During a period of 11 years 653 of the people selected for grand jury duty were sampled, and 72% of them were immigrants. use the sample data to construct a 99% confidence interval estimate of the proportion of grand jury members who were immigrants. given that among the people eligible for jury duty, 68.7% of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants?

99% confidence interval estimate of the proportion of grand jury members who were immigrants would be 0.72±0.045, that is (0.675, 0.765) or (67.5%, 76.5%)

Since 68.7% falls in the confidence interval, there is no significant evidence that the jury selection process was somehow biased against immigrants

Step-by-step explanation:

Confidence Interval can be calculated using p±ME where

and margin of error (ME) can be found using the formula

ME= where

Then ME= ≈ 0.045

99% confidence interval would be 0.72±0.045, that is (0.675, 0.765)

i dont know sorry

step-by-step explanation:

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.