A new soft drink is being market tested. It is estimated that 60% of consumers will like the new drink. A sample of 96 taste tested the new drink. a) Determine the standard error of the proportion b) What is the probability that more than 75% of consumers will indicate they like the drink

Respuesta :

Answer:

a

  [tex]S.E = 0.05[/tex]

b

 [tex]P(P > 0.75) = 0.0013499[/tex]

Step-by-step explanation:

From the question we are told that

      The population  [tex]p = 0.60[/tex]

       The  sample size is  [tex]n = 96[/tex]

       The  sample proportion is  [tex]\r p = 0.75[/tex]

       

Generally the standard error is mathematically represented as

      [tex]S.E = \sqrt{ \frac{p(1-p)}{n } }[/tex]

substituting values    

       [tex]S.E = \sqrt{ \frac{0.60 (1-0.60 )}{96 } }[/tex]

       [tex]S.E = 0.05[/tex]

The  probability that  more  than  75% of consumers will indicate they like the drink is mathematically represented as

       [tex]P(P > 0.75) = P(\frac{\r P - p }{\sqrt{\frac{p(1-p)}{n} } } > \frac{\r p - p }{\sqrt{\frac{p(1-p)}{n} } } )[/tex]

 The  z-score is  evaluated as

              [tex]z = \frac{\r p - p }{\sqrt{\frac{p(1-p)}{n} } }[/tex]

So  

         [tex]P(P > 0.75) = P(Z > \frac{0.75 - 0.60 }{0.05} )[/tex]

         [tex]P(P > 0.75) = P(Z > 3)[/tex]

         [tex]P(P > 0.75) = 0.0013499[/tex]

This value above is obtained from the z-table