The National Collegiate Athletic Association ( NCAA) reportedthat the mean numbe
ID: 2940428 • Letter: T
Question
The National Collegiate Athletic Association ( NCAA) reportedthat the mean number of hours spent per week on coaching andrecruiting by college football assistant coaches during the seasonwas 70.
A random sample of 50 assistant coaches showed the sam-ple meanto be 68.6 hours, with a standard deviation of 8.2 hours. a. Usingthe sample data, construct a 99 percent confidence interval for thepopulation mean.
b. Does the 99 percent confidence interval include thevalue suggested by the NCAA? Interpret this result.
c. Suppose you decided to switch from a 99 to a 95 percentconfidence interval. With-out performing any calculations, will theinterval increase, decrease, or stay the same? Which of the valuesin the formula will change?
Explanation / Answer
The National Collegiate Athletic Association ( NCAA) reported that the mean number of hours spent per week on coaching and recruiting by college football assistant coaches during the season was 70.
A random sample of 50 assistant coaches showed the sam-ple mean to be 68.6 hours, with a standard deviation of 8.2 hours. a. Using the sample data, construct a 99 percent confidence interval for the population mean.
b. Does the 99 percent confidence interval include the value suggested by the NCAA? Interpret this result.
c. Suppose you decided to switch from a 99 to a 95 percent confidence interval. With-out performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change?
A random sample of 50 assistant coaches showed the sam-ple mean to be 68.6 hours, with a standard deviation of 8.2 hours. a. Using the sample data, construct a 99 percent confidence interval for the population mean.
N=50
SAMPLE MEAN = M = 68.6 HRS
STD= 8.2 HRS
STD.ERROR = STD/SQ.RT[N] = 8.2/SQ.RT[50]=1.159655
99 % CONFIDENCE LIMITS REFERS TO VARIATION OF
LIMIT OF VARIATION = L = 2.575*STD.ERROR...........[ 2.575 FROM TABLES CORESPONDING TO 99% AREA UNDER PDF CURVE]
=2.5575*1.159655=2.986112..PLUS OR MINUS FROM SAMPLE MEAN
THAT IS THE INTERVAL IS
68.6-2.986112 TO 68.6+2.986112.
= 65.61389 TO 71.58611......................................ANSWER
=========================================================
b. Does the 99 percent confidence interval include the value suggested by the NCAA? Interpret this result.
WE FIND THAT THE STANDARD MEAN 70 IS WITHIN THIS INTERVAL
HENCE the 99 percent confidence interval include the value suggested by the NCAA.
THIS RESULT INDICATES THAT FOR 99% CONFIDENCE LIMIT , 2.5575 TIMES THE VALUE OF STD.ERROR ON EITHER SIDE OF STD. ERROR IS INCLUDED IN THE TOTAL RANGE OF THE INTERVAL
=========================
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.