74)
Standard Deviation , σ =
4.346
sampling error , E = 0.5
Confidence Level , CL= 95%
alpha = 1-CL = 5%
Z value = Zα/2 = 1.960 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z * σ / E )² = ( 1.960
* 4.346 / 0.5 ) ²
= 290.226
So,Sample Size needed=
291
The answer for number 74 plz. 73. A survey asked couples for the number of years...
NFL Players and heights A.Q. Shipley 73 Evan Boehm 74 Taylor Boggs 75 Justin Bethel 72 Harlan Miller 72 Cariel Brooks 69 Mike Jenkins 70 Jerraud Powers 70 Asa Jackson 70 Shaun Prater 70 Ronald Zamort 70 Brandon Williams 71 Elie Bouka 73 Patrick Peterson 73 Trevon Hartfield 74 Chandler Jones 77 Tristan Okpalaugo 78 Frostee Rucker 75 Corey Peters 75 Red Bryant 76 Ed Stinson 76 Olsen Pierre 76 Robert Nkemdiche 76 Rodney Gunter 77 Josh Mauro 78 Calais Campbell ...
A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.00 years and the standard deviation is 9.99 3 years. a Construct a 99% confidence interval for the mean age of all customers, assuming that the assumptions and conditions or the con interval have been met b) How large is the margin of error? c) How would the confidence interval change i you had assumed that the standard deviation was known to be 10.0...
Of 1000 randomly selected cases of lung cancer, 833 resulted in death within 10 years. Construct a 95% two-sided confidence interval on the death rate from lung cancer. (a) Construct a 95% two-sided confidence interval on the death rate from lung cancer. Round your answers to 3 decimal places. sps i (b) Using the point estimate of p obtained from the preliminary sample, what sample size is needed to be 95% confident that the error in estimating the true value...
Refer to the accompanying data set and construct a 90% confidence interval estimate of the mean pulse rate of adult females then do the same for adult males Compare the results Click the icon to view the pulse rates for adult females and adult males Construct a 90% confidence interval of the mean pulse rate for adult females Pulse Rates - X bpm bpm (Round to one decimal place as needed) Males 85 69 75 65 51 74 57 73...
A survey asked college freshman how far away from their hometown is from the college campus. A random sample of 36 students is taken with a mean distance of 100 miles and a standard deviation of 30 Complete parts (a) and (b) below (a) Find the 95% confidence interval for the population mean Lower bound Upper bound (Round your answer to one decimal place) (b) Suppose you want the estimate to be within 3 miles of the population mean. Determine...
Can you the answer with formula plz
and I can not find Z0.05 in the table
C:/Users/norma/Desktop/STAT255/stat255supplement2020.pdf A) 朗读此页内容ys 99 14 EXTRA PROBLEMS 5. Normally the leaves of the Mimosa pudica are horizontal. However, if one of them is touched lightly, the leaflets will fold. It is reported that , the true mean time from touch to complete closure is 3.0s. An experiment is run to test this value. A random sample of size observations on the elapsed time (seconds)...
Chapter 9 A survey was conducted that asked 1017 people how many books they had read in the past years. Results indicated that x bar =10.8 books and s = 16.6 books. Construct a 90% confidence interval for the mean number of books people read. Construct a 90% confidence interval for the mean number of books people read and interpret the results. (Use ascending order. Round to two decimal places as needed.) 1.) If repeated samples are taken,90% of them...
7.2.13-T before the best A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 23 subjects had a mean wake time of 105,0 min. After treatment the 23 subjects had a mean wake time of 100.4 min and a standard deviation of 20.1 min. Assume that the 23 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate of the mean wake time...
A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 22 subjects had a mean wake time of 103.0 min. After treatment, the 22 subjects had a mean wake time of 98.3 min and a standard deviation of 23.1 min. Assume that the 22 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate of the mean wake time for a population with...
A dinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before reatment, 16 subjects had a mean wake time of 103.0 min. After treatment, the 16 subjects had a mean wake time of 98.2 min and a standard deviation of 24.6 min. Assume that the 16 sample values appear to be from a normall distributed population and construct a 90% confidence interval estimate of the mean wake time for a population with...