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A publisher reports that 54% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200 found that 44% of the readers owned a personal computer. Is there sufficient evidence at the 0.10 level to support the executive's claim

Answer :

akiran007

The null and alternate hypotheses are

H0 : u = 0.44  vs   Ha: u > 0.44

Null hypothesis: 44% of readers own a personal computer.

Alternate Hypothesis : greater than 44% of readers own a personal computer.

This is one tailed test and the critical region for this one tailed test for the significance level 0.1 is  Z >  ±1.28

The given values are

p1= 0.54 , p2= 0.44 ; q2= 1-p2= 0.56

Using z test

Z = p1-p2/√p2(1-p2)/n

Z= 0.54-0.44/ √0.44*0.56/200

z= =0.1/ 0.03509

z=  2.849

Since the calculated value of Z=  2.849 is greater than Z= 1.28  reject the null hypothesis therefore there is sufficient evidence to support the executive's claim.

Null hypothesis is rejected

There is sufficient evidence to support the executive's claim at 0.10 significance level.

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