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You wish to test the following claim (Ha Ha ) at a significance level of α=0.05...

You wish to test the following claim (Ha Ha ) at a significance level of α=0.05 α=0.05 . Ho:p1=p2 Ho:p1=p2 Ha:p1>p2 Ha:p1>p2 You obtain 82.6% successes in a sample of size n1=447 n1=447 from the first population.

You obtain 75.2% successes in a sample of size n2=472 n2=472 from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.

The test statistic's value is pˆ1−pˆ2 √pˉ⋅(1−pˉ)⋅( 1 n1 + 1 n2 ) p̂1-p̂2p̄⋅(1-p̄)⋅(1n1+1n2) =(0.826 - 0.752)/sqrt(0.788*(1-0.788)*(1/447+1/472)) = 2.742.

What is the test statistic for this sample? z-test, paired samples t-test,independent samples t-test?

What is the p-value for this sample? (Report answer accurate to four decimal places.)

p-value =

The p-value is... greater than α α less than (or equal to) α?

This test statistic leads to a decision to...? reject the null, fail to reject the null,accept the null?

As such, the final conclusion is that... ?

There is not sufficient evidence to warrant rejection of the claim that the first population proportion is greater than the second population proportion.

The sample data support the claim that the first population proportion is greater than the second population proportion.

There is sufficient evidence to warrant rejection of the claim that the first population proportion is greater than the second population proportion.

There is not sufficient sample evidence to support the claim that the first population proportion is greater than the second population proportion

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Answer #1

sample #1   ----->              
first sample size,     n1=   447          
number of successes, sample 1 =     x1=   369.222          
proportion success of sample 1 , p̂1=   x1/n1=   0.8260          
                  
sample #2   ----->              
second sample size,     n2 =    472          
number of successes, sample 2 =     x2 =    354.944          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.7520          
                  
difference in sample proportions, p̂1 - p̂2 =     0.8260   -   0.7520   =   0.0740
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.7880          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0270          
Z-statistic = (p̂1 - p̂2)/SE = (   0.074   /   0.0270   ) =   2.742

-------------------

What is the test statistic for this sample? z-test

p-value =   0.0030 [excel function =NORMSDIST(-z)]      

The p-value is...less than (or equal to) α

This test statistic leads to a decision to...reject the null,

As such, the final conclusion is that...

The sample data support the claim that the first population proportion is greater than the second population proportion.

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