Question

7. Elliot has finally figured out that his veggie tacos had WAY too much cilantro. He finds a new recipe and invites 500 loyal customers to come and try it. 394 customers indicate that they like the new veggie taco better than the old recipe. Elliot wants to know if 75% of the population will prefer the new recipe and YOU are the savvy business statistician; he’s found to figure it out! May the odds be ever in your… flavor.

a. Set up the null and the alternative hypotheses.

b. Determine the test statistic.

c. Determine the p-value.

d. At 95% confidence, test to determine if more than 80% of the population will prefer the new recipe.

Answer 1

a)

Ho :   p =    0.75  
H1 :   p >   0.75   (Right tail test)

b)
              
Number of Items of Interest,   x =   394  
Sample Size,   n =    500  
          
Sample Proportion ,    p̂ = x/n =    0.7880  
          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0194  
Z Test Statistic = ( p̂-p)/SE =    (0.788-0.75)/0.0194=   1.96  
          
c)

  
p-Value   =   0.0249   [Excel function =NORMSDIST(-z)

d)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   394          
Sample Size,   n =    500          
                  
Sample Proportion ,    p̂ = x/n =    0.7880          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.018279          
margin of error , E = Z*SE =    1.960   *   0.01828   =   0.0358
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.78800   -   0.03583   =   0.7522
Interval Upper Limit = p̂ + E =   0.78800   +   0.03583   =   0.8238
                  
95%   confidence interval is (   0.752   < p <    0.824   )

As we can see 0.80 is fall into the above interval and hence  more than 80% of the population will not prefer the new recipe.

Please let me know in case of any doubt.

Thanks in advance!

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