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LHhal Exam Name:_ in a sample of 150 hospital emergency admissions with a certain diagnosis, 128 listed vomiting as a present
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Answer #1

4 a)

Here we have to test that

Ho : p=0.92

H:p < 0.92

n = 150

x = 128

128 = 0.8533

Test statistic:

p - po Po+(1-po)

0.8533 -0.92 0.92*(1-0.92) 10

-0.0667 10.022151

z = -3.01              (Round to 2 decimal)

Test statistic = -3.01

alpha = 0.01

Test is one tailed (left tailed) test.

Critical value for alpha = 0.01 is

z critical = -2.33            (From statistical table of z values, value corresponding to 0.0099)

Here test statistic < z critical

So we reject H0.

Conclusion: There sufficient evidence to conclude that population proportion who experience the symptom is less than 0.92

b) Test is one tailed (left tailed) test.

P value = P(z < -3.01)

             = 0.0013                         (From statistical table of z values)

P value = 0.0013

Interpretation of p value:

Here p value indicates strong evidence against null hypothesis.

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