1.Express the confidence interval (168.9,309.7)(168.9,309.7) in
the form of ¯x±MEx¯±ME.
¯x±ME
2.A quality control company was hired to study the length of meter sticks produced by a certain company. The team carefully measured the length of many many meter sticks, and the distribution seems to be slightly skewed to the right with a mean of 100.03 cm and a standard deviation of 0.13 cm.
a) What is the probability of finding a meter stick with a length of more than 100.12 cm
b) What is the probability of finding a group of 22 meter sticks with a mean length of less than 100 cm?
c) What is the probability of finding a group of 34 meter sticks with a mean length of more than 100.06 cm?
d) What is the probability of finding a group of 44 meter sticks with a mean length of between 100.02 and 100.05 cm?
e) For a random sample of 50 meter sticks, what mean length would be at the 92nd percentile?
3.We wish to estimate what percent of adult residents in a
certain county are parents. Out of 400 adult residents sampled, 216
had kids. Based on this, construct a 90% confidence interval for
the proportion ππ of adult residents who are parents in this
county.
Give your answers as decimals, to three places.
_____< ππ <____
1. Here (168.9,309.7)
So sample mean is
And Margin of Error is
So answer here is
1.Express the confidence interval (168.9,309.7)(168.9,309.7) in the form of ¯x±MEx¯±ME. ¯x±ME 2.A quality control company was...
Can anyone please help me with my Confidence Intervals for Proportions homework? Thank you so much! 1. Express the confidence interval 38.6%±9%38.6%±9% in interval form. Express the answer in decimal format (do not enter as percents). 2. We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 26 had kids. Based on this, construct a 95% confidence interval for the proportion, p, of adult residents who are parents in this...