Question

1. A consulting firm has predicted that 32% of the employees at a large firm would take advantage of a new company Credit Union. A survey of 300 employees shows that 132 of them take advantage of the Credit Union. Test the consulting firm’s hypothesis.

2. Alex interviews a random sample of 20 students at CSI to find out on a scale of 0 to 100 how they rate the Governor Cuomo on his handling of the state. She believes that he will get a rating of 70 on average. The average rating is 40 with a standard deviation of 11. Test the hypothesis that Alex’s guess is plausible in the population.

3. In January 2020 of 1,402 New Jersey residents between age 20 and 65 were surveyed about whether they were working. The proportion working was 0.75. In April 2020, 1,074 New Jersey residents between 20 and 65 were surveyed. The proportion working was 0.64. Test the hypothesis that the proportion of New Jersey residents who were working was the same in April as it was in January.

Answer 1

1)

Ho :   p =    0.32                  
H1 :   p ╪   0.32       (Two tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   132                  
Sample Size,   n =    300                  
                          
Sample Proportion ,    p̂ = x/n =    0.4400                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.02693                  
Z Test Statistic = ( p̂-p)/SE = (   0.4400   -   0.32   ) /   0.0269   =   4.4557
                          

p-Value   =   0.0000   [excel formula =2*NORMSDIST(z)]              
Decision:   p-value<α , reject null hypothesis                       
  

2)

Ho :   µ =   70  
Ha :   µ ╪   70   (Two tail test)
          
Level of Significance ,    α =    0.050  
sample std dev ,    s =    11.0000  
Sample Size ,   n =    20  
Sample Mean,    x̅ =   40.0000  
          
degree of freedom=   DF=n-1=   19  
          
Standard Error , SE = s/√n =   11/√20=   2.4597  
t-test statistic= (x̅ - µ )/SE =    (40-70)/2.4597=   -12.1967  
          
critical t value, t* =    ±   2.0930   [Excel formula =t.inv(α/no. of tails,df) ]
          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value≤α, Reject null hypothesis       
Conclusion: There is enough evidence to reject the claim

3)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->              
first sample size,     n1=   1402          
number of successes, sample 1 =     x1=   1051.5          
proportion success of sample 1 , p̂1=   x1/n1=   0.7500          
                  
sample #2   ----->              
second sample size,     n2 =    1074          
number of successes, sample 2 =     x2 =    687.36          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.6400          
                  
difference in sample proportions, p̂1 - p̂2 =     0.7500   -   0.6400   =   0.1100
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.7023          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.01854          
Z-statistic = (p̂1 - p̂2)/SE = (   0.110   /   0.0185   ) =   5.9325
                  
z-critical value , Z* =        1.9600   [excel formula =NORMSINV(α/2)]      
p-value =        0.0000   [excel formula =2*NORMSDIST(z)]      
decision :    p-value<α,Reject null hypothesis               
                  
Conclusion:   There is enough evidence to reject the claim