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Section 7.1 Hypothesis Testing: Hypothesis: Null Hypothesis (H 0 ): Alternative Hypothesis (H 1 ):. a statistical analysis used to decide which of two competing hypotheses should be believed (analogous to a court trial). - PowerPoint PPT Presentation
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Section 7.1Hypothesis Testing:
Hypothesis:
Null Hypothesis (H0):
AlternativeHypothesis (H1):
a statistical analysis used to decide which of two competing hypotheses should be believed (analogous to a court trial)
a statement often involving the value of a parameter in a distribution that we want to decide whether or not to believe; the hypothesis is called simple if it specifies only one possible value and is called composite if it specifies multiple possible values
a statement assumed to be true at the outset of a hypothesis test (comparable to “innocence” in a court trial)
a statement for which sufficient evidence is required before it will be believed (comparable to “guilt” in a court trial)
Critical (Rejection)Region:
Type I Error:
Type II Error:
p-value(probability value):
a set describing the observed results of data collection that will lead to rejecting H0
rejecting H0 and accepting H1 when H0 is true (in a court trial, saying that the defendant is guilty when the defendant is really innocent)
failing to reject H0 when H1 is true (in a court trial, saying that the defendant is innocent when the defendant is really guilty)
denotes the probability of making a Type I error and is called the significance level of the test
denotes the probability of making a Type II error
the probability of observing data as supportive or more supportive of H1 than the actual data, calculated by assuming H0 is true
Tables 7.1-1 and 7.1-2 summarize hypothesis tests about one or two proportions (for sufficiently large sample size(s)).
1. Do Text Exercise 7.1-2 three different ways:
First Way
(a) Let C0 = {x | x 0} be the critical region.
(b) = P{x : x 0; p = 3/5} =
P{X = 0; p = 3/5} =
(2/5)4 = 16/625 = 0.0256
= P{x : x > 0; p = 2/5} = P{X > 0; p = 2/5} =
1 – (3/5)4 = 1 – 81/625 =1 – P{X = 0; p = 2/5} =
544/625 = 0.8704
Second Way
(a) Let C1 = {x | x 1} be the critical region.
(b) = P{x : x 1; p = 3/5} =
P{X = 0; p = 3/5} + P{X = 1; p = 3/5} =
(2/5)4 + (4)(3/5)(2/5)3 = 16/625 + 96/625 = 0.1792
= P{x : x > 1; p = 2/5} = P{X > 1; p = 2/5} =
1 – [(3/5)4 + (4)(2/5)(3/5)3] = 1 – 297/625 =
1 – [P{X = 0; p = 2/5} + P{X = 1; p = 2/5}] =
328/625 = 0.5248
1.-continued
Third Way
(a) Let C2 = {x | x 2} be the critical region.
(b) = P{x : x 2; p = 3/5} =
P{X = 0; p = 3/5} + P{X = 1; p = 3/5} + P{X = 2; p = 3/5} =
(2/5)4 + (4)(3/5)(2/5)3 + (6)(3/5)2(2/5)2 = 16/625 + 96/625 + 216/625 =
= P{x : x > 2; p = 2/5} = P{X > 2; p = 2/5} =
(4)(2/5)3(3/5) + (2/5)4 = 96/625 + 16/625 = 0.1792
P{X = 3; p = 2/5} + P{X = 4; p = 2/5} =
328/625 = 0.5248
2.
(a)
A bowl contains 2 red balls and 3 other balls each of which is either red or white. Let p denote the probability of drawing at random a red ball from the bowl. Consider testing H0: p = 2/5 vs. H1: p > 2/5. Drawing ten balls from the bowl one at a time at random and with replacement, we let X equal the number of red balls drawn, and we define the critical region to be C = {x | x 6}.
Calculate . = P{x : x 6; p = 2/5} =
P{X 6; p = 2/5} =
1 – P{X 5; p = 2/5} = 1 – 0.8338 = 0.1662(from Table II in Appendix B)
2.-continued
(b) Calculate if p = 3/5.
If p = 3/5, = P{x : x 5; p = 3/5} =
P{X 5; p = 3/5} =
P{10 – X 5; 1 – p = 2/5} =
1 – P{10 – X 4; 1– p = 2/5} =
1 – 0.6331 = 0.3669 (from Table II in Appendix B)
(c)
(d)
Calculate if p = 4/5.
Calculate if p = 1.
If p = 4/5, = P{x : x 5; p = 4/5} =
P{X 5; p = 4/5} =
P{10 – X 5; 1 – p = 1/5} = 1 – P{10 – X 4; 1– p = 1/5} =1 – 0.9672 = 0.0328 (from Table II in Appendix B)
If p = 1, = P{x : x 5; p = 1} =
0
3. Do Text Exercise 7.1-6 and calculate if p = 0.15.
(a)
(b)
Let Y = the number of ones (1s) observed in 8000 rolls
The test statistic is z =
The one-sided critical region with = 0.05 is
y / 8000 – 1/6———————— (1/6)(5/6) / 8000
z z0.95 = – z0.05 = – 1.645
With y = 1265, we have z =1265 / 8000 – 1/6———————— = (1/6)(5/6) / 8000
– 2.05
Since z = – 2.05 < – z0.05 = – 1.645, we reject H0. We conclude that the probability of rolling a one (1) is less than 1/6.
Since H0 is rejected. we expect the hypothesized proportion 1/6 not to be in the confidence interval:
1265 / 8000 1.960 (1265 / 8000)(6735 / 8000) / 8000
0.15013 , 0.16612
(c)
If p = 0.15, = P{z : z – 1.645; p = 0.15} = y / 8000 – 1/6
P ———————— – 1.645 ; p = 0.15 = (1/6)(5/6) / 8000 y / 8000 – 0.15 + (0.15 – 1/6)P ———————————— – 1.645 ; p = 0.15 =
(1/6)(5/6) / 8000
Find a two-sided 95% confidence interval, instead of the one-sided interval.
y / 8000 – 0.15 1/6 – 0.15 P ———————— ———————— – 1.645 ; p = 0.15 = (1/6)(5/6) / 8000 (1/6)(5/6) / 8000
y / 8000 – 0.15 + (0.15 – 1/6)P ———————————— – 1.645 ; p = 0.15 =
(1/6)(5/6) / 8000
y / 8000 – 0.15 1/6 – 0.15 P ———————— ———————— – 1.7169 ; p = 0.15 = (0.15)(0.85) / 8000 (0.15)(0.85) / 8000
P(Z 2.46) = 1 – (2.46) =
1 – 0.9931 = 0.0069
4. Do Text Exercise 7.1-16.
(a)
(b)
(c)
Let Y = number of yellow candies observed in n random candies
The test statistic is z =
The two-sided critical region with = 0.05 is
p – 0.2—————— (0.2)(0.8) / n
|z| 1.960
H0 is rejected when p = 5/54, but H0 is not rejected for each of the other 19 samples.
If H0 is true, then when this hypothesis test is performed repeatedly, H0 will be rejected 5% of the time in the long run.
(d)
(e)
If 95% confidence intervals for p are repeatedly obtained, then 95% of the intervals in the long run will contain p.
If the 20 samples are pooled, then, p = 219——1124
We have z = 219 / 1124 – 0.2———————— = (0.2)(0.8) / 1124
– 0.43
Since z = – 0.43, and |z| < z0.025 = 1.960, we fail to reject H0. We conclude that the proportion of yellow candies produced is not different from 0.2.
5. Do Text Exercise 7.1-20.
(a)
(b)
The test statistic is z =
The one-sided critical region with = 0.05 is
p1 – p2————————— p(1 – p)(1/n1 + 1/n2)
z 1.645
p1 = p2 =
p =
135 / 900 = 0.15 77 / 700 = 0.11
212 / 1600 = 0.1325
z = + 2.341
Since z = 2.341, and z z0.05 = 1.645, we reject H0. We conclude that the proportion of babies with low birth weight is higher for developing countries in Africa than for developing countries in the Americas.
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