In this exercise we examine the effect of the sample size on the significance te

Question

In this exercise we examine the effect of the sample size on the significance test for comparing two proportions. In each case suppose that p1 = 0.65 and p2 = 0.45, and take n to be the common value of n1 and n2. Use the z statistic to test H0: p1 = p2 versus the alternative Ha: p1 p2. Compute the statistic and the associated P-value for the following values of n: 40, 50, 60, 80, 380, 480, and 980. Summarize the results in a table. (Test the difference p1 p2. Round your values for z to two decimal places and round your P-values to four decimal places.) n z P-value 40 1.795 Incorrect: Your answer is incorrect. .073 Incorrect: Your answer is incorrect. 50 2.01 Incorrect: Your answer is incorrect. .044 Incorrect: Your answer is incorrect. 60 2.198 Incorrect: Your answer is incorrect. .028 Incorrect: Your answer is incorrect. 80 2.54 Incorrect: Your answer is incorrect. .011 Incorrect: Your answer is incorrect. 380 5.54 Incorrect: Your answer is incorrect. .00001 Incorrect: Your answer is incorrect. 480 6.231 Incorrect: Your answer is incorrect. .00001 Incorrect: Your answer is incorrect. 980 8.85 Incorrect: Your answer is incorrect. .00001 Incorrect: Your answer is incorrect. Explain what you observe about the effect of the sample size on statistical significance when the sample proportions p1 and p2 are unchanged. As sample size increases, the test becomes less significant. As sample size increases, the test becomes more significant. As sample size increases, there is no effect on significance. There is not enough information. Incorrect: Your answer is incorrect.

Explanation / Answer

When size = 40,
Given that,
sample one, n1 =40, p1= x1/n1=0.65
sample two, n2 =40, p2= x2/n2=0.45
null, Ho: p1 = p2
alternate, H1: p1 != p2
zo =(0.65-0.45)/sqrt((0.55*0.45(1/40+1/40))
zo =1.798
| zo | =1.798
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.7979 ) = 0.0722
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When size = 50,
Given that,
sample one, n1 =50, p1= x1/n1=0.65
sample two, n2 =50, p2= x2/n2=0.45
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.65-0.45)/sqrt((0.55*0.45(1/50+1/50))
zo =2.01
| zo | =2.01
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.0101 ) = 0.0444
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When size = 60,
Given that,
sample one, n1 =60, p1= x1/n1=0.65
sample two, n2 =60, p2= x2/n2=0.45
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.65-0.45)/sqrt((0.55*0.45(1/60+1/60))
zo =2.202
| zo | =2.202
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.2019 ) = 0.0277
hence value of p0.01 < 0.0277,here we do not reject Ho

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When size = 80,
Given that,
sample one, n1 =80, p1= x1/n1=0.65
sample two, n2 =80, p2= x2/n2=0.45
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.65-0.45)/sqrt((0.55*0.45(1/80+1/80))
zo =2.543
| zo | =2.543
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.5426 ) = 0.011

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