Answer:
95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].
Step-by-step explanation:
We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. 
A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.
Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;
                         P.Q. =  ~ N(0,1)
  ~ N(0,1)
where,  = sample proportion of males having blood disorder=
 = sample proportion of males having blood disorder=  = 0.25
 = 0.25
  = sample proportion of females having blood disorder =
 = sample proportion of females having blood disorder =  = 0.275
 = 0.275
 = sample of males = 1000
 = sample of males = 1000
 = sample of females = 1000
 = sample of females = 1000
 = population proportion of males having blood disorder
 = population proportion of males having blood disorder
 = population proportion of females having blood disorder
 = population proportion of females having blood disorder
<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>
<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u> <u>)</u><u> is ;</u>
<u>)</u><u> is ;</u>
P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level
                                              of significance are -1.96 & 1.96}  
P(-1.96 <  < 1.96) = 0.95
 < 1.96) = 0.95
P(  <
 <  <
 <  ) = 0.95
 ) = 0.95
P(  < (
 < ( ) <
) <  ) = 0.95
 ) = 0.95
<u>95% confidence interval for</u> ( ) =
) = 
[ ,
,  ]
]
= [  ,
,  ]
 ]
  = [-0.064 , 0.014]
Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].