Download solutions

MATH 321
HYPOTHESIS TESTS
APRIL 6, 2016
We start with a null hypothesis Ho, which we'll assume to be true until we have evidence to the contrary. Exactly what
constitutes contrary evidence is determined by our choice of alternative hypothesis Hi. We reject Ho in favor of Hi if
an appropriate test statistic falls in a critical region. If the test statistic does not fall in the critical region, then we fail to
reject Ho.
1. Let x1 , x2,... , xn, be a random sample from a population with mean p. We will use the sample mean x to test the null
hypothesis Ho : p = Po against different alternative hypotheses. Describe the critical region for each alternative hypothesis
(i.e. which values of x would lead us to reject Ho in favor of Hi ).
a) Hi : p po •
14&k.dt\s
b
0 x
a er kin A
AikkkCiA (21:fir 4-c•--91'A
Aito.
b) Hl : µ >
&AAA
1,7
AA0
c) Hi : p < [to.
svdtte- 4.4v, A.ko .
4.4-6 (i■
When testing a hypothesis, there are two kinds of mistake.
• Type I error is rejecting Ho when it is true;
• Type II error is failing to reject Ho when it is false.
The probability of a type I error is a. The probability of a type II error is /3. Usually you should choose the largest
acceptable value for a since this will minimize O. The critical region is chosen so that the test statistic lands in thecritical
region with probability a (when Ho is true). It may also be useful to find the P-value (or observed significance level)
of your data: this is the smallest value for a that leads you to reject Ho with your data Different disciplines have different
standards for what P-value is evidence against Ho , but generally a P-value of 0.05 (or less) is considered significant while a
P-value of 0.01 (or less) is highly significant.
Example. We can use your height data to test the hypothesis that the mean height of a male Gonzaga undergraduate is 70
inches (because a web site told me that the mean height of a 20 year old American male is about 70 inches). Formally, we'll
test Ho : p = 70 against Hi : p 0 70. We assume that the population is normally distributed and that the class constitutes
a random sample. The sample data give x = 72.08 and s = 2.19 with a sample size of n = 28. Our test statistic is
t
72.08 — 70
2.19/ 28
5.02
The P-value of the test is 2P(T > 5.02) 0.000029 (where T has a t-distribution with 27 df; we multiply by 2 because this
is a two-tailed test and x being either large or small constitutes evidence against Ho).
2. What is the conclusion of the test in the example (i.e. should you reject Ho)?
0&o.Aivk rel colt
kvtl
su, cik
or,r ker,d,
For tests about the mean (H0 : µ = ) test statistics are:
— lio for samples from a pop lation with a known variance a2 (all sample sizes if the population is normal,
• z=
otherwise just for large samples);
• z=
—
• t=
—
f-to for large samples;
3 µ0 for small samples (n < 0 or as big as your t-table goes) from a normally distributed population. The test
statistic has a t distribution with n 1 degrees of freedom.
For tests about a population proportion (Ho : B = 90) use the sample proportion O or the sample total X = me. Test
statistics:
• x (X is binomial with parameters n and 00, works best for small samples);
• z—
— 00
1
80(1-610)
x — n,00
= ,
for large samples (both n80 > 10 and n(1 — 00) > 10).
V7100 (1 — Bo)
For tests about the variance (H0 : a2 ad) the test statistic is
)s2
2 (n — 1for
samples from a ormally distributed population. The test statistic has a chi-square distribution
• X =
2
with n — 1 degrees of freedom.
For tests about the standard deviati n, square everything and do a test for the variance.
3. Newly purchased tires are supposed to be filled to a pressure of 30 lb/in2. Let p denote the true average pressure. We
wish to test the hypothesis Ho : µ = 30 gainst the alternative hypothesis H1 : p, 0 30. Use the given sample mean and
sample standard deviation of a random s. mple of size n = 100 to test Ho against Hi at significance level a = 0.05.
a) x=28.2, s = 8
'\e.4
2to.1-?,0
=
'417)Q
P-u4A%ik-e:. 420, 9-e-jeth 41-P
- 2.2c
(Qv
n
Atli;
\ If\
q
b) x=28.2, s = 4
TedA-%-\-gtk;
c) x = 30.6,
.000 0
b5 LIA
To x,30,
Te V s4 etA.,
c?-uaw-e; ,1343 cm) 1-0 1-6cc- 4o.
s=4
4. A random sample of 10 Black Angus steers is weighed, giving a sample standard deviation of 238 pounds. Test
z k 2.
Ho : a = 250 against H1 : a 250.
C(21
'
-14'.'0"1 t0-5-00 . ICS+ 5.6tk 'Y"
Li; Orz ;62-500
- 0-0)z-6.I 5-6731°.
lic, \,.0-- s-6kA IS \I 6-k-we\ 7,67c , q ot-ck 2Ct, 02f 11 5° --t ' C ' r
1-' re.) ex-c cd- ot-= •oc.
5. It is known that roughly 3 of all people have a dominant right foot and i have a dominant right eye. Do people also kiss
to the right? The article "Human Behavior: Adult Persistence of Head-Turning Asymmetry" reported that in a random
sample of 124 kissing couples, 80 of the couples tended to lean more to the right than left. Does this result suggest that
more than half of all couples lean right when kissing? Does this result provide evidence against the hypothesis that i of all
kissing couples lean right?
60- 62-
o=
,.
12-
a ) 44.1k- kV. 19,-L
2%--.' --- -
-- 7i
- aptoAsi- kA,.ez:1'. Z--
,23
ci-U 4k1/4/4Z-1 4 0006
r,(3'- I" (-A---) "- — 6'015
4-2-i-$0-ir
P-e_.itc4- \ID
&A a` .OS
?-odkue., , 05-7-vIA .44 re)ccis RA 0/ -- . 0C