Question

4 18 of Tiger deviation of Woods drive (in general) is over 350 yards? Assume that the "distances" of all of Woods' drives are normally distributed. Woods' drives are measured, yielding a mean of 356 yards, with a standard 32 yards. To a 500 level of significance, can it be asserted that the average

Answer 1

Ho :µ =350     

Ha :µ >350(Right tail test)

  

Level of Significance , α = 0.05

sample std dev ,  s = 32.0000

Sample Size ,n = 18

Sample Mean,  x̅ =356

  

degree of freedom=DF=n-1=17

  

Standard Error , SE =    s/√n =32.0000/ √ 18=7.542

t-test statistic= (x̅ - µ )/SE = (356.000-350) / 7.542=0.795

  

p-Value=0.2186    [Excel formula =t.dist(t-stat,df) ]

Decision:p-value>α, Do not reject null hypothesis              

Conclusion:     There is not enough evidence to conclude that true mean is over 350 yards at α=0.05