Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6980 subjects randomly selected from an online group involved with ears. There were 1332 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis. A. Upper H 0​: pequals0.2 Upper H 1​: pless than0.2 B. Upper H 0​: pgreater than0.2 Upper H 1​: pequals0.2 C. Upper H 0​: pequals0.2 Upper H 1​: pgreater than0.2 D. Upper H 0​: pless than0.2 Upper H 1​: pequals0.2 E. Upper H 0​: pequals0.2 Upper H 1​: pnot equals0.2 F. Upper H 0​: pnot equals0.2 Upper H 1​: pequals0.2 The test statistic is zequals nothing. ​(Round to two decimal places as​ needed.) The​ P-value is nothing. ​(Round to three decimal places as​ needed.) Because the​ P-value is ▼ greater than less than the significance​ level, ▼ reject fail to reject the null hypothesis. There is ▼ insufficient sufficient evidence to support the claim that the return rate is less than​ 20%.

Answer 1

Null hypothesis     

               
                      
Alternative hypothesis

                   
                      
We have for given example,                      
Population proportion value is =0.2                  
x=1332                  
                      
n=6980                  
Level of significance =   0.01                  
Estimate for sample proportion   
                      
Z test statistic formula for proportion


=-1.92      

P value = 0.028......................by using Z table or by using Excel command NORMSDIST(-1.92)

P value = 0.028 > 0.01

Because the​ P-value is greater than the significance​ level, fail to reject the null hypothesis.
There is insufficient evidence to support the claim that the return rate is less than​ 20%.