Lilly was working on a report on Italian and Spanish scientists. She decided to find a 96 percent confidence interval for the di
fference in mean age at the time of significant scientific discoveries for Italian versus Spanish scientists. She found the ages at the time of science discovery of all the members of both groups and found the 96 percent confidence interval based on a t-distribution using a calculator. The procedure she used is not appropriate in this context because The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used.
The sample sizes for the two groups are not equal.
Age at the time of science discovery occurs at different intervals in the two countries, so the distribution of ages cannot be the same.
Age at the time of science discovery is likely to be skewed rather than bell shaped, so the assumptions for using this confidence interval formula are not valid.
Age at the time of science discovery is likely to have a few large outliers, so the assumption for using this confidence interval formula is not valid.
A t statistic was used to conduct a test of the null hypothesis H0: µ = 11 against the alternative Ha: µ ? 11, with a p-value equal to 0.042. A two-sided confidence interval for µ is to be considered. Of the following, which is the largest level of confidence for which the confidence interval will NOT contain 11?
A. A 90% confidence level
B. A 92% confidence level
C. A 96% confidence level
D. A 97% confidence level
E. A 98% confidence level
In all cases, the entire population is measured so that the actual medium discrepancy could be measured as well as an interval of trust cannot be used.
This issue would be that she calculated the ages with all representatives of both classes, such that she measured a whole population. It's not necessary.
In question 2:
When the p-value is 0.042. At 90% trust and 92% trust level 11 (p-value below 0.10 and 0.08) are not included. however the biggest confidence level of 92%. Consequently, the largest trust level where the 11 is Not included in the trust interval is 92% trust.
Find the Roots (Zeros) f(x)=x^3-6x^2+13x-20. f(x)=x3−6x2+13x−20 f ( x ) = x 3 - 6 x 2 + 13 x - 20. Set x3−6x2+13x−20 x 3 - 6 x 2 + 13 x - 20 equal to 0 0 .
