Question

2.9 - A current test for detecting a genetic condition in unborn babies during the first trimester of pregnancy has a detection rate of between 82% and 87%. A new test will be implemented if it exceeds the 87% detection rate of the current test. In a trial using a random sample of 200 women whose babies were later verified to have the condition, 178 of the cases were detected by the new test. Should the new test replace the current test? -- Show working out.

a) Write appropriate hypotheses.

b) Check the assumptions and conditions.

c) Perform the test and find the P-value.

d) State your conclusion in plain English. Use a significance level of 5%.

e) Construct a 90% confidence interval for the true detection rate, and comment on your interval in relation to your conclusion from part d

Answer 1

a)

Ho :   p =    0.87
H1 :   p >   0.87

b)

assumptions

population from which samples came are normally distrubuted and samples are indepedent

conditions:

n=200 ≤5% of populations

np = 200* 0.87>5

n(1-p) = 200*(1-0.87) >5

all conditions veriied

c)

Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   178                  
Sample Size,   n =    200                  
                          
Sample Proportion ,    p̂ = x/n =    0.89                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0238                  
Z Test Statistic = ( p̂-p)/SE = (   0.8900   -   0.87   ) /   0.0238   =   0.841
                          
  
p-Value   =   0.2002   [Excel function =NORMSDIST(-z)              

d)

p value>α ,do not reject null hypothesis   

there is not enough evidence to conclude that new test should replace the current test at α=0.05

e)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   178          
Sample Size,   n =    200          
                  
Sample Proportion ,    p̂ = x/n =    0.890          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0221          
margin of error , E = Z*SE =    1.645   *   0.0221   =   0.0364
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.890   -   0.0364   =   0.8536
Interval Upper Limit = p̂ + E =   0.890   +   0.0364   =   0.9264
                  
so, confidence interval is (   0.854   < p <    0.926   )  

since, interval contain the null hypothesis 0.87 , so test is not significant

which is same result as concluded from part d)