This exercise deals with statistical inference and drawing conclusions based on
ID: 3223016 • Letter: T
Question
This exercise deals with statistical inference and drawing conclusions based on sample evidence. There are two parts, both deal with experimental results but there is a slight difference in how the experiment is conducted between the two applications l. LB, the designer and manufacturer of golf balls claims that its new ball, labeled the LDB, will travel farther than any golf ball on the market. To substantiate its claim, LB had Alliance Testing Services (ATS) design and conduct an experiment. In the experiment, ATS randomly selected 48 golfers. Twenty four gofers hit 10 LDB's and the other 24 golfers hit 10 of the leading contender, the Elkin Max (EM) The average distance (in yards) for each golfer is shown in the table below. Does the LDB travel farther than the leading contender? Follow appropriate hypothesis testing procedures (i.e. state hypotheses, etc). Solve both manually and with Minitab. You can use the means and standard deviations from Minitab in your calculations. Since the data is in a table, you can copy and paste it right into Minitab. This file (Golf Ball Data.MTW) is also in Minitab format on the course site (Canvas).Explanation / Answer
The mean average distance of LDB (223.1,250.1,187.3,213.5,200,214.6,240.1,179.1,161.7,201.5,193.3,233.9,191.8,196.3,186.5,194.7,232.4,237.7,247.3,210.2,180.3,212.3,224.2,209.3)
is 209.2167
The mean average distance of Elkin Max
(215.5,225.7,194.3,194.9,202.7,218.5,234.1,177.4,172.7,193.4,193.5,223.4,195,187.5,183.8,198.2,216.2,219.8,229.3,212.6,182.6,217.5,218.8,212.7)
is 205.0042
The standard deviation of average distance of LDB is 23.39158
The standard deviation of average distance of Elkin Max is 17.44857
Null Hypothesis H0: The average distance of LDB is equal to average distance of Elkin Max.
Alternative Hypothesis H1: The average distance of LDB is greater than the average distance of Elkin Max.
We will conduct one-tail t-test to validate the null hypothesis.
Standard Error for the test, SE = sqrt[ (s12/n1) + (s22/n2) ]
Where s1 = 23.39158, s2 = 17.44857, n1 = n2 = 24
SE = sqrt[ (23.391582/24) + (17.448572/24) ] = 5.956854
Degree of freedom for the test , DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (23.391582/24 + 17.448572/24)2 / { [ (23.391582 / 24)2 / (24 - 1) ] + [ (17.448572 / 24)2 / (24 - 1) ] }
= 42 (Rounding to the nearest integer)
t-stat = Difference in mean / SE
= (209.2167 - 205.0042)/ 5.956854 = 0.7071686
p-value for t = 0.7071686 at df = 42 is 0.2416
As, the p-value is > 0.05, we fail to reject the null hypothesis and there is not enough statistical evidence to prove that the average distance of LDB is greater than the average distance of Elkin Max.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.