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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.

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 α 2   = 2.093 . The calculator function is invT(0.975, 19).

13 . E B M =   t α 2 ( s n ) = ( 2.093 ) ( 0.3 20 ) =   0.140
98.4 ± 0.14 = (98.26, 98.54).
The calculator function Tinterval answer is (98.26, 98.54).

14 . t α 2 = 2.861. The calculator function is invT(0.995, 19).
E B M = t α 2 ( s n ) = ( 2.861 ) ( 0.3 20 ) = 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 α 2   = 2.045
E B M =   z t ( s n ) = ( 2.045 ) ( 0.3 30 ) =   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 = 280 500 = 0.56
q = 1 p = 1 0.56 = 0.44
s = p q n = 0.56 ( 0.44 ) 500 = 0.0222

17 . Because you are using the normal approximation to the binomial, z α 2 = 1.96 .
Calculate the error bound for the population ( EBP ):
E B P =   z a 2 p q n = 1.96 ( 0.222 ) = 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 α 2 = 1.64
E B P =   z a 2 p q n = 1.64 ( 0.0222 ) = 0.0364
0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965)

19 . z α 2 = 2.58
E B P =   z a 2 p q n = 2.58 ( 0.0222 ) =   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 α 2 = 1.96 for a 95% confidence interval
n =   z 2 p q E B P 2 =   1.96 2 ( 0.5 ) ( 0.5 ) 0.04 2 =   0.9604 0.0016 = 600.25
You need 601 subjects (rounding upward from 600.25).

21 . n =   n 2 p q E B P 2 =   1.96 2 ( 0.6 ) ( 0.4 ) 0.04 2 =   0.9220 0.0016 = 576.24
You need 577 subjects (rounding upward from 576.24).

22 . n =   n 2 p q E B P 2 =   1.96 2 ( 0.5 ) ( 0.5 ) 0.03 2 =   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.

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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