Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Answer:
a.
• The population is normally distributed
• The 7 subjects represent a random sample from this population
b. True
c. 74.16%
Step-by-step explanation:
on using data [649, 832, 41 8, 530, 384, 899, 755] (please correct if wrong)
sample mean: 638.143
s: 201.72
Sample Mean = 638.143
SD = 201.72
Sample Size (n) = 7
Standard Error (SE) = SD/root(n) = 76.243
alpha (a) = 1-0.9 = 0.1
we use t-distribution as population standard deviation is unknown t(a/2, n-l ) = 1.9432
Margin of Error (ME) = SE = 148.1554
90% confidence interval is given by: Sample Mean +/- (Margin of Error) 638.143 +/- 148.1554 = (489.9876 , 786.2984)
• The population is normally distributed
• The 7 subjects represent a random sample from this populatio
(b) true, this is the definition of Cl
(c) (41 8, 432) has width = 14
hence, Margin of error = 7
sorted data is 384, 418, 530, 649, 755, 832, 899
Here 418 and 432 are 2nd and 6th entry of sample size of 7
Since median is given by (1 + n/2 + z(alpha/2)*sqrt(n) /2 )th entry on the right, we conclude that 1 + (3.5 + this gives z = 1.1339 ==> alpha = 0.2584
Hence this Cl is of 74.16% (approximately)
Note that your course might have given a different formula for calculating Cl medians from sample.
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.