a)
Ho : p = 0.5
H1 : p > 0.5
(Right tail test)
Level of Significance, α =
0.01
Number of Items of Interest, x =
15
Sample Size, n = 17
1.764705882
Sample Proportion , p̂ = x/n =
0.8824
Standard Error , SE = √( p(1-p)/n ) =
0.1213
Z Test Statistic = ( p̂-p)/SE = ( 0.8824
- 0.5 ) / 0.1213
= 3.1530
p-Value =
0.0008 [Excel function
=NORMSDIST(-z)
b)
Decision: p-value<α , reject null hypothesis
There is enough evidence to support the claim.
c)
np=0.5*17=8.5
n(1-p)=8.5
At least 5 expected successes and 5 expected failures in sample.
Normality assumption appear to be adequate
15.2 A study reported in the American Journal of Public Health (Science News)-the first to follow...