During a period of 11 years 737737 of the people selected for grand jury duty were sampled, and 6868% 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, 69.469.4% of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants?

The formula to set out the lower and the upper margin of a confidence interval when given the proportion (no standard deviation) is

Lower margin = p - Z* √[(pq) ÷ n]
Upper margin  = p + Z* √[(pq) ÷ n]

Where:
p is the sample proportion
q is 1 - p
Z* is the z-score for the confidence level
n is the number size
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p = (68% of 737) ÷ 737 = 501.16 ÷ 737 = 0.68
q = 1 - p = 1 - 0.68 = 0.32
Z* = 2.58 (refer to the table attached below)
n = 737 

substituting these values into the formula, we have

lower margin = 0.68 - (2.58) √[(0.68×0.32) ÷ 737]
lower margin = 0.68 - (2.58) √(0.0002952510176...)
lower margin = 0.68 - 0.04433180431
lower margin = 0.6357 (rounded to four decimal places)
lower margin = 63.57%

upper margin = 0.68 + 0.04433180431
upper margin = 0.7243 (rounded to four decimal places)
upper margin = 72.43%

The confidence interval is between 63.57% and 72.43%. In other words, we can say that between 63.57% and 72.43% of jury members are immigrant. 

The claim of 69.46% is within the confidence interval, hence we can conclude that the selection of grand jury duty is not biased against the immigrant. 

50 % is the answer !