Question

A golf association requires that golf balls have a diameter that is 1.68 inches. To determine if golf balls conform to the​ standard, a random sample of golf balls was selected. Their diameters are shown in the accompanying data table. Do the golf balls conform to the​ standards? Use the alpha=.01 level of significance.

1.683	1.676	1.681
1.685	1.678	1.686
1.684	1.685	1.673
1.685	1.682	1.674

A) Find the test statistic

B) Find the P value

C) What can be concluded? Reject or not? Sufficient evidence?

In 1990, the club wondered how much water gathers in their holes. In 1990, the concentration of water in the golf holes was .15 milligrams per liter. A random sample of 10 precipitation dates in 2009 results in the following data recorded from the holes.  

.076	.076	.074	.232	.118
.178	.119	.246	.306	.112

D) Construct a 95% confidence interval about the water in the holes at the golf club.

E) Does the sample evidence suggest that water concentrations in the holes changed from 1990 to 2009?

A random sample size of n=200 golfers were asked if they enjoyed golf in the winter. Of the 200 asked, 104 responded that they did. Determine if more than half of all golfers enjoy golf in the winter at the α=0.05 level of significance.

F) Calculate null and alternative hypotheses

H0= ___ ___ .5

H1= ____ ____ .5

G) Calculate the P value

H) State the conclusion. Reject or not? Is there sufficient evidence at the alpha=.05 level of significance to conclude that more than half of the golfers like to golf in the winter?

Answer 1

1)

Ho :   µ =   1.68                  
Ha :   µ ╪   1.68       (Two tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   0.0046                  
Sample Size ,   n =    12                  
Sample Mean,    x̅ = ΣX/n =    1.6810                  
                          
degree of freedom=   DF=n-1=   11                  
                          
Standard Error , SE = s/√n =   0.004612237   / √    12   =   0.0013      
t-test statistic= (x̅ - µ )/SE = (   1.681   -   1.68   ) /    0.001   =   0.751

b)

p-Value   =   0.4684   [Excel formula =t.dist(t-stat,df) ]  

c)

Decision:   p-value>α, Do not reject null hypothesis       
Conclusion: There is not sufficient evidence......................
=============================================================

D)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   0.0824
Sample Size ,   n =    10
Sample Mean,    x̅ = ΣX/n =    0.1537

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   2.2622   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.082   / √   10   =   0.0261
margin of error , E=t*SE =   2.2622   *   0.026   =   0.059
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    0.15   -   0.059   =   0.0948
Interval Upper Limit = x̅ + E =    0.15   -   0.059   =   0.2126
95%   confidence interval is (   0.09   < µ <   0.21   )

E)

Ho :   µ =   0.15
Ha :   µ ╪   0.15

since, confidence interval do contain null hypothesis value, so, do not reject Ho

sample evidence do not suggest that water concentrations in the holes changed from 1990 to 2009

==========================

F)

Ho :   p =    0.5      
H1 :   p >   0.5       (Right tail test)

G)

Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   104                  
Sample Size,   n =    200                  
                          
Sample Proportion ,    p̂ = x/n =    0.5200                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0354                  
Z Test Statistic = ( p̂-p)/SE = (   0.5200   -   0.5   ) /   0.0354   =   0.5657
                          
  
p-Value   =   0.2858   [Excel function =NORMSDIST(-z)              

H)

Decision:   p value>α ,do not reject null hypothesis       
There is not enough evidence    at the alpha=.05 level of significance to conclude that more than half of the golfers like to golf in the winter