Answers to “Check Your Progress” Questions

Answers to “Check Your Progress” Questions Below are the answers to the Check Your Progress questions that can be found in the StudyCube study guide. We recommend answering these questions on your own and then checking them. Ex a m 2 Re v i e w 1. E 2. C 3. E 4. A 5. D 6. A 7. C 8. A Ex a m 1 Re v i e w 1. C 2. E 3. A 4. B 5. D 6. D 7. C 8. C 9. A Ex a m 3 Re v i e w 1. C 2. A 3. D 4. D 5. C 6. B Exam 1 Review 1. (c) TIME is not an example of discrete data because there are infinite possible measurable outcomes within a certain interval. An example of this is a short sprint could take one runner 10.31 seconds, but another runner 10.47 seconds. All of the other answer choices have finite, specific integer values. 2. (e) To calculate the z-score, use the formula z x   , where x is the data value you are interested in, µ is the population mean, and σ is the population standard deviation. Using the information provided, z 70, 000  65, 000  1.41, 3, 550 meaning the employee’s salary is 1.41 standard deviations above the mean. 1 3. (a) To calculate the percentage of employees with salaries between $66,000 and $70,000, we must first find the z-scores associated with these values. The z-score for $66,000 is z x    66,000  65, 000  0.28. 3,550  70,000  65, 000  1.41. 3,550 The z-score for $70,000 is z x   The percentage of employees with salaries between these z-values is the probability (area) between these z-values: P(1.41) – P(0.28) = 0.9207 – 0.6103 = 0.3104. Therefore, 31.04% of employees make between $66,000 and $70,000. 4. (b) Begin by calculating the standard deviation of the sample mean. Recall that the equation for the standard deviation of the sample mean is calculated as x   n , where σ is the population standard deviation and n is the sample size. The population standard deviation is given as 4.8 lbs and the sample size is 40. Therefore, the standard deviation of the sample mean is 4.8 40  0.76 lbs. Next, we are asked to calculate the probability that the sample mean is between 44.7 lbs and 46.7 lbs (within plus or minus 1 lb of the population mean, 45.7 lbs). Calculate the z-scores for each of these values. For 44.7 lbs, z 44.7  45.7   1.32. 0.76 For 46.7 lbs, z 46.7  45.7  1.32. 0.76 2 Using the z-table, P(-1.32 < Z < 1.32) = P(Z < 1.32) – P(Z < -1.32) = 0.9066 – 0.0934 = 0.8132. Therefore, the probability of having a sample mean between 44.7 lbs and 46.7 lbs is equal to 0.8132. 5. (d) Remember that the level of significance = 1 – confidence coefficient. For a 90% confidence interval, the confidence coefficient is equal to 0.90. Therefore, α = 1 – 0.90 = 0.10. 6. (d) Recall that the equation for an interval estimate of a population proportion is p  z  /2 pq . n The sample proportion, p , of freshmen who have attended more than one football game is 54/100 = 0.54. Since q = 1 – p , q = 1 – 0.54 = 0.46. The sample size, n, is 100. Finally, the critical z-value for a 90% confidence interval is 1.645. Plugging this information into the equation, 0.54  1.645 (0.54)(0.46)  0.54  0.082  (0.458, 0.622) freshmen. 100 Therefore, we can be 90% confident that the proportion of UCF freshmen who attended more than one football game during their freshman year is between 0.458 and 0.622. 7. (c) For a population proportion, n ( z /2 )2 pq , E2 where z is the critical z-value corresponding with the desired confidence level, p is the sample proportion, q = 1 – p , and E is the desired maximum margin of error. For a 95% confidence interval, the critical z-value = 1.96. Since we are not given sample statistics, we use a value of p = q = 0.5 in order to obtain the largest estimate of n. Finally, E = 3% or 0.03. 3 Plugging these values into the equation, n (1.96)2 (0.5)(0.5)  1,067.11 people. (0.03) 2 Delta Air Lines would need to sample 1,068 people in order to predict with 95% probability the proportion of consumers who prefer direct flights over flights with layovers with a margin of error less than or equal to 3%. 8. (c) Assume that the probability distribution of the hypothesized proportion is normally distributed; therefore, we are calculating the test statistic z. Remember that z p  p0 , p 0q 0 n where p is the sample proportion, p0 is the hypothesized proportion (given in the null hypothesis), q0 = 1 – p0, and n is the sample size. The sample proportion, p , is equal to 87/150 = 0.58, the hypothesized proportion p0 = 0.70, q0 = 1 – 0.70 = 0.30, and n = 150. Plugging these values into the equation, z 0.58  0.70 (0.70)(0.30) 150   0.12  3.243. 0.037 9. (a) Remember that the p-value is the probability that, assuming the null hypothesis is true, we attain the observed sample statistic or greater for a positive z-score or that we attain the observed sample statistic or less for a negative z-score with a two-tailed hypothesis test. Therefore, it is the probability of obtaining a sample proportion less than 87/150 = 0.58. (Since the sample proportion is less than the hypothesized proportion, we know that it will be a negative z-score.) We calculated the z-score in question 8 as z = -3.243. Since we are using a twotailed test, the p-value is equal to double the probability that the z-score is less than -3.243, since we are evaluating the lower tail rejection region. 2*P(z < -3.243) = 2*0.0006 = 0.0012. 4 Exam 2 Review 1. (e) Since we are asked to find the positive population mean difference, we must form a confidence interval for µ2 - µ1, since the sample of apartments in NYC cost more than the sample of students in Boston. Since both of our samples are large, the equation for a confidence interval is x2  x1  z /2 s1 2 s2 2  . n1 n 2 The standard deviation of sample 1, s1, is $240, the standard deviation of sample 2, s2, is $415, the sample size for Boston is 50, and the sample size for New York City is 45. The sample mean for sample 1 is $3,100, and the mean of sample 2 is $3,750. Finally, the critical z-value for a 90% confidence interval is 1.645. Plugging these values into the equation, the interval estimate is  2402   4152  (3, 750  3,100) 1.645      $650  $116.08 = ($533.92, $766.08).  50   45  Therefore, we can be 90% confident that the actual mean difference in the price of apartments in Boston and New York City is between $533.92 and $766.08. 2. (c) Remember that the test statistic z for a two-sample hypothesis test of the population mean difference with large samples is given as z ( x1  x2 )  D0 s12 s22  n1 n2 . The means of samples 1 and 2 are 198 and 203 calories, respectively. The standard deviations of samples 1 and 2 are 5.2 and 3.7 calories, and the hypothesized difference, D0, is 0. Plugging these values into the equation, z (198  203)  0  5.2   3.7   45   45      2 2  5  5.25. 0.95 5 3. (e) We are using a lower-tailed hypothesis test with a significance level of α = 0.05. The critical z-value, zα, for a 95% confidence level one-tailed hypothesis test is  1.645. Since we are using a lower-tail test, a test statistic is in the rejection region if z < -1.645. We would only use the absolute value sign if we were conducting a two-tailed hypothesis test. 4. (a) Since we are using small sample sizes (n < 30), we are interested in finding the critical t-value. Remember that the critical t-value depends on the degrees of freedom and the area in one or two tails. Since we are conducting a one-tailed hypothesis test with α = 0.05, the area in one tail = 0.05 and df = n1 + n2 – 2 = 15 + 12 – 2 = 25. Using the t-distribution table, we identify the critical t-value as 1.708. Because we are using an upper-tailed-tailed test, reject H0 if t > 1.708. 5. (d) The null hypothesis is that the mean of the difference values is equal to 0. Since the question states that the researchers wanted to determine if the treatment significantly INCREASED the number of WBC over the placebo, we are using an upper tail test. The alternative hypothesis is that the mean of the difference values is greater than 0. 6. (a) The MSTR predicts the population variance if the null hypothesis is true. This value is calculated as k SST  MSTR  k 1 n ( x j j  x )2 j 1 k 1 , where n j represents the number of observations in a given sample j, x j is the mean of sample j, x is the overall sample mean, and k denotes the number of treatments. The overall sample mean is equal to the average of the sample means for k treatments. This value is equal to (2.94  4.24  5.23)  4.14 drinks/week. 3 Since there are three colleges sampled, k = 3. The sample means are given as 2.94, 4.24, and 5.23 drinks/week. Plugging our data into the equation, MSTR  7(2.94  4.14) 2  8(4.24  4.14) 2  6(5.23  4.14) 2 17.2886   8.6443. 3 1 2 6 7. (c) The MSE provides an estimate of the population variance that is not dependent on the assumption that the population means are equal. It is given by the equation k SSE  MSE = k 1 s 2 j ( n j  1) j 1 Nk , where s j2 is the variance of a given sample j, nj is the number of observations in a sample j, N is the total number of observations collected, and k is the number of treatments. The sample variances are given as 1.30, 1.46, and 2.02 drinks/week. The total number of observations n = 7 + 8 + 6 = 21. Since we are sampling three college campuses, k = 3. Plugging these values into the equation, MSE  1.30(7  1)  1.46(8  1)  2.02(6  1)  1.5622. 21 3 8. (a) Remember that the equation for a simple regression equation is yˆ  ˆ0  ˆ1x. The y-intercept, ˆ0 , is given in the output as the estimate for “intercept”, or -157.6479. The slope, ̂1 , is given in the output as the estimate for “slope”, or 6.6569. Therefore, the estimated regression equation is yˆ   157.6479  6.6569 x. You can predict concession sales by simply plugging in the desired value of x into the regression equation and solving for yˆ . For x = 41,000 (where game attendance is in thousands), yˆ  157.6479  6.6569(41)  115.285, or $115,285. Exam 3 Review 1. (c) Remember that an outlier is an observation that lies outside of the general pattern of the scatterplot. In this graph, point C is an outlier. Since the data point deviates significantly from a line of best fit we can draw between the other values, this represents an outlier that would decrease the magnitude of the correlation coefficient. 7 2. (c) Recall that the standard deviation of the residuals is equal to the square root of the MSE. This value is given in the table as 110.476. Therefore, the standard deviation of the residuals is 110.476  $10.51. 3. (b) A hypothesis test for the regression slope is used to determine if an independent variable is significantly related to the dependent variable by testing the coefficient (slope) of each independent variable. Using the p-value decision rule, we see that the p-values for the two independent variables, distance traveled and gas price, are 0.0001 and 0.0219, respectively. Since both values of p < α = 0.05, we can reject the null hypothesis, that the coefficient is equal to 0, and conclude with 95% confidence that both distance traveled and gas price are significantly affecting cost of transportation. 4. (b) SSR and SSE can be found directly in the Excel regression output. The SSR, the regression sum of squares, can be found under the column “SS’ and the row “Regression.” The SSE, the sum of squared errors (or residuals), can be found under the column “SS” and the row “Residual.” In the data output, the SSR is 54,380 and the SSE is 1,878. 5. (c) The R-square can be found by dividing the SSR by SST. SSR and SST are directly stated in the regression output as 54,380 and 56,259. Dividing the two results in: 54380/56259 = 0.9666. 6. (a) MSE(full) can be found by dividing SSE(full) by n – p – 1. Since the first regression equation has just one independent variable (IV), this will be the reduced equation and the second equation with three IVs will be the full equation. The SSE for the full equation is therefore 620.65. Because the second equation uses a sample size of 10 and there are three IVs being tested, the denominator will be 10 – 3 – 1 = 6. Therefore, MSE(full)  620.65  103.44. 6 8 7. (e) The equation to find the F-test statistic is as follows: F (SSE(reduced)-SSE(full))/number of extra terms MSE(full) Since the original regression equation has just one independent variable (IV), this will be the reduced and the second equation with three IVs will be the full equation. Therefore, SSE(reduced) is 719.07 and SSE(full) is 620.65. The number of extra terms is the difference in number of IVs between the two models. In this case, it will be 3 – 1 = 2. MSE(full) is equal to SSE(full) divided by n – p – 1. Because the second regression equation (full) uses a sample size of 10 and there are three IVs being tested, the denominator will be 10 – 3 – 1 = 6. Therefore, 620.65  103.44 6 8. (719.07-620.65)/2 F  0.4757 103.44 MSE(full)  After finding the F-test statistic, we need to find the F-critical value by using an F-table. The F-critical value associated with an alpha level of 0.05, a numerator df of 2 and denominator df of 6 is 5.1433. Since the F-ratio 0.4757 < 5.1433, the addition of variables x2 and x3 is NOT statistically significant. 9