Review 6 Use the traditional method to test the given hypothesis. Assume that the samples are independent and that they have been randomly selected ) A researcher finds that of,000 people who said that they attend a religious service at ) least once a week, 3 stopped to help a person with car trouble. Of,00 people interviewed who had not attended a religious service at least once a month, stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are equal. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent and that they have been randomly selected. ) Independent samples from two different populations yield the following data. x = 383, x = 448, s = 44, s = 8. The sample size is 77 for both samples. Find the 95 percent confidence interval for! -!. A) -79 <! -! < -5 B) -7 <! -! < -58 C) -7 <! -! < -59 D) -66 <! -! < -64 ) Test the indicated claim about the means of two populations. Assume that the two samples are independent and that they have been randomly selected. 3) A researcher wishes to determine whether people with high blood pressure can reduce 3) their blood pressure by following a particular diet. Use the sample data below to test the claim that the treatment population mean! is smaller than the control population mean!. Test the claim using a significance level of 0.0. Treatment Group Control Group n = 85 n = 75 x = 89. x = 03.7 s = 38.7 s = 39. Construct a confidence interval for! d, the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. 4) A coach uses a new technique in training middle distance runners. The times for 9 different athletes to run 800 meters before and after this training are shown below. Athlete A B C D E F G H I Time before training (seconds) 5. 0.9 08.0.4 07.5 9..3 0.8.3 Time after training (seconds) 6.0 9. 05..9 09. 5. 8.5 0.7 0.9 Construct a 99% confidence interval for the mean difference of the "before" minus "after" times. A) -0.8 <! d < 3.6 B) -0.76 <! d < 3.0 C) -0.54 <! d <.98 D) -0.85 <! d < 3.9 4)
Use the traditional method of hypothesis testing to test the given claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations. 5) The table below shows the weights of seven subjects before and after following a 5) particular diet for two months. Subject A B C D E F G Before 57 60 56 83 90 57 6 After 50 5 54 88 76 59 49 Using a 0.0 level of significance, test the claim that the diet is effective in reducing weight. Test the indicated claim about the variances or standard deviations of two populations. Assume that the populations are normally distributed. Assume that the two samples are independent and that they have been randomly selected. 6) Test the claim that populations A and B have different variances. Use a significance level 6) of 0.0. Sample A Sample B n = 8 n = 4 x = 9. x = 3.7 s = 4.7 s = 5.89.Construct a scatter diagram for the given data. 7) x 0.7 0.07 0.6 0.34-0.6 0.3 0.5 0.08 y 0.5 0.85 0.35 0.46 0.0 0.53 0. -0.03 y 7) - x - A) y B) y - x - x - -
C) y D) y - x - x - - Find the value of the linear correlation coefficient r. 8) x 6 53 64 5 5 54 58 y 58 76 5 64 64 74 6 A) -0.775 B) 0.754 C) 0 D) -0.08 8) Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary. 9) x 6 8 0 8 36 9) y 4 3 0 30 A) y^ = -.79 + 0.950x B) y^ = -3.79 + 0.897x C) y^ = -3.79 + 0.80x D) y^ = -.79 + 0.897x Solve the problem. 0) For the data below, determine the value of the linear correlation coefficient r between y and x. Test whether the correlation is significant. Use a significance level of 0.0. 0) x..7 4.4 6.6 9.5 y.6 4.7 9.9 4.5 39.0 A) 0.985 B) 0.93 C) 0.873 D) 0.990 Find the total variation for the paired data. ) The equation of the regression line for the paired data below is y^ = 6.886 + 4.33937x. Find the explained variation, the unexplained variation, the total variation, and the coefficient of determination. x 9 7 3 4 7 y 43 35 6 3 0 8 A) 6,53.37 B) 6,693.7 C) 3.479 D) 6,544.86 ) 3
Construct the indicated prediction interval for an individual y. ) The equation of the regression line for the paired data below is y^ = 6.89 + 4.3394x and the standard error of estimate is s e =.649. Find the 99% prediction interval of y for x =. x 9 7 3 4 7 y 43 35 6 3 0 8 A) 5. < y < 65.4 B) 59.9 < y < 64.9 C) 56.4 < y < 68.5 D) 53.6 < y < 7.3 ) 4
Answer Key Testname: REVIEW 6 ) H 0 : p = p. H : p p. Test statistic: z =.93. Critical values: z =.96, -.96. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two proportions are equal. ) B 3) H 0 :! =!. H :! <!. Test statistic t = -.365. Critical value: t = -.377. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the treatment population mean! is smaller than the control population!. 4) A 5) Test statistic t =.954. Critical value: t = 3.43. Fail to reject H 0 :! d = 0. There is not sufficient evidence to support the claim that the diet is effective in reducing weight. 6) H 0 : σ = σ. H : σ σ. Test statistic: F =.57. Critical value: F =.84. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that populations A and B have different variances. 7) A 8) A 9) B 0) D ) D ) A 5
Problem The data in the problem can be summarized as The alternative hypothesis indicates that we will use a two-tailed test. The pooled sample proportion (p. 457) p x n The test statistic (p. 457) + x 3+ 53 + n 000 + 00 00 = = = 0.04. The critical value (from Table A) zα = z.050 =.96. We fail to reject the null-hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two proportions are equal.
Problem We use the formula for confidence interval from page 470. ( ) ( µ µ ) ( ) x x E < < x x + E where the margin error E is given by the formula s s E = tα +. n n The number of degrees of freedomdf = min( n, n ) = 76. Table A3 contains only values of df equal to 00 or 00. We take the closest valuedf = 00. From table A3 we find t α =.97. Therefore 44 + 8 E =.97 7 77 Because x x = 65we have the confidence interval as The correct answer is B. 7 < x x < 58. Problem 3 The null hypothesis H 0 : µ = µ. The alternative hypothesis H : µ < µ. The test is left-tailed. The test statistic x x 89. 03.7 t = =.365 s s 38.7 39. + + n n 85 75 The critical value from Table A3 (area in one tail 0.0, df = 74) is.377 We fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the treatment population mean is smaller than the control population mean.
Problem 4 The differences in pairs are 0.8,.8,.9, 0.5,.6, 3.9,.8, 0.,.4 The mean of differences is d.. The standard deviation of differences is sd.86 s.86 The margin error (page 485) E = t d α = 3.355.04. n 3 The confidence interval The correct answer is A. d E < µ < d + E or -0.8< µ < 3.6 H µ >. The test is right- Problem 5 The null hypothesis H : 0 0 µ d =. The alternative hypothesis (the claim) : d 0 tailed. The differences are 7, 9,, -5, 4, -,. 7 + 9 + 5 + 4 + 37 The meand = = 5.857. 7 7 The standard deviation sd 7.033. The test statisticst d d d µ d 5.857 0 = n = 7.989. s 7.033 d The critical value from Table A3 for the area in one tail 0.0 and 6 degrees of freedom is 3.43. We fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the diet is effective in reducing weight.
The null hypothesis The alternative hypothesis H : σ = σ. 0 Problem 6 H : σ σ. 4.7 The test is two tailed. s 5.89 The test statistics F = =.5705 s The critical value can be found in Table A5 on page 779. The area in one tail is 0.05, the numerator degrees of freedom is 40 and the denominator degrees of freedom is 7. The critical value is.836. We fail to reject the null hypothesis. There is not sufficient evidence to support the claim that populations A and B have different variances. Problem 7 Above we see the scatterplot constructed by Statdisk. The closest answer is C.
Problem 8 We use the formula 0- on page 50. The correct answer is B. n xy ( x)( y) ( ) ( ) ( ) ( ) r = = n x x n y y 7 64636 395 49 = 0.775. 7 437 395 7 89053 49 Problem 9 We use formula 0- on page 54 to compute the slope of the regression line. ( ) ( )( ) n( x ) ( x) n xy x y 5 944 98 69 b = = 0.897 5 580 98 To find the y-intercept of the regression line we use formula 0-3. The correct answer is C. b = y bx= 69 0.897 98 0 5 3.78 Problem 0 First we compute the correlation coefficient. Because we use values x instead of x formula 0- takes the form n xy ( x)( y) 4 ( ) ( ) ( ) ( ) r = = n x x n y y 5 485.0 6.9 79.7 =.990. 5 047.5634 6.9 5 43.9 79.7 The correct answer to the first part of the problem is C. Next we follow the procedure described on page 57. The null-hypothesis H : 0 0 ρ = (There is no linear correlation). The alternative hypothesis H : 0 ρ (There is linear correlation). The test statistics r 0.990 t = =.554. 0.099 / 3 ( r ) ( n ) From table A-3 we see that the critical value for two-tailed test at a significance level 0.0 is 5.84. Because the test statistics is larger than
the critical value we reject the null-hypothesis. There is a significant correlation between values of x and y. Problem We will use formula 0-4 on page 559. First we need to compute the predicted values according to the formula yˆ = 6.886 + 4.33937x The results of computations are presented in the table below. x 9 7 3 4 7 y 43 35 6 3 0 8 ŷ 45.379 36.55845 4.866 9.0097 3.54034 0.649 79.955 The mean of y-values is y = 45.8574. The explained variation is ( yˆ y) 653.36578. The unexplained variation is ( y yˆ ) 3.47940. The total variation is 653.36578 + 3.47940 = 6544.8459. The closest answer is D. The coefficient of determination is explained variation 653.36578 r = = 0.99794. total variation 6544.8459 Problem We use the formulas for prediction interval on page 56. yˆ E < y< yˆ + E, nx ( 0 x) E = tα se + +, and n n x x s e = ˆ ( y y). n ( ) ( ) First we compute the standard error of estimate. From the previous problem we know that the unexplained variation ( y yˆ ) is 3.47940, whence 3.47940 s e =.649 5
Next from Table A-3 keeping in mind that the number of degrees of freedom is n = 5 we find that t 0.005 = 4.03. Finally x =, x = 9.486, x= 64, x = 93, and y = 6.89 + 4.3394 58.3 ˆ 0 Plugging in these numbers we get that the margin of error is 7( 9.486) E = 4.03.649 + + 7. 7 7 93 64 The prediction interval is 5.<y<65.4. The correct answer is B.