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8 . The 99% confidence interval, because it includes all but one percent of the distribution. The 95% confidence interval will be narrower, because it excludes five percent of the distribution.
9 . The t -distribution will have more probability in its tails (“thicker tails”) and less probability near the mean of the distribution (“shorter in the center”).
10 . Both distributions are symmetrical and centered at zero.
11 . df = n – 1 = 20 – 1 = 19
12 . You can get the
t -value from a probability table or a calculator. In this case, for a
t -distribution with 19 degrees of freedom, and a 95% two-sided confidence interval, the value is 2.093, i.e.,
${t}_{\frac{\alpha}{2}}=2.093\text{.}$ The calculator function is invT(0.975, 19).
13 .
$EBM=\text{}{t}_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right)=\left(2.093\right)\left(\frac{0.3}{\sqrt{20}}\right)=\text{}0.140$
98.4 ± 0.14 = (98.26, 98.54).
The calculator function Tinterval answer is (98.26, 98.54).
14 .
${t}_{\frac{\alpha}{2}}=\mathrm{2.861.}$ The calculator function is invT(0.995, 19).
$$EBM={t}_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right)=(2.861)\left(\frac{0.3}{\sqrt{20}}\right)=0.192$$
98.4 ± 0.19 = (98.21, 98.59). The calculator function Tinterval answer is (98.21, 98.59).
15 .
df =
n – 1 = 30 – 1 = 29.
${t}_{\frac{\alpha}{2}}=2.045$
$EBM=\text{}{z}_{t}\left(\frac{s}{\sqrt{n}}\right)=\left(2.045\right)\left(\frac{0.3}{\sqrt{30}}\right)=\text{}0.112$
98.4 ± 0.11 = (98.29, 98.51). The calculator function Tinterval answer is (98.29, 98.51).
16 .
${p}^{\prime}=\frac{280}{500}=0.56$
${q}^{\prime}=1-{p}^{\prime}=1-0.56=0.44$
$s=\sqrt{\frac{pq}{n}}=\sqrt{\frac{0.56(0.44)}{500}}=0.0222$
17 . Because you are using the normal approximation to the binomial,
${z}_{\frac{\alpha}{2}}=1.96$ .
Calculate the error bound for the population (
EBP ):
$EBP=\text{}{z}_{\frac{a}{2}}\sqrt{\frac{pq}{n}}=1.96\left(0.222\right)=0.0435$
Calculate the 95% confidence interval:
0.56 ± 0.0435 = (0.5165, 0.6035).
The calculator function 1-PropZint answer is (0.5165, 0.6035).
18 .
${z}_{\frac{\alpha}{2}}=1.64$
$EBP=\text{}{z}_{\frac{a}{2}}\sqrt{\frac{pq}{n}}=1.64\left(0.0222\right)=0.0364$
0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965)
19 .
${z}_{\frac{\alpha}{2}}=2.58$
$EBP=\text{}{z}_{\frac{a}{2}}\sqrt{\frac{pq}{n}}=2.58\left(0.0222\right)=\text{}0.0573$
0.56 ± 0.05 = (0.5127, 0.6173).
The calculator function 1-PropZint answer is (0.5028, 0.6172).
20 .
EBP = 0.04 (because 4% = 0.04)
${z}_{\frac{\alpha}{2}}=1.96$ for a 95% confidence interval
$n=\text{}\frac{{z}^{2}pq}{EB{P}^{2}}=\text{}\frac{{1.96}^{2}\left(0.5\right)(0.5)}{{0.04}^{2}}=\text{}\frac{0.9604}{0.0016}=600.25$
You need 601 subjects (rounding upward from 600.25).
21 .
$n=\text{}\frac{{n}^{2}pq}{EB{P}^{2}}=\text{}\frac{{1.96}^{2}\left(0.6\right)(0.4)}{{0.04}^{2}}=\text{}\frac{0.9220}{0.0016}=576.24$
You need 577 subjects (rounding upward from 576.24).
22 .
$n=\text{}\frac{{n}^{2}pq}{EB{P}^{2}}=\text{}\frac{{1.96}^{2}\left(0.5\right)(0.5)}{{0.03}^{2}}=\text{}\frac{0.9604}{0.0009}=1067.11$
You need 1,068 subjects (rounding upward from 1,067.11).
23 .
H
_{0} :
p = 0.58
H
_{a} :
p ≠ 0.58
24 .
H
_{0} :
p ≥ 0.58
H
_{a} :
p <0.58
25 .
H
_{0} :
μ ≥ $268,000
H
_{a} :
μ <$268,000
26 . H _{a} : μ ≠ 107
27 . H _{a} : p ≥ 0.25
28 . a Type I error
29 . a Type II error
30 . Power = 1 – β = 1 – P (Type II error).
31 . The null hypothesis is that the patient does not have cancer. A Type I error would be detecting cancer when it is not present. A Type II error would be not detecting cancer when it is present. A Type II error is more serious, because failure to detect cancer could keep a patient from receiving appropriate treatment.
32 . The screening test has a ten percent probability of a Type I error, meaning that ten percent of the time, it will detect TB when it is not present.
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