0.2 Practice tests (1-4) and final exams  (Page 17/36)

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.

8.2: confidence interval, single population mean, standard deviation unknown, student’s t

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 . $EBM=t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=\left(2.093\right)\left(\frac{0.3}{\sqrt{20}}\right)=0.140$
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. $t_{\frac{\alpha }{2} }=2.045$
$EBM=z_t\left(\frac{s}{\sqrt{n}}\right)=\left(2.045\right)\left(\frac{0.3}{\sqrt{30}}\right)=0.112$
98.4 ± 0.11 = (98.29, 98.51). The calculator function Tinterval answer is (98.29, 98.51).

8.3: confidence interval for a population proportion

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 ):
$EBP=z_{\frac{a}{2}}\sqrt{\frac{pq}{n}}=1.96\left(0.222\right)=0.0435$
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 . $z_{\frac{\alpha }{2}}=1.64$
$EBP=z_{\frac{a}{2}}\sqrt{\frac{pq}{n}}=1.64\left(0.0222\right)=0.0364$
0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965)

19 . $z_{\frac{\alpha }{2}}=2.58$
$EBP=z_{\frac{a}{2}}\sqrt{\frac{pq}{n}}=2.58\left(0.0222\right)=0.0573$
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
$n=\frac{z^{2}pq}{EB{P}^2}=\frac{{1.96}^2\left(0.5\right)(0.5)}{{0.04}^2}=\frac{0.9604}{0.0016}=600.25$
You need 601 subjects (rounding upward from 600.25).

21 . $n=\frac{n^{2}pq}{EB{P}^2}=\frac{{1.96}^2\left(0.6\right)(0.4)}{{0.04}^2}=\frac{0.9220}{0.0016}=576.24$
You need 577 subjects (rounding upward from 576.24).

22 . $n=\frac{n^{2}pq}{EB{P}^2}=\frac{{1.96}^2\left(0.5\right)(0.5)}{{0.03}^2}=\frac{0.9604}{0.0009}=1067.11$
You need 1,068 subjects (rounding upward from 1,067.11).

9.1: null and alternate hypotheses

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

9.2: outcomes and the type i and type ii errors

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.