In this exercise we examine the effect of the sample size on the significance test for comparing two proportions. In each case suppose that p̂1 = 0.65 and p̂2 = 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
50
60
80
380
480
980
Explain what you observe about the effect of the sample size on statistical significance when the sample proportions p̂1 and p̂2 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.
SOLUTION:
From given data,
In this exercise we examine the effect of the sample size on the significance test for comparing two proportions. In each case suppose that p̂1 = 0.65 and p̂2 = 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.
p̂1 = 0.65
p̂2 = 0.45
The value of z for the sample size as 40 is obtained as shown below:
z = (p̂1 - p̂2) / sqrt ( (p̂1 (1-p̂1) / n ) + (p̂2 (1-p̂2) / n ) )
z = (0.65 - 0.45) / sqrt ( (0.65 (1-0.65) / 40 ) + (0.45 (1-0.45) / 40 ) )
z = 0.2 / sqrt ( (0.65 (1-0.65) / 40 ) + (0.45 (1-0.45) / 40 ) )
z = 0.2 / 0.10897247
z = 1.83
The p-value for two tailed ,
P(z=1.83) = 2*P(z > 1.83)
= 2* (1-P(z < 1.83))
= 2* (1- 0.96638)
= 2*0.03362
= 0.06724
The value of z for the sample size as 50 is obtained as shown below:
z = (p̂1 - p̂2) / sqrt ( (p̂1 (1-p̂1) / n ) + (p̂2 (1-p̂2) / n ) )
z = (0.65 - 0.45) / sqrt ( (0.65 (1-0.65) / 50 ) + (0.45 (1-0.45) / 50 ) )
z = 0.2 / sqrt ( (0.65 (1-0.65) / 50 ) + (0.45 (1-0.45) / 50 ) )
z = 0.2 / 0.09746794
z = 2.05
The p-value for two tailed ,
P(z=2.05) = 2*P(z > 2.05)
= 2* (1-P(z < 2.05))
= 2* (1- 0.97982)
= 2*0.02018
= 0.04036
The value of z for the sample size as 60 is obtained as shown below:
z = (p̂1 - p̂2) / sqrt ( (p̂1 (1-p̂1) / n ) + (p̂2 (1-p̂2) / n ) )
z = (0.65 - 0.45) / sqrt ( (0.65 (1-0.65) / 60 ) + (0.45 (1-0.45) / 60 ) )
z = 0.2 / sqrt ( (0.65 (1-0.65) / 60 ) + (0.45 (1-0.45) / 60 ) )
z = 0.2 / 0.08897565
z =2.24
The p-value for two tailed ,
P(z=2.24) = 2*P(z > 2.24)
= 2* (1-P(z < 2.24))
= 2* (1- 0.98745)
= 2*0.01255
= 0.0251
The value of z for the sample size as 80 is obtained as shown below:
z = (p̂1 - p̂2) / sqrt ( (p̂1 (1-p̂1) / n ) + (p̂2 (1-p̂2) / n ) )
z = (0.65 - 0.45) / sqrt ( (0.65 (1-0.65) / 80 ) + (0.45 (1-0.45) / 80 ) )
z = 0.2 / sqrt ( (0.65 (1-0.65) / 80 ) + (0.45 (1-0.45) / 80 ) )
z = 0.2 / 0.07705517
z =2.59
The p-value for two tailed ,
P(z=2.59) = 2*P(z > 2.59)
= 2* (1-P(z <2.59))
= 2* (1- 0.99520)
= 2*0.0048
= 0.0096
The value of z for the sample size as 380 is obtained as shown below:
z = (p̂1 - p̂2) / sqrt ( (p̂1 (1-p̂1) / n ) + (p̂2 (1-p̂2) / n ) )
z = (0.65 - 0.45) / sqrt ( (0.65 (1-0.65) / 380 ) + (0.45 (1-0.45) / 380 ) )
z = 0.2 / sqrt ( (0.65 (1-0.65) / 380 ) + (0.45 (1-0.45) / 380 ) )
z = 0.2 / 0.03535533
z =5.65
The p-value for two tailed ,
P(z=5.65) = 2*P(z > 5.65)
= 2*0
=0
The value of z for the sample size as 480 is obtained as shown below:
z = (p̂1 - p̂2) / sqrt ( (p̂1 (1-p̂1) / n ) + (p̂2 (1-p̂2) / n ) )
z = (0.65 - 0.45) / sqrt ( (0.65 (1-0.65) / 480 ) + (0.45 (1-0.45) / 480 ) )
z = 0.2 / sqrt ( (0.65 (1-0.65) / 480 ) + (0.45 (1-0.45) / 480 ) )
z = 0.2 / 0.03145764
z =6.35
The p-value for two tailed ,
P(z=6.35) = 2*P(z > 6.35)
= 2*0
=0
The value of z for the sample size as 980 is obtained as shown below:
z = (p̂1 - p̂2) / sqrt ( (p̂1 (1-p̂1) / n ) + (p̂2 (1-p̂2) / n ) )
z = (0.65 - 0.45) / sqrt ( (0.65 (1-0.65) / 980 ) + (0.45 (1-0.45) / 980 ) )
z = 0.2 / sqrt ( (0.65 (1-0.65) / 980 ) + (0.45 (1-0.45) / 980 ) )
z = 0.2 / 0.02201576
z =9.08
The p-value for two tailed ,
P(z=9.08) = 2*P(z > 9.08)
= 2*0
=0
Please thumbs-up / vote up this answer if it was helpful. In case of any problem, please comment below. I will surely help. Down-votes are permanent and not notified to us, so we can't help in that case.
In this exercise we examine the effect of the sample size on the significance test for...
The null and alternative hypotheses for a test are given, as well as some information about the actual sample and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. Ho = p1 = p2 vs. ha p1 < p2 Sample: p̂1 = 0.3, n= 20 p̂2 = 0.167, n=12 Randomization statistic: p̂1 - p̂2 Answer _______________
The null and alternative hypotheses for a test are given, as well as some information about the actual sample and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. Ho: p1 = p2 vs. Ha: p1>p2 Sample: p̂1 = 0.3, n=20, p̂2 = 0.467 n=12 Randomization statistic p̂1 - p̂2 P.S. The answer is not -0.167
Gun Murders - Texas vs New York - Significance Test In 2011, New York had much stricter gun laws than Texas. For that year, the proportion of gun murders in Texas was greater than in New York. Here we test whether or not the proportion was significantly greater in Texas. The table below gives relevant information. Here, the p̂'s are population proportions but you should treat them as sample proportions. The standard error (SE) is given to save calculation time...
The sample size needed to estimate the difference between two population proportions to within a margin of error m with a significance level of α can be found as follows. In the expression m=z∗p1(1−p1)n1+p2(1−p2)n2−−−−−−−−−−−−−−−−−−−−√ we replace both n1 and n2 by n (assuming that both samples have the same size) and replace each of p1, and p2, by 0.5 (because their values are not known). Then we solve for n, and get n=(z∗)22E2. Finally, increase the value of n to...
Conduct a test at the a = 0.10 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether p 1 greater than p 2. The sample data are x1 = 127, n1 = 247, x2 = 142, and n2 = 312 Choose the correct null and alternative hypotheses below. A. H0 : p1 = p2...
I need help with the last part of this problem. I posted the first parts that I already have the answers to also but only need help with the final part - Interpret the P-value. Thanks. In 1950, an organization surveyed 1100 adults and asked, "Are you a total abstainer from, or do you on occasion consume, alcoholic beverages?" Of the 1100 adults surveyed, 352 indicated that they were total abstainers. In a recent survey, the same question was asked...
According to the Department of Education: 27% of adults over the age of 65 in the United States have earned at least a bachelor's degree 36% of adults 25 to 34 years old in the United States have earned at least a bachelor's degree Suppose these proportions are true for the respective populations. A researcher is planning on the following samples: A random sample of 200 adults over the age of 65 will be selected and their highest degree will...
For one binomial experiment, n1 = 75 binomial trials produced r1 = 60 successes. For a second independent binomial experiment, n2 = 100 binomial trials produced r2 = 85 successes. At the 5% level of significance, test the claim that the probabilities of success for the two binomial experiments differ. (a) Compute the pooled probability of success for the two experiments. (Round your answer to three decimal places.) (b) Check Requirements: What distribution does the sample test statistic follow? Explain....
I need help with the last part of this problem. I posted the first parts that I already have the answers to also but only need help with the final part. Thanks. In 1950, an organization surveyed 1100 adults and asked, "Are you a total abstainer from, or do you on occasion consume, alcoholic beverages?" Of the 1100 adults surveyed, 352 indicated that they were total abstainers. In a recent survey, the same question was asked of 1100 adults and...
(1 point) The sample size needed to estimate the difference between two population proportions to within a margin of error E with a significance level of α can be found as follows. In the expression E=z∗p1(1−p1)n1+p2(1−p2)n2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√ we replace both n1 and n2 by n (assuming that both samples have the same size) and replace each of p1, and p2, by 0.5 (because their values are not known). Then we solve for n, and get n=(z∗)22E2. Finally, increase the value of...