During the last decade, the average score of the quantitative portion of the com
ID: 3209014 • Letter: D
Question
During the last decade, the average score of the quantitative portion of the compare the Graduate (GRE General Test is known to be 588. In order to GRE scores of DePaul MBA students with this national average, the scores of 30 DePaul were selected and tabulated. Their average turns out to be 602 with a standard deviation of 27. Verify if the average score of DePaul MBA students is the same as the national average at the 5% significance level via the confidence interval approach. via the critical value approach. via the p-value approach. Use the Excel to calculate the p-value. The Illinois Wildlife Society has been tagging deer population in Illinois to determine their annual growth rate. Based on this effort, the society knows that the annual growth rate is 3%. To verify the growth rate for this year, the society took a sample of 1,000 deer and found 37 newly born ones. Do you think the society's 3% growth rate can be supported by this sample data? Use the confidence interval approach at the 10% significance level to answer this question.Explanation / Answer
Given that,
population mean(u)=588
standard deviation, =27
sample mean, x =602
number (n)=30
null, Ho: =588
alternate, H1: !=588
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 602-588/(27/sqrt(30)
zo = 2.84004
| zo | = 2.84004
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.84004 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.84004 ) = 0.00451
hence value of p0.05 > 0.00451, here we reject Ho
ANSWERS
---------------
null, Ho: =588
alternate, H1: !=588
test statistic: 2.84004
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00451
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