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STANDARD VIEW PRINTER VERSIO Chapter 5, Section 1, Exercise 008 The following is a set of hypotheses, some information from o
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Given that:

The following is a set of hypothesis,some information from one or more samples and a standard error from a randomization distribution

n=600,\hat{p}=0.219,SE=0.02

This is the left tailed test .

The null and alternative hypothesis is

H_0 : p =0.27

Hp <0.27

\hat p = 0.219

p_0 = 0.27

1 - p_0 = 1-0.27=0.73

standard error SE=[\sqrt {p_0 * (1 - p_0 ) / n}] =0.02

Test statistic = z

=\frac{ \hat p - p_0}{[\sqrt{ p_0 * (1 - p_0 ) / n}]}

=\frac{ 0.219-0.27 }{[\sqrt{ 0.27 * (1 - 0.27 ) / 600}]}

=\frac{ -0.051 }{[\sqrt{ 0.27 * 0.73 / 600}]}

=\frac{ -0.051 }{[\sqrt{ 0.1971 / 600}]}

=\frac{ -0.051 }{\sqrt{ 0.00032}}

=\frac{ -0.051 }{0.0178}

=-2.86

Z=-2.86

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