Question

A husband and wife, Ed and Rina, share a digital music player that has a feature that randomly selects which song to play. A total of 3476 songs have been loaded into the player, some by Ed and the rest by Rina. They are interested in determining whether they have each loaded different proportions of songs into the player. Suppose that when the player was in the random-selection mode, 35 of the first 51 songs selected were songs loaded by Rina. Let p denote the proportion of songs that were loaded by Rina.

(part a) State the null and alternative hypotheses to be tested. How strong is the evidence that Ed and Rina have each loaded a different proportion of songs into the player? Make sure to check the conditions for the use of this test. (Round your test statistic to TWO decimal places and your P-value to FOUR decimal places. Assume a 95% confidence level.)
Hypotheses:

A.) H₀: p = 0.5
H_a: p > 0.5

B.) H₀: p = 0.5
H_a: p < 0.5    

C.) H₀: p = 0.5
H_a: p ≠ 0.5

Z = _____

P-value = _____

Conclusion:
A.) There is strong evidence that the proportion of songs downloaded by Ed and Rina differs from 0.5

B.) There is not enough evidence to conclude that the proportion of songs downloaded by Ed and Rina differs from 0.5.    

(part b) Are the conditions for the use of the large sample confidence interval met?

A.) Yes, the conditions are met.

B.) No, the conditions are not met.    

If so, estimate with 95% confidence the proportion of songs that were loaded by Rina. (If the conditions are not met, enter NONE. Round your answers to four decimal places.)
_____ to _____

Answer 1

Ho :   p =    0.5                  
H1 :   p ╪   0.5       (Two tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   35                  
Sample Size,   n =    51                  
                          
Sample Proportion ,    p̂ = x/n =    0.6863                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0700                  
Z Test Statistic = ( p̂-p)/SE = (   0.6863   -   0.5   ) /   0.0700   =   2.66
                          
  
p-Value   =   0.0078   [excel formula =2*NORMSDIST(z)]              
Decision:   p-value<α , reject null hypothesis

                      
Conclusion:
A.) There is strong evidence that the proportion of songs downloaded by Ed and Rina differs from 0.5

.........

A.) Yes, the conditions are met.

.................

margin of error , E = Z*SE =    1.960   *   0.0650   =   0.1273
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.686   -   0.1273   =   0.5589
Interval Upper Limit = p̂ + E =   0.686   +   0.1273   =   0.8136
                  
95%   confidence interval is (   0.5589   < p <    0.8136   )

................

THANKS

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