A husband and wife, Ed and Rina, share a digital music player that has a feature

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, 37 of the first 56 songs selected were songs loaded by Rina. Let p denote the proportion of songs that were loaded by Rina. 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: Conclusion: There is strong evidence that the proportion of songs downloaded by Ed and Rina differs from 0.5. There is not enough evidence to conclude that the proportion of songs downloaded by Ed and Rina differs from 0.5. Are the conditions for the use of the large sample confidence interval 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.)

Explanation / Answer

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.5
Ha:   p   =/=   0.5
As we see, the hypothesized po =   0.5      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.660714286      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.06681531      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    2.405351177   [ANSWER, Z]  
      
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As this is a    2   tailed test, then, getting the p value,  
          
Pvalue =    0.016156931   [ANSWER, P VALUE]

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Note that              
              
p^ = point estimate of the population proportion = x / n =    0.660714286          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.063269678          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.12400629          
lower bound = p^ - z(alpha/2) * sp =   0.536707996          
upper bound = p^ + z(alpha/2) * sp =    0.784720575          
              
Thus, the confidence interval is              
              
(   0.536707996   ,   0.784720575   ) [ANSWER, CONFIDENCE INTERVAL]

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