Answer:
The company should take a sample of 148 boxes.
Step-by-step explanation:
Hello!
The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.
They estimated a "pilot" proportion of p'=0.20
And using a 90% confidence level the CI should have a margin of error of 2% (0.02).
The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"
[p' ± ]
Where
p' is the sample proportion/point estimator of the population proportion
is the margin of error (d) of the confidence interval.
So
n= 147.28 ≅ 148 boxes.
I hope it helps!
Answer: All of the numbers in the sequence are going up twice the amount as before. The next three terms will be 208, 416, 832
Answer:
A point estimate for the true population proportion is and the 90% confidence interval is (0.8025, 0.8575)
Step-by-step explanation:
We have a large sample of people of size n = 506. Let X be the random variable that represents the number of adults who feel that education is one of the top issues facing California. We are interested in the unknown true proportion p of adults who feel that education is one of the top issues facing California. A point estimate for the true population proportion is given by = X/n = 418/506 = 0.83, the standard deviation is given by . Therefore, the 90% confidence interval for the true population proportion is , i.e., , i.e., , (0.8025, 0.8575).
Four. One joining the tops, one joining the bottoms, and one joining the top and bottom of each.
Answer: x < 8
P/s: if -1.x > -a => x < -a/-1
Ok done. Thank to me >:3