A) Variance = 164.99, standard deviation = 12.84
B) Variance = 233.75, standard deviation = 15.89
C) Variance = 128.75, standard deviation = 11.35
D) Variance = 134.75, standard deviation = 11.61
The management of the Seaside Golf Club regularly monitors the golfers on its course
for speed of play. Suppose a random sample of golfers was taken in 2011 and another
random sample of golfers was selected in 2006. The results of the two samples are as
follows:
Based on the sample results, can the management of the Seaside Golf Club conclude
that average speed of play was different in 2012 than in 2011? Conduct the appropriate
hypothesis test at the 0.10 level of significance. Assume that the management of the
club is willing to accept the assumption that the populations of playing times for each
year are approximately normally distributed with equal variances.
A) Because the calculated value of t = -2.03 is less than the lower tail critical value of t
= – 1.6686, reject the null hypothesis. Based on these sample data, at the α = 0.10 level
of significance there is sufficient evidence to conclude that the average speed of play is
different in 2012 than in 2011.
B) Because the calculated value of t = 1.84 is greater than the upper tail critical value of
t = 1.6686, reject the null hypothesis. Based on these sample data, at the α = 0.10 level
of significance there is sufficient evidence to conclude that the average speed of play is
different in 2012 than in 2011.