Question

You work for a bubble gum company testing new flavors to bring to market. Your marketing department claims in an add they made that 8 out of 10 people prefer the new flavor "Crazy-Tasks". You conducted a poll and found that 28 out of 60 people prefer the new flavor "Crazy-Tasks". Test the claim by using a confidence interval at a 95% confidence level.

1. Would this claim involve a mean or a proportion and why?
2. What would be your point-estimator?
   (present your answer as a decimal rounded to the 3rd decimal place)
3. The null and alternative hypothesis in symbols would be:
4. Explain what the value 0.467 means in respect to the data and explain what the value 0.8 means in respect to the data.
5. State your null hypothesis as a complete sentence:
6. State your alternative hypothesis as a complete sentence:
7. What type of test would this require?
8. What is the 95% confidence interval? (Round to three decimal places)
9. What would be your formal conclusion?
10. Explain why you came to your formal conclusion:
11. What practical conclusion can we make about the claim?

Answer 1

Ho :   p =    0.8
H1 :   p ╪   0.8
claim is involving the proportion

point estimate =  p̂ = x/n =    0.467  

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   28          
Sample Size,   n =    60          
                  
Sample Proportion ,    p̂ = x/n =    0.467          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0644          
margin of error , E = Z*SE =    1.960   *   0.0644   =   0.1262
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.467   -   0.1262   =   0.3404
Interval Upper Limit = p̂ + E =   0.467   +   0.1262   =   0.5929
                  
95%   confidence interval is (   0.340 < p <    0.593 )
0.8 does not lie in the CI , so reject Ho

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

0.467 is point estimate of test statistics

0.8 is the null nypothesis proportion

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

Standard Error ,    SE = √( p(1-p)/n ) =    0.0516                  
Z Test Statistic = ( p̂-p)/SE = (   0.4667   -   0.8   ) /   0.0516   =   -6.4550
                          
critical z value =    ±    1.960   [excel formula =NORMSINV(α/2)]              
                          
p-Value   =   0.0000   [excel formula =2*NORMSDIST(z)]              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to say that proportion is not equal to 0.8   

Please revert back in case of any doubt.

Please upvote. Thanks in advance.