Question

Each of a group of 20 intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut strings. After several weeks of playing with the two rackets, each player will be asked to state a preference for one of the two types of strings. Let p denote the proportion of all such players who would prefer gut to nylon, and let X be the number of players in the sample who prefer gut. Because gut strings are more expensive, consider the null hypothesis that at most 50% of all such players prefer gut. We simplify this to H0: p = .5, planning to reject H0 only if sample evidence strongly favors gut strings.

a. Is a significance level of exactly .05 achievable? If not, what is the largest a smaller than .05 that is achievable?
b. If 60% of all enthusiasts prefer gut, calculate the probability of a type II error using the significance level from part (a). Repeat if 80% of all enthusiasts prefer gut.
c. If 13 out of the 20 players prefer gut, should H0 be rejected using the significance level of (a)?

Answer 1

Answer:

Find answers below

Step-by-step explanation:

H0: P <= 0.5

Ha: P > 0.5

, the number who prefer gut strings is <= a number or the test tends towards the  left-tailed.

{0,1,2,3,4,5} ;  

{15,16,17,18,19,20} is a right-tailed test and not appropriate for H0:

{ 0,1,2,3,17,18,19,20} is two-tailed and not appropriate for H0:

b)

Does the region specify a level .05 test? No

 

P = proportion who prefer gut strings to nylon

P = X /20

Assume alpha = 0.05

z(alpha) = -1.645

Reject if (x/20 - 0.5) / sqrt[ (0.5)(0.5)/20 ] < -1.645

Reject if (x/20 - 0.5) <  < (-1.645) sqrt ( (0.5)(0.5)/20 )

Reject if x/20   < (-1.645) sqrt ( (0.5)(0.5)/20 ) + 0.5

Reject if x/20   < 0.316

Reject if x   < (0.316)(20) = 6.32

{0,1,2,3,4,5,6} is the region for the best level 0.05 test

c)

According to (a),  reject H0 if x <= 5

P( Type II error) = P( do not reject H0/ when Ha is true)

P( Type II error) = P( x > 5/ P=0.6)

x ---p(x)

6  0.004854  0.998388  

7  0.014563    

8  0.035497  

9  0.070995  

10  0.117142  

11  0.159738  

12  0.179706  

13  0.165882  

14  0.124412  

15  0.074647  

16  0.034991  

17  0.012350  

18  0.003087  

19  0.000487  

20  0.000037  

add: 0.9984 --  proba bility of a type II error

Assuming P=0.8

P( Type II error) = P( x > 5/ P=0.8)

6  0.000002  1.000000  

7  0.000013  

8  0.000087  

9  0.000462  

10  0.002031  

11  0.007387  

12  0.022161  

13  0.054550  

14  0.109100  

15  0.174560  

16  0.218199  

17  0.205364  

18  0.136909  

19  0.057646  

20  0.011529  

add: 1.0000  probability of a type II error

d)

P( x <= 13) =  

0  0.000001  

1  0.000019  

2  0.000181  

3  0.001087  

4  0.004621  

5  0.014786  

6  0.036964  

7  0.073929  

8  0.120134  

9  0.160179  

10  0.176197  

11  0.160179  

12  0.120134  

13  0.073929  

add: 0.9423 < 0.10 ,  H0 cannot be rejected