1)
Pop 1 | Pop 2 | |
sample mean x = | 5.090 | 4.310 |
std deviation s= | 2.290 | 1.650 |
sample size n= | 40 | 40 |
Point estimate =x1-x2= | 0.780 | |
std error =√(S21/n1+S22/n2)= | 0.4463 |
Point estimate of differnce =x1-x2 = | 0.780 | ||
for 95 % CI & 39 df value of t= | 2.023 | from excel: t.inv(0.975,39) | |
margin of error E=t*std error = | 0.903 | ||
lower bound=mean difference-E = | -0.1228 | ||
Upper bound=mean differnce +E = | 1.6828 | ||
from above 95% confidence interval for population mean =(-0.12,1.68) |
2)
because npo(1-po) =23.5 is greater than 10 , the sample size is less than or equal to 5%, of the population size, and the sample can be reasonably assumed to be random , the requirements for testing the hypothesis is satisfied. |
null Hypothesis: Ho: p= | 0.027 | |
alternate Hypothesis: Ha: p > | 0.027 |
sample success x = | 28 | ||
sample size n = | 894 | ||
std error σp =√(p*(1-p)/n) = | 0.0054 | ||
sample prop p̂ = x/n=28/894= | 0.0313 | ||
z =(p̂-p)/σp=(0.031-0.027)/0.005= | 0.80 | ||
p value = | 0.212 | (from excel:1*normsdist(-0.8) |
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