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17. (a) Yes. The value 3.5 is greater than the upper confidence bound of 3.45. Quantities greater than the upper
confidence bound will have P-values less than 0.05. Therefore P < 0.05.
(b) No, we would need to know the 99% upper confidence bound to determine whether P < 0.01.
18. (a) No, we would need to know the 99% lower confidence bound in order to determine whether P < 0.01.
(b) Yes, since 50 is less than the 98% lower confidence bound, we know that P < 0.02. Therefore P < 0.05.
19.
Yes, we can compute
the P-value exactly.
√
√ Since the 95% upper confidence bound is 3.45, we know that
3.40 + 1.645s/ n = 3.45. Therefore s/ n = 0.0304. The z-score is (3.40 − 3.50)/0.0304 = −3.29. The
P-value is 0.0005, which is less than 0.01.
20.
No, the confidence bound itself tells us only that P < 0.02. To determine the P-value more precisely,
we would
√
need to know the standard deviation of the sample mean. This is approximately equal to s/ n where s is the
sample standard deviation and n is the sample size. We know that s = 5, but we do not know n.
Section 6.3
1.
X = 25, n = 300, pb = 25/300 = 0.083333.
The null and alternate hypotheses are H0 : p ≤ 0.05 versus H1 : p > 0.05.
p
z = (0.083333 − 0.5)/ 0.05(1 − 0.05)/300 = 2.65.
Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 2.65,
so P = 0.0040. The claim is rejected.
2.
3.
No. n = 100 and p0 = 0.05, so np0 = 5, which is less than 10.
X = 25, n = 400, pb = 25/400 = 0.0625.
The null and alternate hypotheses are H0 : p ≤ 0.05 versus H1 : p > 0.05.
p
z = (0.0625 − 0.05)/ 0.05(1 − 0.05)/400 = 1.15.
Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 1.15,
so P = 0.1251. The company cannot conclude that more than 5% of their subscribers would pay for the channel.
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SECTION 6.3
4.
n = 304, pb = 0.62.
The null and alternate hypotheses are H0 : p ≤ 0.5 versus H1 : p > 0.5.
p
z = (0.62 − 0.5)/ 0.5(1 − 0.5)/304 = 4.18.
Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 4.18,
so P ≈ 0. It can be concluded that more than half of patients prefer a physician with high technical skills.
5.
X = 330, n = 600, pb = 330/600 = 0.55.
The null and alternate hypotheses are H0 : p ≤ 0.50 versus H1 : p > 0.50.
p
z = (0.55 − 0.50)/ 0.50(1 − 0.50)/600 = 2.45.
Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 2.45,
so P = 0.0071. It can be concluded that more than 50% of all likely voters favor the proposal.
6.
X = 18, n = 50, pb = 18/50 = 0.36.
The null and alternate hypotheses are H0 : p ≥ 0.50 versus H1 : p < 0.50.
p
z = (0.36 − 0.50)/ 0.50(1 − 0.50)/50 = −1.98.
Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −1.98,
so P = 0.0239. It can be concluded that fewer than half of the incinerators meet the standard.
7.
X = 470, n = 500, pb = 470/500 = 0.94.
The null and alternate hypotheses are H0 : p ≥ 0.95 versus H1 : p < 0.95.
p
z = (0.94 − 0.95)/ 0.95(1 − 0.95)/500 = −1.03.
Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −1.03,
so P = 0.1515. The claim cannot be rejected.
8.
X = 12, n = 300, pb = 12/300 = 0.04.
The null and alternate hypotheses are H0 : p ≥ 0.08 versus H1 : p < 0.08.
p
z = (0.04 − 0.08)/ 0.08(1 − 0.08)/300 = −2.55.
Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −2.55,
so P = 0.0054. The machine can be qualified.
9.
X = 15, n = 400, pb = 15/400 = 0.0375.
The null and alternate hypotheses are H0 : p ≥ 0.05 versus H1 : p < 0.05.
p
z = (0.0375 − 0.05)/ 0.05(1 − 0.05)/400 = −1.15.
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CHAPTER 6
Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −1.15,
so P = 0.1251. It cannot be concluded that the new process has a lower defect rate.
10.
X = 37, n = 50, pb = 37/50 = 0.74.
The null and alternate hypotheses are H0 : p ≤ 0.60 versus H1 : p > 0.60.
p
z = (0.74 − 0.60)/ 0.60(1 − 0.60)/50 = 2.02.
Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 2.02,
so P = 0.0217. It can be concluded that more than 60% of the measurements are satisfactory.
11.
X = 17, n = 75, pb = 17/75 = 0.22667.
The null and alternate hypotheses are H0 : p ≥ 0.40 versus H1 : p < 0.40.
p
z = (0.22667 − 0.40)/ 0.40(1 − 0.40)/75 = −3.06.
Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −3.06,
so P = 0.0011. It can be concluded that less than 40% of the fuses have burnout amperages greater than 15 A.
12. (a) One-tailed. The alternate hypothesis is of the form p < p0 .
(b) H0 : p ≥ 0.4
(c) Yes, P = 0.001, which is less than 0.02.
(d) The sample proportion is 0.304167. Therefore the P-value for the null hypothesis p ≥ 0.45 will be less than
the P-value for the null hypothesis p ≥ 0.40, which is 0.001. It follows that the null hypothesis p ≥ 0.45 can
be rejected at the 2% level.
(e) X = 73, n = 240, pb = 73/240 = 0.304167.
The null and alternate hypotheses are H0 : p ≤ 0.25 versus H1 : p > 0.25.
p
z = (0.304167 − 0.25)/ 0.25(1 − 0.25)/240 = 1.94.
Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 1.94.
Thus P = 0.0262.
(f) X = 73, n = 240, p̃ = (73 + 2)/(240 + 4) = 0.307377, z.05 = 1.645.
p
The confidence interval is 0.307377 ± 1.645 0.307377(1 − 0.307377)/(240 + 4), or (0.2588, 0.3560).
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SECTION 6.4
13. (a) Sample p = pb = 345/500 = 0.690.
(b) The null and alternatephypotheses are H0 : p ≥ 0.7 versus H1 : µ < 0.7. n = 500. From part (a), pb = 0.690.
z = (0.690 − 0.700)/ 0.7(1 − 0.7)/500 = −0.49.
(c) Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −0.49.
Thus P = 0.3121.
Section 6.4
1. (a) X = 100.01, s = 0.0264575, n = 3. There are 3 − 1 = 2 degrees of freedom.
The null and alternate hypotheses are H0 : µ = 100 versus H1 : µ 6= 100.
√
t = (100.01 − 100)/(0.0264575/ 3) = 0.6547.
Since the alternate hypothesis is of the form µ 6= µ0 , the P-value is the sum of the areas to the right of t = 0.6547
and to the left of t = −0.6547.
From the t table, 0.50 < P < 0.80. A computer package gives P = 0.580.
We conclude that the scale may well be calibrated correctly.
(b) The t-test cannot be performed, because the sample standard deviation cannot be computed from a sample of
size 1.
2. (a) Reject the assumptions: the value 221.03 is an outlier.
(b) The assumptions appear to be met.
(c) Reject the assumptions: the values are in increasing order, so this does not appear to be a simple random
sample.
3. (a) H0 : µ ≤ 35 versus H1 : µ > 35
(b) X = 39, s = 4, n = 6. There are 6 − 1 = 5 degrees of freedom.
From part (a), the null and alternate hypotheses are H0 : µ ≤ 35 versus H1 : µ > 35.
√
t = (39 − 35)/(4/ 6) = 2.449.
Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 2.449.
From the t table, 0.025 < P < 0.050. A computer package gives P = 0.029.
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CHAPTER 6
(c) Yes, the P-value is small, so we conclude that µ > 35.
4. (a) X = 20.75, s = 3.93, n = 26. There are 26 − 1 = 25 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≤ 20 versus H1 : µ > 20.
√
t = (20.75 − 20)/(3.93/ 25) = 0.973.
Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 0.973.
From the t table, 0.10 < P < 0.25. A computer package gives P = 0.170.
We cannot conclude that the mean concentration is greater than 20 mg/kg.
(b) X = 20.75, s = 3.93, n = 26. There are 26 − 1 = 25 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≥ 25 versus H1 : µ < 25.
√
t = (20.75 − 25)/(3.93/ 25) = −5.514.
Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −5.514.
From the t table, P < 0.0005. A computer package gives P = 4.96 × 10−6.
We conclude that the mean concentration is less than 25 mg/kg.
5. (a) X = 61.3, s = 5.2, n = 12. There are 12 − 1 = 11 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≤ 60 versus H1 : µ > 60.
√
t = (61.3 − 60)/(5.2/ 12) = 0.866.
Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 0.866.
From the t table, 0.10 < P < 0.25. A computer package gives P = 0.202.
We cannot conclude that the mean concentration is greater than 60 mg/L.
(b) X = 61.3, s = 5.2, n = 12. There are 12 − 1 = 11 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≥ 65 versus H1 : µ < 65.
√
t = (61.3 − 65)/(5.2/ 12) = −2.465.
Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −2.465.
From the t table, 0.01 < P < 0.025. A computer package gives P = 0.0157.
We conclude that the mean concentration is less than 65 mg/L.
6.
X = 90.3, s = 1.290349, n = 5. There are 5 − 1 = 4 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≤ 90 versus H1 : µ > 90.
√
t = (90.3 − 90)/(1.290349/ 5) = 0.5199.
Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 0.5199.
From the t table, 0.25 < P < 0.40. A computer package gives P = 0.315.
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SECTION 6.4
We cannot conclude that the mean octane rating is greater than 90%.
7. (a)
3.8
4
4.2
(b) Yes, the sample contains no outliers.
X = 4.032857, s = 0.061244, n = 7. There are 7 − 1 = 6 degrees of freedom.
The null and alternate hypotheses are H0 : µ = 4 versus H1 : µ 6= 4.
√
t = (4.032857 − 4)/(0.061244/ 7) = 1.419.
Since the alternate hypothesis is of the form µ 6= µ0 , the P-value is the sum of the areas to the right of t = 1.419
and to the left of t = −1.419.
From the t table, 0.20 < P < 0.50. A computer package gives P = 0.2056.
It cannot be concluded that the mean thickness differs from 4 mils.
(c)
3.9
4
4.1
4.2
4.3
(d) No, the sample contains an outlier.
8. (a) H0 : µ ≤ 85 versus H1 : µ > 85
(b) X = 90.55, s = 2.901551, n = 6. There are 6 − 1 = 5 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≤ 85 versus H1 : µ > 85.
√
t = (90.55 − 85)/(2.901551/ 6) = 4.685.
Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 4.685.
From the t table, 0.001 < P < 0.005. A computer package gives P = 0.0027.
(c) Yes, the P-value is small, so we conclude that µ > 85.
9.
X = 1.88, s = 0.21, n = 18. There are 18 − 1 = 17 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≥ 2 versus H1 : µ < 2.
√
t = (1.88 − 2)/(0.21/ 18) = −2.424.
Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −2.424.
From the t table, 0.01 < P < 0.025. A computer package gives P = 0.013.
We can conclude that the mean warpage is less than 2 mm.
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10.
CHAPTER 6
X = 7.22, s = 2.328519, n = 5. There are 5 − 1 = 4 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≥ 8 versus H1 : µ < 8.
√
t = (7.22 − 8)/(2.328519/ 5) = −0.7490.
Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −0.7490.
From the t table, P ≈ 0.25. A computer package gives P = 0.248.
We cannot conclude that the mean amount is less than 8%.
11.
X = 1.25, s = 0.624500, n = 4. There are 4 − 1 = 3 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≥ 2.5 versus H1 : µ < 2.5.
√
t = (1.25 − 2.5)/(0.624500/ 4) = −4.003.
Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −4.003.
From the t table, 0.01 < P < 0.025. A computer package gives P = 0.014.
We can conclude that the mean amount absorbed is less than 2.5%.
12. (a) One-tailed. The alternate hypothesis is of the form µ > µ0 .
(b) H0 : µ ≤ 5.5
(c) Yes, the P-value is 0.002, which is less than 0.01.
√
(d) X = 5.92563, n = 5, s/ n = 0.07046. There are 5 − 1 = 4 degrees of freedom.
The null and alternate hypotheses are H0 : µ ≥ 6.5 versus H1 : µ < 6.5.
t = (5.92563 − 6.5)/0.07046 = −8.152.
Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −8.152.
From the t table, 0.001 < P < 0.0005. A computer package gives P = 0.00062.
√
(e) X = 5.92563, n = 5, s/ n = 0.07046. There are 5 − 1 = 4 degrees of freedom.
t4,.005 = 4.604. The confidence interval is 5.92563 ± 4.604(0.07046), or (5.6012, 6.2500).
√
√
13. (a) StDev = (SE Mean) N = 1.8389 11 = 6.0989.
(b) t10,.025 = 2.228. The lower 95% confidence bound is 13.2874 − 2.228(1.8389) = 9.190.
(c) t10,.025 = 2.228. The upper 95% confidence bound is 13.2874 + 2.228(1.8389) = 17.384.
(d) t = (13.2874 − 16)/1.8389 = −1.475.
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SECTION 6.5
Section 6.5
1.
X = 8.5, sX = 1.9, nX = 58, Y = 11.9, sY = 3.6, nY = 58.
The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0.
p
z = (8.5 − 11.9 − 0)/ 1.92 /58 + 3.62/58 = −6.36. Since the alternate hypothesis is of the form µX − µY < ∆,
the P-value is the area to the left of z = −6.36.
Thus P ≈ 0.
We can conclude that the mean hospital stay is shorter for patients receiving C4A-rich plasma.
2.
X = 20.95, sX = 14.5, nX = 482, Y = 22.79, sY = 15.6, nY = 614.
The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0.
p
z = (20.95 − 22.79 − 0)/ 14.52 /482 + 15.62/614 = −2.02. Since the alternate hypothesis is of the form
µX − µY < ∆, the P-value is the area to the left of z = −2.02.
Thus P = 0.0217.
We can conclude that the mean weight of male spotted flounder is greater than that of females.
3.
X = 5.92, sX = 0.15, nX = 42, Y = 6.05, sY = 0.16, nY = 37.
The null and alternate hypotheses are H0 : µX − µY = 0 versus H1 : µX − µY 6= 0.
p
z = (5.92−6.05−0)/ 0.152 /42 + 0.162/37 = −3.71. Since the alternate hypothesis is of the form µX − µY 6= ∆,
the P-value is the sum of the areas to the right of z = 3.71 and to the left of z = −3.71.
Thus P = 0.0001 + 0.0001 = 0.0002.
We can conclude that the mean dielectric constant differs between the two types of asphalt.
4.
X = 27.2, sX = 1.2, nX = 40, Y = 28.1, sY = 2.0, nY = 40.
The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0.
p
z = (27.2 − 28.1 − 0)/ 1.22 /40 + 2.02/40 = −2.44. Since the alternate hypothesis is of the form µX − µY < ∆,
the P-value is the area to the left of z = −2.44.
Thus P = 0.0073.
We can conclude that the mean mileage is greater with premium fuel.
5.
X = 40, sX = 12, nX = 75, Y = 42, sY = 15, nY = 100.
The null and alternate hypotheses are H0 : µX − µY > 0 versus H1 : µX − µY ≤ 0.
p
z = (40 − 42 − 0)/ 122/75 + 152/100 = −0.98. Since the alternate hypothesis is of the form µX − µY ≤ ∆,
the P-value is the area to the left of z = −0.98.
Thus P = 0.1635.
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CHAPTER 6
We cannot conclude that the mean reduction from drug B is greater than the mean reduction from drug A.
6. (a) The 60 counts may not be independent if they are taken in consecutive time periods.
(b) X = 73.8, sX = 5.2, nX = 60, Y = 76.1, sY = 4.1, nY = 60.
The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0.
p
z = (73.8 − 76.1 − 0)/ 5.22 /60 + 4.12/60 = −2.69. Since the alternate hypothesis is of the form µX − µY < ∆,
the P-value is the area to the left of z = −2.69.
Thus P = 0.0036.
We can conclude that machine 2 is faster.
7. (a) X = 7.79, sX = 1.06, nX = 80, Y = 7.64, sY = 1.31, nY = 80.
Here µ1 = µX and µ2 = µY . The null and alternate hypotheses are H0 : µX − µY ≤ 0 versus H1 : µX − µY > 0.
p
z = (7.79 − 7.64 − 0)/ 1.062/80 + 1.312/80 = 0.80. Since the alternate hypothesis is of the form µX − µY > ∆,
the P-value is the area to the right of z = 0.80.
Thus P = 0.2119.
We cannot conclude that the mean score on one-tailed questions is greater.
(b) The null and alternate hypotheses are H0 : µX − µY = 0 versus H1 : µX − µY 6= 0.
The z-score is computed as in part (a): z = 0.80.
Since the alternate hypothesis is of the form µX − µY 6= ∆, the P-value is the sum of the areas to the right of
z = 0.80 and to the left of z = 0.80.
Thus P = 0.2119 + 0.2119 = 0.4238.
We cannot conclude that the mean score on one-tailed questions differs from the mean score on two-tailed
questions.
8. (a) X = 495.6, sX = 19.4, nX = 50, Y = 481.2, sY = 14.3, nY = 50.
The null and alternate hypotheses are H0 : µX − µY ≤ 0 versus H1 : µX − µY > 0.
p
z = (495.6−481.2−0)/ 19.42 /50 + 14.32/50 = 4.22. Since the alternate hypothesis is of the form µX − µY > ∆,
the P-value is the area to the right of z = 4.22.
Thus P ≈ 0.
We can conclude that the mean speed for the new chips is greater.
(b) X = 495.6, sX = 19.4, nX = 50, Y = 391.2, sY = 17.2, nY = 60.
The null and alternate hypotheses are H0 : µX − µY ≤ 100 versus H1 : µX − µY > 100.
p
z = (495.6 − 391.2 − 100)/ 19.42/50 + 17.22/60 = 1.25. Since the alternate hypothesis is of the form
µX − µY > ∆, the P-value is the area to the right of z = 1.25.
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SECTION 6.5
Thus P = 0.1056.
The data do not provide convincing evidence for this claim.
9. (a) X = 625, sX = 40, nX = 100, Y = 640, sY = 50, nY = 64.
The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0.
p
z = (625 − 640 − 0)/ 402 /100 + 502/64 = −2.02. Since the alternate hypothesis is of the form µX − µY < ∆,
the P-value is the area to the right of z = −2.02.
Thus P = 0.0217.
We can conclude that the second method yields the greater mean daily production.
(b) The null and alternate hypotheses are H0 : µX − µY ≥ −10 versus H1 : µX − µY < −10.
p
z = (625 − 640 + 10)/ 402 /100 + 502/64 = −0.67. Since the alternate hypothesis is of the form µX − µY < ∆,
the P-value is the area to the right of z = −0.67.
Thus P = 0.2514.
We cannot conclude that the mean daily production for the second method exceeds that of the first by more
than 10 tons.
10.
X = 91.1, sX = 6.23, nX = 50, Y = 90.7, sY = 4.34, nY = 40.
The null and alternate hypotheses are H0 : µX − µY ≤ 0 versus H1 : µX − µY > 0.
p
z = (91.1 − 90.7 − 0)/ 6.232/50 + 4.342/40 = 0.36. Since the alternate hypothesis is of the form µX − µY > ∆,
the P-value is the area to the right of z = 0.36.
Thus P = 0.3594. We cannot conclude that the mean hardness of welds cooled at 10◦ C is greater than that of
welds cooled at 30◦ C.
11.
X 1 = 4387, s1 = 252, n1 = 75, X 2 = 4260, s2 = 231, n2 = 75.
The null and alternate hypotheses are H0 : µ1 − µ2 ≤ 0 versus H1 : µ1 − µ2 > 0.
p
z = (4387 − 4260 − 0)/ 2522/75 + 2312/75 = 3.22. Since the alternate hypothesis is of the form µX − µY > ∆,
the P-value is the area to the right of z = 3.22.
Thus P = 0.0006. We can conclude that new power supplies outlast old power supplies.
12. (a) One-tailed. The alternate hypothesis is of the form µX − µY < ∆.
(b) H0 : µX − µY ≥ 0
(c) Yes, the P-value is equal to 0.047, which is less than 0.050.
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CHAPTER 6
p
(d) z = (3.94−4.43−0)/ 2.652 /135 + 2.382/180 = −1.70. Since the alternate hypothesis is of the form µX − µY < ∆,
the P-value is the area to the left of z = −1.70.
Thus P = 0.0446, and the result is similar to that of the t test.
√
0.232 + 0.182 is used for z, then z = −1.68 and P = 0.0465. The result is still
Note that if the denominator
similar to that of the t test.
(e) X = 3.94, sX = 2.65, nX = 135, Y = 4.43, sY = 2.38, nY = 180, z.005 = 2.58.
p
The confidence interval is 3.94 − 4.43 ± 2.58 2.652/135 + 2.382/180, or (−1.235, 0.255).
√
Alternatively, the confidence interval can be computed as 3.94−4.43±2.58 0.232 + 0.182, or (−1.244, 0.264).
√
√
13. (a) (i) StDev = (SE Mean) N = 1.26 78 = 11.128.
√
√
(ii) SE Mean = StDev/ N = 3.02/ 63 = 0.380484.
√
(b) z = (23.3 − 20.63 − 0)/ 1.262 + 0.3804842 = 2.03. Since the alternate hypothesis is of the form µX − µY 6= ∆,
the P-value is the sum of the areas to the right of z = 2.03 and to the left of z = −2.03.
Thus P = 0.0212 + 0.0212 = 0.0424, and the result is similar to that of the t test.
√
√
(c) X = 23.3, sX / nX = 1.26, Y = 20.63, sY / nY = 0.380484, z.01 = 2.33.
√
The confidence interval is 23.3 − 20.63 ± 2.33 1.262 + 0.3804842, or (−0.3967, 5.7367).
Section 6.6
1. (a) H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0
(b) X = 960, nX = 1000, pbX = 960/1000 = 0.960, Y = 582, nY = 600, pbY = 582/600 = 0.970,
pb = (960 + 582)/(1000 + 600) = 0.96375.
The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0.
0.960 − 0.970
z= p
= −1.04.
0.96375(1 − 0.96375)(1/1000 + 1/600)
Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −1.04.
Thus P = 0.1492.
(c) Since P = 0.1492, we cannot conclude that machine 2 is better. Therefore machine 1 should be used.
2. (a) H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0
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SECTION 6.6
(b) X = 150, nX = 180, pbX = 150/180 = 0.83333, Y = 233, nY = 270, pbY = 233/270 = 0.86296,
pb = (150 + 233)/(180 + 270) = 0.85111.
The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0.
0.83333 − 0.86296
− 0.86.
z= p
0.85111(1 − 0.85111)(1/180 + 1/270)
Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −0.86.
Thus P = 0.1949.
(c) No. Since P = 0.1949, we cannot conclude that the proportion from vendor B is greater than that from vendor A.
3.
X = 133, nX = 400, pbX = 133/400 = 0.3325, Y = 50, nY = 100, pbY = 50/100 = 0.5,
pb = (133 + 50)/(400 + 100) = 0.366.
The null and alternate hypotheses are H0 : pX − pY = 0 versus H1 : pX − pY 6= 0.
0.3325 − 0.5
z= p
= −3.11.
0.366(1 − 0.366)(1/400 + 1/100)
Since the alternate hypothesis is of the form pX − pY 6= 0, the P-value is the sum of the areas to the right of
z = 3.11 and to the left of z = −3.11.
Thus P = 0.0009 + 0.0009 = 0.0018.
We can conclude that the response rates differ between public and private firms.
4.
X = 48, nX = 60, pbX = 48/60 = 0.8, Y = 44, nY = 60, pbY = 44/60 = 0.73333,
pb = (48 + 44)/(60 + 60) = 0.76667.
The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0.
0.8 − 0.73333
= 0.86.
z= p
0.76667(1 − 0.76667)(1/60 + 1/60)
Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 0.86.
Thus P = 0.1949.
We cannot conclude that the model with one hidden layer has the greater success rate.
5.
X = 57, nX = 100, pbX = 57/100 = 0.57, Y = 135, nY = 200, pbY = 135/200 = 0.675,
pb = (57 + 135)/(100 + 200) = 0.64.
The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0.
0.57 − 0.675
z= p
= −1.79.
0.64(1 − 0.64)(1/100 + 1/200)
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CHAPTER 6
Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −1.79.
Thus P = 0.0367.
We can conclude that awareness of the benefit increased after the advertising campaign.
6.
X = 19, nX = 161, pbX = 19/161 = 0.11801, Y = 22, nY = 95, pbY = 22/95 = 0.23158,
pb = (19 + 22)/(161 + 95) = 0.16016.
The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0.
0.11801 − 0.23158
= −2.39.
z= p
0.16016(1 − 0.16016)(1/161 + 1/95)
Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −2.39.
Thus P = 0.0084.
We can conclude that the larger particles are more likely to contain coliform bacteria.
7.
X = 20, nX = 1200, pbX = 20/1200 = 0.016667, Y = 15, nY = 1500, pbY = 15/1500 = 0.01,
pb = (20 + 15)/(1200 + 1500) = 0.012963.
The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0.
0.016667 − 0.01
z= p
= 1.52.
0.012963(1 − 0.012963)(1/1200 + 1/1500)
Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 1.52.
Thus P = 0.0643.
The evidence suggests that heavy packaging reduces the proportion of damaged shipments, but may not be
conclusive.
8.
X = 260, nX = 300, pbX = 260/300 = 0.86667, Y = 370, nY = 400, pbY = 370/400 = 0.925,
pb = (260 + 370)/(300 + 400) = 0.9.
The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0.
0.86667 − 0.925
= −2.55.
z= p
0.9(1 − 0.9)(1/300 + 1/400)
Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −2.55.
Thus P = 0.0054.
We can conclude that the proportion of items rated acceptable or better is greater on Wednesday than on
Monday.
9.
X = 22, nX = 41, pbX = 22/41 = 0.53659, Y = 18, nY = 31, pbY = 18/31 = 0.58065,
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283
SECTION 6.6
pb = (22 + 18)/(41 + 31) = 0.55556.
The null and alternate hypotheses are H0 : pX − pY = 0 versus H1 : pX − pY 6= 0.
0.53659 − 0.58065
= −0.37.
z= p
0.55556(1 − 0.55556)(1/41 + 1/31)
Since the alternate hypothesis is of the form pX − pY 6= 0, the P-value is the sum of the areas to the right of
z = 0.37 and to the left of z = −0.37.
Thus P = 0.3557 + 0.3557 = 0.7114.
We cannot conclude that the proportion of wells that meet the standards differs between the two areas.
10.
X = 76, nX = 100, pbX = 76/100 = 0.76, Y = 128, nY = 200, pbY = 128/200 = 0.64,
pb = (76 + 128)/(100 + 200) = 0.68.
The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0.
0.76 − 0.64
z= p
= 2.10.
0.68(1 − 0.68)(1/100 + 1/200)
Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 2.10.
Thus P = 0.0179.
We can conclude that drug A is more effective than drug B.
11.
X = 285, nX = 500, pbX = 285/500 = 0.57, Y = 305, nY = 600, pbY = 305/600 = 0.50833,
pb = (285 + 305)/(500 + 600) = 0.53636.
The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0.
0.57 − 0.50833
= 2.04.
z= p
0.53636(1 − 0.53636)(1/500 + 1/600)
Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 2.04.
Thus P = 0.0207.
We can conclude that the proportion of voters favoring the proposal is greater in county A than in county B.
12. (a) X = 4243, nX = 82, 486, pbX = 4243/82, 486 = 0.05144, Y = 10, 701, nY = 219, 170, pbY = 10, 701/219, 170 = 0.04883,
pb = (4243 + 10, 701)/(82, 486 + 219, 170) = 0.04954.
The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0.
0.05144 − 0.04883
z= p
= 2.95.
0.04954(1 − 0.04954)(1/82, 486 + 1/219, 170)
Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 2.95.
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Thus P = 0.0016.
We can conclude that accidents involving drivers aged 15–24 are more likely to occur in driveways than accidents involving drivers aged 25–64.
(b) No, the observed difference is only 0.2%. Although this is statistically significant, it is not practically significant.
13.
No, because the two samples are not independent.
14. (a) Two-tailed. The alternate hypothesis is of the form p1 − p2 6= 0.
(b) H0 : p1 − p2 = 0
(c) Yes, P = 0.024, which is less than 0.05.
15. (a) 101/153 = 0.660131.
(b) 90(0.544444) = 49.
(c) X1 = 101, n1 = 153, pb1 = 101/153 = 0.660131, X2 = 49, n2 = 90, pb2 = 49/90 = 0.544444,
pb = (101 + 49)/(153 + 90) = 0.617284.
0.660131 − 0.544444
z= p
= 1.79.
0.617284(1 − 0.617284)(1/153 + 1/90)
(d) Since the alternate hypothesis is of the form pX − pY 6= 0, the P-value is the sum of the areas to the right of
z = 1.79 and to the left of z = −1.79.
Thus P = 0.0367 + 0.0367 = 0.0734.
Section 6.7
1. (a) X = 3.05, sX = 0.34157, nX = 4, Y = 1.8, sY = 0.90921, nY = 4.
The number of degrees of freedom is
2
0.341572 0.909212
+
4
4
ν=
= 3, rounded down to the nearest integer.
2
2
(0.909212/4)2
(0.34157 /4)
+
4−1
4−1
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