umay aduicoursesTYO4119123take D Question 4 5 pts Two survey

umay aduicourses/TYO4//119123take D Question 4 5 pts Two surveys are taken. One estimates the mean number of shots it takes to complote a golf course is 86, while the other suggests it is 93 a) Suppose the first sunvey had 16 subjects, a mean of 86, and a sample standard devation of 5 shots. Find the 99% confidence interval for the true mean using this survey b) Suppose the second surey had 20 subjects, a mean of 93, and a sample standard deviation of 9 shots. Find the 99% confidence interval for the true mean using this survey (c) Let\'s assume these two surveys are truly random and are not eluenced by weather conditions or other confounding vaniables. What shouid we look for in data of the second sample to try to explain the difurence in the means? Hent one word or one sentence

Solution

a)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    86          
t(alpha/2) = critical t for the confidence interval =    2.946712883          
s = sample standard deviation =    5          
n = sample size =    16          
df = n - 1 =    15          
Thus,              
Margin of Error E =    3.683391104          
Lower bound =    82.3166089          
Upper bound =    89.6833911          
              
Thus, the confidence interval is              
              
(   82.3166089   ,   89.6833911   ) [ANSWER]

*******************

b)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    93          
t(alpha/2) = critical t for the confidence interval =    2.860934606          
s = sample standard deviation =    9          
n = sample size =    20          
df = n - 1 =    19          
Thus,              
Margin of Error E =    5.757519833          
Lower bound =    87.24248017          
Upper bound =    98.75751983          
              
Thus, the confidence interval is              
              
(   87.24248017   ,   98.75751983   ) [ANSWER]

*********************

c)

We can look at its standard deviation or variance.