Statistics 13 Solution for Homework #5

Statistics 13 Solution for Homework #56.42 p-value = P(z 2 2.17) + P(z I -2.17) = (-5 - .4850)2 = .0300 (using Table IV, Appendix A)6.46 a. From Exercise 8.25, z = -85.52. The y-value isp = P(z 5 -85.52) + P(z 2 85.52)= (.5 - .5) -+ (.5 - .5) = 0 (using Table lV, Appendix A).b. The y-value is y = 0. Since this is less than a= .lo, Ho is rejected. There is sufficientevidence to indicate the mean response for the population of all New York City publicschool children different from 3 at a= .lo. .6.50 a. If chickens are more apt to peck at white string, then they are less apt to peck at bluestring. Let ,u= mean number of pecks at a blue string. To determine if chickens are moreapt to peck at white string than blue string (or less apt to peck at blue string), we test:The test statistic isZ-PO 1.13-7.5z = - fl.. 2.21= -24.46The rejection regio~i requires ff= .05 in tlie lower tail of the z distribution. From TableIV, Appendix A, 2.0~ = 1.645. The rejection region is z < -1.645.Since the observed value of the test statistic falls in the rejection region (z = -24.46 < --I .645),Ho is rejected. There is sufficient eviderice to indicate the chickens are less apt to peck at bluestring at a= .05.b. In Exercise 7.17 b, we concluded that tlic birds were more apt to peck at white string.The mean nurnbcr of pccks for white string is 7.5. Since 7.5 is riot in the 99%confidence interval for the mean number of pecks at blue string, it is not a likely valuefor the true mean for blue string.6.60 a. We ~iiust assume that a random sample was draw11 from a normal population.b. The hypotheses are:'I Ile test slati>~.ii: is I1 .S9.The y-value is ,038.There is evidence to rejcct kIO for a> .038. There is evidence to indicate tlie mean isgreater than I000 for a> .038.

c. The hypotheses are:The test statistic is t = 1.89.'There is no evidence to reject Ho for a= .05. There is insufficient evidence to indicatethe mean is different than 1000 for a= .05.'fliere is evidence to reject Ho for a > .076. There is evidence to indicate the mean isdifferent than I000 for a > .076.6.62 a. To determine if the average number of books read by all students who participate in theextensive reading program exceeds 25, we test:b. The rejection region requires a=.05 in the upper tail of the t distribution with df = n - 1= 14 - 1 = 13. From Table V1, Appendix A, toj = 1.771. The rejection region is t >1.77 1.X-,D, - 31.64-25c. 'I'he test statistic is t = - S 10.485= 2.37d. Since the observed value of the test statistic falls in the rejection region(t = 2.37 > 1.771), Ho is rejected. There is sufficient evidence to indicate that theaverage number of books read by all students who participate in the extensive readingprogram exceeds 25 at a = .05.e. The conditions required for this test is a random sample from the target population andthe population from which the sample is selected is approximately normal.6.66 a. To deter~iiine if the mean repellency percentage of the new mosquito repellent is lessthan 95, we test:Z-Po 83 - 95The test statistic is t = ----- ------ = -1.79s/& 15/&

The rejection region requires a= . I0 in the lower tail of the t distribution. From 'TableV1, Appendix A, with df = n - I = 5 - I = 4, t = 1.533. The rejection region is t 1.28.Since the observed value of the test statistic falls in the rejection region (1.89 > 1.28),Ho is rejected. There is sufficient evidence to indicate that the proportion is greater than.65 at a= .lo.P - po - p - po - .74 -.90The test statistic is z L - ----- . = -5.33The rejection region requlres d 2 = ,0512 = .025 in each tail of the z distribution. FromTable lV, Appendix A. = 1 96. The re-iection region is z .: - 1.96 or z > I .96.Since the observed value of the test statistic falls in the rejection region (-5.33 < -1.96),Ho is rejected. There is sufficient evidence to indicate that the proportion is differentfrom .90 at a= .05.

d. For confidence coefficient .95, a= .05 and d2 = .025. From Table IV, Appendix A,z 025 = 1.96. The confidence interval is:e. For confidence coefficient .99, a= .0l and d2 = .005. From Table IV, Appendix A,zoos = 2.58. The confidence interval is:x 646.78 a. The point estimate forp is j = - =- = .604n 106b. Ho: p = .70Ha: p # .70p - po - ,604 - .70c. The test statistic is z = -2.16d. 'The rejection region requires d 2 = .01/2 -= .005 in each tail of the z distribution. FromTable IV, Appendix A, zoos = 2.58. The rejection region is z < -2.58 or z > 2.58.e. First, check to see if the normal approximation will be adequate:Since the interval is co~npletely in the interval (0, I), the normal approxi~nation will beadequate. ,Since the observed value of the test statistic does not fall in the rejection region(z = -2.16 Q: -2.58), Ho is not rejected. There is insufficient evidence to indicate theproportion of consumers who believe "Made in the USA" means 100% of labor andmaterials are from the United States is different from .70 at a= .O1.x 356.82 ' a. ?.he point estimate for p is p = - = ---- = .030n 1,165First, check to see if the normal approximation will be adequate:Since the interval is completely in the interval (0, I), the normal approximation will beadequate.

To determine if the true driver cell phone use rate differs from .02, we test:,030 - .02The test statistic is z = ----- - 2.44The rejection region requires d 2 = .05/2 = .025 in each tail of the z distribution. Fro111Table IV, Appendix A, 2.025 = 1.96. The rejection region is 2 < -1.96 or z > 1.96.Since the observed value of the test statistic falls in the rejection region(z = 2.44 > 1.96), Ho is rejected. There is sufficient evidence to indicate the true drivercell phone use rate differs from .02 at a= .05.b. I11 Exercise 7.49, the confidence interval forp is (.020, .040). Since .02 is contained inthis interval, there is no evidence to reject Ho. This does not agree with the conclusionfor this test. However, if the calculations were carried out to 4 decimal places, theinterval would be (.0202, .0398). Using this interval, .02 is not contained in the interval.This agrees with the test in part a.6.86 For the Top of the Core:First, check to see if the normal approximation is adequate:Since the interval falls conlpletelq in the interval (0,l ), the normal tlistribution will bcadequate.To determine if the coat index exceeds .5, we test:The test statistic ia z -The rejection region requires a = .05 in the upper tai I of the z distribution. From Table IV,Appendix A, zo5 = 1.645. The rejection region is z > 1.645.Since thc observed value of the test statistic falls 111 the rejection region (z - 4.80 > 1.645), fhis rejected. 'fhere is sufficient evidence to indicate that the coat index exceeds .5 at LY= .05For tllc Rliddle of the Core:First, check to see if the ~~onnal approxi~nation is adequate:puk 30,) ail + 7 -3 5 5 3u- -v-11 -('5)('5) 3 .5 & 1 76 =, t.324, ,676)

Since the interval falls completely in the interval (0,l ), the normal distribution will beadequate.To determine if the coat index differs from .5, we test:p-po .479-.5 1The test statistic is z = - . - -.36The rejection region requires ~$2 = .05/2 = .025 in each tail of the z distribution. From TableIV, Appendix A, 2025 = I .96. The rejection region is z < -1.96 6r z > 1.96.Since the observed value of the test statistic does rlot fall in the rejection region (z = -.36 P:-1.96), Ho is not rejected. There is insufficient evidence to indicate that the coat index differsfrom .5 at a= .05.For the Bottom of the Core:First, check to see if the normal approxi~natioll is adequate:Since the interval falls co~npletely in the interval (0,l ), the normal distribution will beadequate.To determine if the coat index is less than .5, we test:The test statistic is z =The rejection region requires a= .05 in lhe lower tail of'the z distribution. From Table IV.Appendix A, zoj - 1.645. The rejectiorl region is z < -1.645.Since the observed value of the test statictic fklls in the re-jection region (z = -2.56 < -1.645),Ho is rejected. There is sufficient evidence to indicate that the coat index is less than .5 ata= .05.