HW13 Solutions

9.1

1.  = 31,250
    
   (5,275 - 5,240)  1.96 * 12.5
    
   35  24.5
    
   (10.5, 59.5)
    
2. H_o: ( - ) = 0
   H_a: ( - )  0
    
   = 0.05
    
   z = 35/12.5 = 2.80
    
   p-value = 0.0051
    
   Reject the null hypothesis. Conclude that the two means are different.
    
3. The p-value would be cut in half, to 0.0026. We would still reject the null.
    
4. H_o: ( - ) = 25
   H_a: ( - )  25
    
   = 0.05
    
   z = (35 - 25)/12.5 = 0.80
    
   p-value = 0.4237
    
   Fail to reject the null hypothesis. There is insufficient evidence to conclude that the two means are differ by anything other than 25.

9.4

We must assume that the two populations have equal variances. If the sample size is small we must also assume the populations are normally distributed. We also must assume the samples are randomly selected.

9.10 

(the mean for sample 1 is 43.6 and the standard deviation is 5.8. For sample 2 the mean is 53.6 and the standard deviation is 5.4)
 

1. H_o: ( - )  10
   H_a: ( - ) > 10
    
   = 0.01
    
    = 31.3 
    
   z = 0
    
   p-value = 0.5
    
   Fail to reject the null hypothesis. There is insufficient evidence to conclude that the difference between the two means is greater than 10.
    
2. (53.6 - 43.6)  2.46 * 2.01
    
   10 4.92
    
   (5.08, 14.92)

9.12

1. H_o: ( - )  0
   H_a: ( - ) > 0
    
   where is the mean for women and is the mean for mean.
    
2. We would reject the null and conclude that women have a higher recall score.
    
3. H_o: ( - )  0
   H_a: ( - ) > 0
    
   where is the mean for younger adults and is the mean for older adults.
    
4. We would reject the null and conclude that younger adults have higher mean scores than older adults
   

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