hw12sol_fall03

# hw12sol_fall03 - < AMSBm 01> 7.68 1 Let m be the mean...

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Unformatted text preview: < AMSBm, 01> 7.68 1. Let m be the mean for Method A and #2 be the mean for Method B. Null hypothesis Ho : n1 —- M2 = 0 Alternative hypothesis H1 : n1 —— M < 0 2. Level of signiﬁcance: (1 = 0.05. 3. Criterion: The null hypothesis speciﬁes 6 2 n1 —,u0 = 0. Since the samples are small, but we can assume that the populations are normal with the same variance, we use the two—sample t statistic t_ (X1 —— X2) — 6 71171201.} + n2 — (111— 1)sl2+ (n2 —1)sg n1+n2 Since the alternative hypothesis is left—sided, 6 < 0, we reject the null hypothesis when t < —t.05 or t < —1.734 since tog, = 1.734 for 18 degrees of freedom. 4. Calculations: Here n1 2 10 and n2 2 10, and we ﬁrst calculate i1 = 70, 81 = 3.3665, i2 = 74, and 82 = 5.3955. Then t: 70—74 10-10-18=_1.9893 9(3.3665)2 + 9(5.3955)2 20 5. Decision: Since —1.989 < —t_05 = —1.734, we reject the null hypothesis at level of signiﬁcance a = .05. Thus, method B is more effective in terms of mean achievement score. 7.72 1. Null hypothesis Ho : ,a = 0 Alternative hypothesis H1 : a > 0 2. Level of signiﬁcance: 0 = 0.01. 3. Criterion: The number of pairs is moderately small so we must assume that each difference has a normal distribution. We use the paired thistatistic __ D1—6 SD/ﬁ t Since a = .01 and the alternative hypothesis is one-sided, we reject the null hypothesis if t > tm. There are 15 degrees of freedom so If > tel = 2.602. 4. Calculations: The sample mean of the differences is 4.0625 and the variance is 16.996. t: 4.0625—0 _3.94 ,/16.996/16 _ 5. Decision We reject the null hypothesis at level of signiﬁcance .01. Thus, the physical exercise program is effective. The P—value is less than .005 so the evidence against the null hypothesis is strong. 0.01 0.0007 1 I \ (a) Rejection region (b) P—value for Problem 7.72 9.6 (a) The sample proportion is 8/60 = .133. Using Table 9(a), the lower con- ﬁdence bounds for sample proportions .12 and .14 are .05 and .067 re— spectively, and the upper conﬁdence bounds are .23 and .26 respectively. Interpolating between the two lower bounds gives the lower conﬁdence bound for p: .013 .05 + 0—2(.067 — .05) = .06. Similarly, the upper conﬁdence bound for p is .013 . —— .2 — . = .2 . 23 + ‘02 ( 6 23) 5 Thus, the 95% conﬁdence interval for p is .06 < p < .25. (b) Using the large sample formula with x/n = .133 and 20/2 = 1.96 gives the 95% conﬁdence interval (.133)(.867) 60 (.133)(.867) .133 —~ 1.96 60 < p < .133 + 1.96 \ 01‘ .047 < p < .219. 9.12 Using the formula for unknown p with E = .035 and 20/2 = 2.575 gives ._1(2-575)2_13532 '4 .035 " “ The required sample size is 1354. 9.20 1. Null hypothesis H0 : p = .2 Alternative hypothesis H1 : p aé .2 2. Level of signiﬁcance: (1 = 0.05. 3. Criterion: Using a normal approximation for the binomial distribution, we reject the null hypothesis when X—n o Z=—p < —z.025 orZ>z_025. “130(1 — Po) Since a = .05 and 2.025 = 1.96, the null hypothesis must be rejected if [Z] > 1.96. 4. Calculations: p0 = .2, X = 11, and n = 104 so ‘\ 11 — 1 4 .2 104(.20)(.80) 5. Decision: Since the observed value ——2.40 < —z.025 = —1.96, we reject the null hypothesis at the 5% level of signiﬁcance. The evidence against the null hypothesis is quite strong since the P-value is .0164 0.025 0.025 -1.96 1.56 z Z (a) Rejection Region (b) P—value for Problem 9.20 9.24 Since the sample size is small, we cannot use the large sample statistic. In this case, the null hypothesis is p = .60, the alternative is p < .60, and the signiﬁcance level is .05. Thus, we need to ﬁnd the largest value ma such that the probability of ma or fewer successes in a sample of size 13 when p = .60 is less than or equal to .05. We then reject the null hypothesis when X S :20. Using Table 1 we see that (ca = 5. Since X = 4, we reject the null hypothesis. 9.30 (a) We use the X2 statistic with 1 degree of freedom to test the null hypotlm : that p] = m against the alternative that p1 # p2. Thus, we reject 1hr null hypothesis at the 5% level when X2 > X205 = 3.841. In Table 9.4, tin expected frequency and the contribution to the X2 statistic of each cell 4m given in the parentheses and brackets respectively. Table 9.4. Exercise 9.30. Total The X2 statistic is 3.002 2 3.002 10.00 3.002 240.00 + 10.00 + 3.002 240.00 + 1.875. N Thus, we cannot reject the null hypothesis. (1)) Using the Z statistic, we reject at the .05 level when lZl > 1.96. In this case, 13 + 7 ——-———-— = .04 250 + 250 ’ f): and 13/250 — 7/250 Z = ._____.—_ (.04)(.96)(2/250) = 1.369. Thus, we cannot reject the null hypothesis. 9.40 To test the null hypothesis that handicap and performance are independent against the alternative that they are not independent at the 5% level, we use the X2 statistic with 4 degrees of freedom and reject the null hypothesis when X2 > X205 2 9.488. In Table 9.6, the expected frequency and the contribution to the X2 statistic of each cell are given in the parentheses and brackets respectively. Table 9.6. Exercise 9.40. 21 64 17 102 (20.34) (63.48) (18.18) [.02] [.00] [.08] 16 49 14 79 I [.00] [.00] [ 00] 29 93 28 150 Neither (29.91) (93.35) (26.74) [.03] [.00] [.06] m- The X2 statistic is .662 .522 ' 1.182 2 + + 20.34 63.48 18.18 .252 .172 .082 + 15.75 + 49.17 + 14.08 .912 .352 1.262 29.91 + 93.35 + 26.74 .196. + ll Thus, we cannot reject the null hypothesis. ...
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