Question

A 2016 poll sampled 523 adults who were planning a vacation during the next six months and found that 141 were expected to travel by airplane. A similar survey question in 2019 found that of 477 adults who were planning a vacation in the next six months, 81 were expecting to travel by airplane.

(i) Express in words for this study, what is meant by the probability of making a Type I and a Type II error.

(ii) Use the p-value method to test for a significant difference. Assume a significance level of 1%. Write the null and alternative hypotheses.

(iii) Repeat part (ii) using the critical value method.

(iv) Repeat part (ii) using the confidence interval approach.

Answer 1

1)

Type 1 error:

A type I error is the probability rejection of a true null hypothesis.

Type2 error:

A type II error is the probability of non-rejection of a false null hypothesis.

2)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->   experimental          
first sample size,     n1=   523          
number of successes, sample 1 =     x1=   141          
proportion success of sample 1 , p̂1=   x1/n1=   0.2696          
                  
sample #2   ----->   standard          
second sample size,     n2 =    477          
number of successes, sample 2 =     x2 =    81          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.170          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2696   -   0.1698   =   0.0998
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2220          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0263          
Z-statistic = (p̂1 - p̂2)/SE = (   0.100   /   0.0263   ) =   3.7924
                  
  
p-value =        0.000149176   [excel formula =2*NORMSDIST(z)]      
decision :    p-value<α,Reject null hypothesis   

3)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->   
first sample size,     n1=   523          
number of successes, sample 1 =     x1=   141          
proportion success of sample 1 , p̂1=   x1/n1=   0.2696          
                  
sample #2   ----->   
second sample size,     n2 =    477          
number of successes, sample 2 =     x2 =    81          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.170          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2696   -   0.1698   =   0.0998
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2220          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0263          
Z-statistic = (p̂1 - p̂2)/SE = (   0.100   /   0.0263   ) =   3.7924
                  
z-critical value , Z* =        2.5758   [excel formula =NORMSINV(α/2)]      

decision :    z stats>Z critical,Reject null hypothesis

4)

level of significance, α =   0.01              
Z critical value =   Z α/2 =    2.576   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0259          
margin of error , E = Z*SE =    2.576   *   0.0259   =   0.0668
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    0.100   -   0.0668   =   0.0330
upper limit = (p̂1 - p̂2) + E =    0.100   +   0.0668   =   0.1666
                  
so, confidence interval is (   0.0330   < p1 - p2 <   0.1666   )  

Since it does not contain zero value.

  

Thanks in advance!

revert back for doubt

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