the mean number of dreams recalled per week given in the statement of part (a). Do the data support the researchers’ belief? μE= mean number of dreams early birds recall μN = mean number of dreams night owls recall. H0: μE= μN vs. Ha: μE< μN
6. 1. Problem states that independent random samples were taken. 2. Normal population distributions or large samples. Since these are not normal, we need to note that nE= 100 and nN= 100 are both large in order to perform the t-test. t = –2.52 df = 192 P-value = 0.006 Because the P-value is small (or less than an selected and stated by the student), reject H0. There is convincing evidence that the mean number of dreams night owls recall is greater for than the mean number of dreams early birds recall.
7. 2010 #5 A large pet store buys the identical species of adult tropical fish from two different suppliers – Buy-Rite Pets and Fish Friends. Several of the managers at the pet store suspect that the lengths of the fish from Fish Friends are consistently greater than the lengths of the fish from Buy-Rite Pets. Random samples of 8 adult fish of the species from Buy-Rite Pets and 10 adult fish of the same species from Fish Friends were selected and the lengths of the fish, in inches, were recorded, as shown in the table below. Do the data provide convincing evidence that the mean length of the adult fish of the species from Fish Friends is greater than the mean length of the adult fish of the same species from Buy-Rite Pets?
8. Let μBrepresent the population mean length of all adult fish of this species from Buy-Rite Pets, and let μFrepresent the population mean length of all adult fish of this species from Fish Friends. H0: μB= μFvs. Ha: μB< μF
9. The appropriate test is a two-sample t-test. The first condition is that the samples are independent random samples from the two populations. This was stated in the question. The second condition is that the population distributions of fish lengths are normal. The following dotplots reveal no obvious departures from normality, so it appears reasonable to proceed with the two-sample t-test. df = 15.99999 p-value = 0.3996.
10. Because this p-value is larger than any conventional significance level (such as α = 0.10 or α = 0.05 ), we fail to reject H0. The sample data do not provide convincing evidence to conclude that the mean length of the adult fish of the species from Fish Friends is greater than the mean length of the adult fish of the same species from Buy-Rite Pets.
11. Chapter 11
12. 2010B # 5 An advertising agency in a large city is conducting a survey of adults to investigate whether there is an association between highest level of educational achievement and primary source for news. The company takes a random sample of 2,500 adults in the city. The results are shown in the table below.
13. a) If an adult is to be selected at random from this sample, what is the probability that the selected adult is a college graduate or obtains news primarily from the internet? Using the addition rule, the probability that the randomly selected adult is a college graduate or obtains news primarily from the internet is: P(college graduate or internet) = P(college graduate) + P(internet) – P(college graduate and internet)
14. b) If an adult who is a college graduate is to be selected at random from this sample, what is the probability that the selected adult obtains news primarily from the internet? Reading values from the table, the conditional probability that the selected adult receives news primarily from the internet given that he or she is a college graduate is:
15. c) When selecting an adult at random from the sample of 2,500 adults, are the events “is a college graduate” and “obtains news primarily from the internet” independent? Justify your answer. These events are not independent. One way to establish this is to note that the unconditional probability equals P(obtains news primarily from the internet) = 687/2500 = 0.275 , but the conditional probability equals P(obtains news primarily from the internet / is a college graduate) = 0.354 . Because these two probabilities are not equal, the events “is a college graduate” and “obtains news primarily from the internet” are not independent.
16. d) The company wants to conduct a statistical test to investigate whether there is an association between educational achievement and primary source for news for adults in the city. What is the name of the statistical test that should be used? What are the appropriate degrees of freedom for this test? Chi-square test of association (or independence), with degrees of freedom = (# of rows – 1) × (# of columns –1) = (5 –1) × (3 –1) = 8 .
17. Chapter 12
18. 2010B #6 A real estate agent is interest in developing a model to estimate the prices of houses in a particular part of a large city. She takes a random sample of 25 recent sales and, for each house, records the price (in thousands of dollars), the size of the house (in square feet), and whether or not the house has a swimming pool. This information, along with regression output for a linear model using size to predict price, is show n below and on the next page.
19. a) Interpret the slope of the least squares regression line in the context of the study. The slope coefficient is 0.165. This means that for each additional square foot of size, the predicted price of the house increases by 0.165 thousand dollars, which is $165. In other words, this model predicts that the average price of a house increases by $165 for each additional square foot of a house’s size.
20. b) The second house in the table has a residual of 49. Interpret this residual value in the context of the study. The residual value of 49 for this house indicates that its actual price is 49 thousand dollars higher than the model would predict for a house of its size.
21. The real estate agent is interested in investigating the effect of having a swimming pool on the price of a house. c) Use the residual from all 25 houses to estimate how much greater the price for a house with a swimming pool would be, on average, than the price for a house of the same size without a swimming pool. The average residual value for the eight houses with a swimming pool is: The average residual value for the 17 houses with no swimming pool is:
22. The residual averages suggest that the regression line tends to underestimate the price of homes with a swimming pool by about 18.6 thousand dollars and to overestimate the price of homes with no pool by about 8.8 thousand dollars. The difference between these two residual averages is 18.6 − (−8.8) = 27.4 thousand dollars. This suggests that, for two houses of the same size, the house with a swimming pool would be estimated to cost $27,400 more than the house with no swimming pool.
23. To further investigate the effect of having a swimming pool on the price of a house, the real estate agent creates two regression models, one for houses with a swimming pool and one for houses without a swimming pool. Regression output for these two models is shown below.
24. d) The conditions for inference have been checked and verified, and a 95 percent confidence interval for the true difference in the two slopes is (–0.099, 0.110). Based on this interval, is there a significant difference in the two slopes? Explain your answer. No, this confidence interval does not indicate a significant difference (at the 95 percent confidence level, equivalent to the 5 percent significance level) between the two slope coefficients because the interval includes the value zero.
25. e) Use the regression model for houses with a swimming pool and the regression model for houses without a swimming pool to estimate how much greater the price for a house with a swimming pool would be than the price for a house of the same size without a swimming pool. How does this estimate compare with your result from part (c)? If the two population regression lines do in fact have the same slope, the impact of a swimming pool is the (constant) vertical distance between the two lines. However, because the two fitted lines do not have the same slope, the distance between the two fitted lines depends on the size of the house. Using the available information, there are two acceptable approaches to estimating the impact of having a swimming pool.
26. Approach 1: Use the two fitted lines to predict the price of a house with and without a pool for a particular house size. For example, using the value of size = 2,250 square feet (which is near the middle of the distribution of house sizes), we find: Predicted price for a 2,250 square-foot house with a swimming pool = −11.602 + 0.166 × 2,250 = 361.898 thousand dollars. Predicted price for a 2,250 square-foot house with no swimming pool = −27.382 + 0.160 × 2,250 = 332.618 thousand dollars. The difference in these predicted prices is 361.898 − 332.618 = 29.280 thousand dollars, which is an estimate of the impact of a swimming pool on the predicted price of a 2,250 square-foot house. This is quite similar to the estimate based on residuals in part (c).
27. Approach 2: Because the slopes of the two sample regression lines were judged not to be significantly different, another acceptable approach would be to use the difference in the intercepts of the two fitted lines as an estimate of the vertical distance between the two population regression lines. The difference in the intercepts of the two fitted lines is −11.602 − (−27.382) = 15.780 thousand dollars, which is an estimate of the impact of a swimming pool on the predicted price of a house, assuming this difference does not change with the size of the house. This is quite different from the estimate based on residuals in part (c).