n2 . 89 x2 103 σ2 . 10 The confidence interval is□< (m- (Round to four...
Independent random sampling from two normally distributed populations gives the results below. Find a 99% confidence interval estimate of the difference between the means of the two populations ni 70 X1-377 ƠI :19 n2-34 x2 334 ơ2-29 The confidence interval is < (m-μ2) (Round to four decimal places as needed)
Use a 95% confidence level. Are these results very different from the confidence interval 25.1 hg< μ< 28.3 hg with only 15 sample values = 26.7 hg and s=28 □hg<μ«Ohg (Round to one decimal place as needed.) Clear All
Determine the margin of error for the confidence interval for the proportion: 0.509 <u<0.845 The margin of error is
Use a calculator to solve the equation on the interval 0 3x <2. Round the answer to o 18) x2 - 3 sin(2x) - 2x val sx21. Round the answer to one decimal place if necessary.
Solve the inequality log2 (x2 – 3x + 4) <1. Give the answer as an interval.
Express the confidence interval 0.666<p<0.888 in the form DIE +E=0+0
Please answer: Express the confidence interval 0.111<p<0.333 in the form p+E. ÔTE= +
Short Answer: 1) Find the margin of sampling error for the confidence interval $365 < p < $575:
#4. A 95% confidence interval for the lives in minutes) of Kodak AA batteries is 4304 <470. Assume that the result is based on a sample of size 100. a) What is the value of sample mean? b) What is the population standard deviation o? c) Construct the 99%confidence interval: d) If the confidence interval 432 < u < 468 is obtained from the same sample data, what is the level of confidence? #5. A 95% confidence interval is 48...
1.1.11 Express the confidence interval (0.054,0.152) in the form of p-E<p<P+E. [<p< (Type integers or decimals.)